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# Proper divisors

(Redirected from Proper Divisors)
Proper divisors
You are encouraged to solve this task according to the task description, using any language you may know.

The   proper divisors   of a positive integer N are those numbers, other than N itself, that divide N without remainder.

For N > 1 they will always include 1,   but for N == 1 there are no proper divisors.

Examples

The proper divisors of     6     are   1, 2, and 3.
The proper divisors of   100   are   1, 2, 4, 5, 10, 20, 25, and 50.

1. Create a routine to generate all the proper divisors of a number.
2. use it to show the proper divisors of the numbers 1 to 10 inclusive.
3. Find a number in the range 1 to 20,000 with the most proper divisors. Show the number and just the count of how many proper divisors it has.

Show all output here.

## 11l

Translation of: Python
`F proper_divs(n)   R Array(Set((1 .. (n + 1) I/ 2).filter(x -> @n % x == 0 & @n != x))) print((1..10).map(n -> proper_divs(n))) V (n, leng) = max(((1..20000).map(n -> (n, proper_divs(n).len))), key' pd -> pd)print(n‘ ’leng)`
Output:
```[[], , , [1, 2], , [1, 2, 3], , [1, 2, 4], [1, 3], [1, 2, 5]]
15120 79
```

## 360 Assembly

Translation of: Rexx

This program uses two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible.

`*        Proper divisors           14/06/2016PROPDIV  CSECT         USING  PROPDIV,R13        base register         B      72(R15)            skip savearea         DC     17F'0'             savearea         STM    R14,R12,12(R13)    prolog         ST     R13,4(R15)         "         ST     R15,8(R13)         "          LR     R13,R15            "         LA     R10,1              n=1LOOPN1   C      R10,=F'10'         do n=1 to 10         BH     ELOOPN1         LR     R1,R10             n         BAL    R14,PDIV           pdiv(n)         ST     R0,NN              nn=pdiv(n)         MVC    PG,PGT             init buffer          LA     R11,PG             pgi=0         XDECO  R10,XDEC           edit n         MVC    0(3,R11),XDEC+9    output n         LA     R11,7(R11)         pgi=pgi+7         L      R1,NN              nn         XDECO  R1,XDEC            edit nn         MVC    0(3,R11),XDEC+9    output nn         LA     R11,20(R11)        pgi=pgi+20         LA     R5,1               i=1LOOPNI   C      R5,NN              do i=1 to nn         BH     ELOOPNI         LR     R1,R5              i         SLA    R1,2               *4         L      R2,TDIV-4(R1)      tdiv(i)         XDECO  R2,XDEC            edit tdiv(i)         MVC    0(3,R11),XDEC+9    output tdiv(i)         LA     R11,3(R11)         pgi=pgi+3         LA     R5,1(R5)           i=i+1         B      LOOPNIELOOPNI  XPRNT  PG,80              print buffer         LA     R10,1(R10)         n=n+1         B      LOOPN1ELOOPN1  SR     R0,R0              0         ST     R0,M               m=0         LA     R10,1              n=1LOOPN2   C      R10,=F'20000'      do n=1 to 20000         BH     ELOOPN2         LR     R1,R10             n         BAL    R14,PDIV           nn=pdiv(n)         C      R0,M               if nn>m         BNH    NNNHM         ST     R10,II             ii=n         ST     R0,M               m=nnNNNHM    LA     R10,1(R10)         n=n+1         B      LOOPN2ELOOPN2  MVC    PG,PGR             init buffer         L      R1,II              ii         XDECO  R1,XDEC            edit ii         MVC    PG(5),XDEC+7       output ii         L      R1,M               m         XDECO  R1,XDEC            edit m         MVC    PG+9(4),XDEC+8     output m         XPRNT  PG,80              print buffer         L      R13,4(0,R13)       epilog          LM     R14,R12,12(R13)    "         XR     R15,R15            "         BR     R14                exit*------- pdiv   --function(x)----->number of divisors---PDIV     ST     R1,X               x         C      R1,=F'1'           if x=1         BNE    NOTONE         LA     R0,0               return(0)         BR     R14NOTONE   LR     R4,R1              x         N      R4,=X'00000001'    mod(x,2)         LA     R4,1(R4)           +1         ST     R4,ODD             odd=mod(x,2)+1         LA     R8,1               ia=1         LA     R0,1               1         ST     R0,TDIV            tdiv(1)=1         SR     R9,R9              ib=0         L      R7,ODD             odd         LA     R7,1(R7)           j=odd+1LOOPJ    LR     R5,R7              do j=odd+1 by odd         MR     R4,R7              j*j         C      R5,X               while j*j<x          BNL    ELOOPJ         L      R4,X               x         SRDA   R4,32              .         DR     R4,R7              /j         LTR    R4,R4              if mod(x,j)=0         BNZ    ITERJ         LA     R8,1(R8)           ia=ia+1         LR     R1,R8              ia         SLA    R1,2               *4 (F)         ST     R7,TDIV-4(R1)      tdiv(ia)=j         LA     R9,1(R9)           ib=ib+1         L      R4,X               x         SRDA   R4,32              .         DR     R4,R7              j         LR     R2,R9              ib         SLA    R2,2               *4 (F)         ST     R5,TDIVB-4(R2)     tdivb(ib)=x/jITERJ    A      R7,ODD             j=j+odd         B      LOOPJELOOPJ   LR     R5,R7              j         MR     R4,R7              j*j         C      R5,X               if j*j=x         BNE    JTJNEX         LA     R8,1(R8)           ia=ia+1         LR     R1,R8              ia         SLA    R1,2               *4 (F)         ST     R7,TDIV-4(R1)      tdiv(ia)=jJTJNEX   LA     R1,TDIV(R1)        @tdiv(ia+1)         LA     R2,TDIVB-4(R2)     @tdivb(ib)         LTR    R6,R9              do i=ib to 1 by -1         BZ     ELOOPILOOPI    MVC    0(4,R1),0(R2)      tdiv(ia)=tdivb(i)         LA     R8,1(R8)           ia=ia+1         LA     R1,4(R1)           r1+=4         SH     R2,=H'4'           r2-=4         BCT    R6,LOOPI           i=i-1ELOOPI   LR     R0,R8              return(ia)         BR     R14                return to caller*        ----   ----------------------------------------TDIV     DS     80FTDIVB    DS     40FM        DS     FNN       DS     FII       DS     FX        DS     FODD      DS     FPGT      DC     CL80'... has .. proper divisors:'PGR      DC     CL80'..... has ... proper divisors.'PG       DC     CL80' 'XDEC     DS     CL12         YREGS         END    PROPDIV`
Output:
```  1 has  0 proper divisors:
2 has  1 proper divisors:  1
3 has  1 proper divisors:  1
4 has  2 proper divisors:  1  2
5 has  1 proper divisors:  1
6 has  3 proper divisors:  1  2  3
7 has  1 proper divisors:  1
8 has  3 proper divisors:  1  2  4
9 has  2 proper divisors:  1  3
10 has  3 proper divisors:  1  2  5
15120 has  79 proper divisors.
```

## Action!

Calculations on a real Atari 8-bit computer take quite long time. It is recommended to use an emulator capable with increasing speed of Atari CPU.

`BYTE FUNC GetDivisors(INT a INT ARRAY divisors)  INT i,max  BYTE count   max=a/2  count=0  FOR i=1 TO max  DO    IF a MOD i=0 THEN      divisors(count)=i      count==+1    FI  ODRETURN (count) PROC Main()  DEFINE MAXNUM="20000"  INT i,j,count,max,ind  INT ARRAY divisors(100)  BYTE ARRAY pdc(MAXNUM+1)   FOR i=1 TO 10  DO    count=GetDivisors(i,divisors)    PrintF("%I has %I proper divisors: [",i,count)    FOR j=0 TO count-1    DO      PrintI(divisors(j))      IF j<count-1 THEN        Put(32)      FI    OD    PrintE("]")  OD   PutE() PrintE("Searching for max number of divisors:")   FOR i=1 TO MAXNUM  DO    pdc(i)=1  OD  FOR i=2 TO MAXNUM  DO    FOR j=i+i TO MAXNUM STEP i    DO      pdc(j)==+1    OD  OD   max=0 ind=0   FOR i=1 TO MAXNUM  DO    count=pdc(i)    IF count>max THEN      max=count ind=i    FI  OD  PrintF("%I has %I proper divisors%E",ind,max)RETURN`
Output:
```1 has 0 proper divisors: []
2 has 1 proper divisors: 
3 has 1 proper divisors: 
4 has 2 proper divisors: [1 2]
5 has 1 proper divisors: 
6 has 3 proper divisors: [1 2 3]
7 has 1 proper divisors: 
8 has 3 proper divisors: [1 2 4]
9 has 2 proper divisors: [1 3]
10 has 3 proper divisors: [1 2 5]

Searching for max number of divisors:
15120 has 79 proper divisors
```

The first part of the task is to create a routine to generate a list of the proper divisors. To ease the re-use of this routine for other tasks, such as Abundant, Deficient and Perfect Number Classification [], Abundant Odd Number [], and Amicable Pairs [], we define this routine as a function of a generic package:

`generic   type Result_Type (<>) is limited private;   None: Result_Type;   with function One(X: Positive) return Result_Type;   with function Add(X, Y: Result_Type) return Result_Type       is <>;package Generic_Divisors is   function Process    (N: Positive; First: Positive := 1) return Result_Type is            (if First**2 > N or First = N then None       elsif (N mod First)=0 then 	(if First = 1 or First*First = N 	   then Add(One(First), Process(N, First+1))	   else Add(One(First), 		    Add(One((N/First)), Process(N, First+1))))      else Process(N, First+1)); end Generic_Divisors;`

Now we instantiate the generic package to solve the other two parts of the task. Observe that there are two different instantiations of the package: one to generate a list of proper divisors, another one to count the number of proper divisors without actually generating such a list:

`with Ada.Text_IO, Ada.Containers.Generic_Array_Sort, Generic_Divisors; procedure Proper_Divisors is begin   -- show the proper divisors of the numbers 1 to 10 inclusive.   declare      type Pos_Arr is array(Positive range <>) of Positive;      subtype Single_Pos_Arr is Pos_Arr(1 .. 1);      Empty: Pos_Arr(1 .. 0);        function Arr(P: Positive) return Single_Pos_Arr is ((others => P));       package Divisor_List is new Generic_Divisors	(Result_Type => Pos_Arr, None => Empty, One => Arr, Add =>  "&");       procedure Sort is new Ada.Containers.Generic_Array_Sort	(Positive, Positive, Pos_Arr);   begin      for I in 1 .. 10 loop	 declare	    List: Pos_Arr := Divisor_List.Process(I);	 begin	    Ada.Text_IO.Put	      (Positive'Image(I) & " has" & 		 Natural'Image(List'Length) & " proper divisors:");	    Sort(List);	    for Item of List loop	       Ada.Text_IO.Put(Positive'Image(Item));	    end loop;	    Ada.Text_IO.New_Line;	 end;      end loop;   end;    -- find a number 1 .. 20,000 with the most proper divisors   declare      Number: Positive := 1;      Number_Count: Natural := 0;      Current_Count: Natural;       function Cnt(P: Positive) return Positive is (1);       package Divisor_Count is new Generic_Divisors	(Result_Type => Natural, None => 0, One => Cnt, Add =>  "+");    begin      for Current in 1 .. 20_000 loop	 Current_Count := Divisor_Count.Process(Current);	 if Current_Count > Number_Count then	    Number := Current;	    Number_Count := Current_Count;	 end if;      end loop;      Ada.Text_IO.Put_Line	(Positive'Image(Number) & " has the maximum number of" & 	   Natural'Image(Number_Count) & " proper divisors.");   end;end Proper_Divisors; `
Output:
``` 1 has 0 proper divisors:
2 has 1 proper divisors: 1
3 has 1 proper divisors: 1
4 has 2 proper divisors: 1 2
5 has 1 proper divisors: 1
6 has 3 proper divisors: 1 2 3
7 has 1 proper divisors: 1
8 has 3 proper divisors: 1 2 4
9 has 2 proper divisors: 1 3
10 has 3 proper divisors: 1 2 5

15120 has the maximum number of 79 proper divisors.```

## ALGOL 68

### As required by the Task

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32
`# MODE to hold an element of a list of proper divisors            #MODE DIVISORLIST = STRUCT( INT divisor, REF DIVISORLIST next ); # end of divisor list value                                       #REF DIVISORLIST nil divisor list = REF DIVISORLIST(NIL); # resturns a DIVISORLIST containing the proper divisors of n      ## if n = 1, 0 or -1, we return no divisors                        #PROC proper divisors = ( INT n )REF DIVISORLIST:     BEGIN         REF DIVISORLIST result   := nil divisor list;         REF DIVISORLIST end list := result;         INT abs n  = ABS n;         IF abs n > 1 THEN             # build the list of divisors backeards, so they are  #             # returned in ascending order                        #             INT root n = ENTIER sqrt( abs n );             FOR d FROM root n BY -1 TO 2 DO                 IF abs n MOD d = 0 THEN                     # found another divisor                      #                     result := HEAP DIVISORLIST                            := DIVISORLIST( d, result );                     IF end list IS nil divisor list THEN                         # first result                           #                         end list := result                     FI;                     IF d * d /= n THEN                         # add the other divisor to the end of    #                         # the list                               #                         next OF end list := HEAP DIVISORLIST                                          := DIVISORLIST( abs n OVER d, nil divisor list );                         end list         := next OF end list                     FI                 FI             OD;             # 1 is always a proper divisor of numbers > 1        #             result := HEAP DIVISORLIST                    := DIVISORLIST( 1, result )         FI;         result     END # proper divisors # ; # returns the number of divisors in a DIVISORLIST                 #PROC count divisors = ( REF DIVISORLIST list )INT:     BEGIN        INT result := 0;        REF DIVISORLIST divisors := list;        WHILE divisors ISNT nil divisor list DO            result +:= 1;            divisors := next OF divisors        OD;        result     END # count divisors # ; # find the proper divisors of 1 : 10                              #FOR n TO 10 DO    REF DIVISORLIST divisors := proper divisors( n );    print( ( "Proper divisors of: ", whole( n, -2 ), ": " ) );    WHILE divisors ISNT nil divisor list DO        print( ( " ", whole( divisor OF divisors, 0 ) ) );        divisors := next OF divisors    OD;    print( ( newline ) )OD; # find the first/only number in 1 : 20 000 with the most divisors  #INT max number         = 20 000;INT max divisors      :=      0;INT has max divisors  :=      0;INT with max divisors :=      0;FOR d TO max number DO    INT divisor count = count divisors( proper divisors( d ) );    IF divisor count > max divisors THEN        # found a number with more divisors than the previous max  #        max divisors       := divisor count;        has max divisors   := d;        with max divisors  := 1    ELIF divisor count = max divisors THEN        # found another number with that many divisors             #        with max divisors +:= 1    FIOD;print( ( whole( has max divisors, 0 )       , " is the "       , IF with max divisors < 2 THEN "only" ELSE "first" FI       , " number upto "       , whole( max number, 0 )       , " with "       , whole( max divisors, 0 )       , " divisors"       , newline       ) )`
Output:
```Proper divisors of:  1:
Proper divisors of:  2:  1
Proper divisors of:  3:  1
Proper divisors of:  4:  1 2
Proper divisors of:  5:  1
Proper divisors of:  6:  1 2 3
Proper divisors of:  7:  1
Proper divisors of:  8:  1 2 4
Proper divisors of:  9:  1 3
Proper divisors of: 10:  1 2 5
15120 is the first number upto 20000 with 79 divisors
```

### Faster Proper Divisor Counting

Alternative version that uses a sieve-like approach for faster proper divisor counting.

Note, in order to run this with Algol 68G under Windows (and possibly Linux) the heap size must be increased, see ALGOL_68_Genie#Using_a_Large_Heap.

`BEGIN # count proper divisors using a sieve-like approach              #    # find the first/only number in 1 : 20 000 and 1 : 64 000 000 with #    # the most divisors                                                #    INT max number            := 20 000;    TO 2 DO        INT max divisors      := 0;        INT has max divisors  := 0;        INT with max divisors := 0;        [ 1 : max number ]INT pdc; pdc[ 1 ] := 0; FOR i FROM 2 TO UPB pdc DO pdc[ i ] := 1 OD;        FOR i FROM 2 TO UPB pdc DO            FOR j FROM i + i BY i TO UPB pdc DO pdc[ j ] +:= 1 OD        OD;        FOR d TO max number DO            INT divisor count = pdc[ d ];            IF divisor count > max divisors THEN                # found a number with more divisors than the previous max  #                max divisors       := divisor count;                has max divisors   := d;                with max divisors  := 1            ELIF divisor count = max divisors THEN                # found another number with that many divisors             #                with max divisors +:= 1            FI        OD;        print( ( whole( has max divisors, 0 )               , " is the "               , IF with max divisors < 2 THEN "only" ELSE "first" FI               , " number upto "               , whole( max number, 0 )               , " with "               , whole( max divisors, 0 )               , " divisors"               , newline               )             );        max number := 64 000 000    ODEND`
Output:
```15120 is the first number upto 20000 with 79 divisors
61261200 is the only number upto 64000000 with 719 divisors
```

## ALGOL-M

Algol-M's maximum allowed integer value of 16,383 prevented searching up to 20,000 for the number with the most divisors, so the code here searches only up to 10,000.

` BEGIN % COMPUTE P MOD Q % INTEGER FUNCTION MOD (P, Q);INTEGER P, Q;BEGIN    MOD := P - Q * (P / Q);END; % COUNT, AND OPTIONALLY DISPLAY, PROPER DIVISORS OF N %INTEGER FUNCTION DIVISORS(N, DISPLAY);INTEGER N, DISPLAY;BEGIN    INTEGER I, LIMIT, COUNT, START, DELTA;    IF MOD(N, 2) = 0 THEN      BEGIN        START := 2;        DELTA := 1;      END    ELSE  % ONLY NEED TO CHECK ODD DIVISORS %      BEGIN        START := 3;        DELTA := 2;      END;    % 1 IS A DIVISOR OF ANY NUMBER > 1 %    IF N > 1 THEN COUNT := 1 ELSE COUNT := 0;    IF (DISPLAY <> 0) AND (COUNT <> 0) THEN WRITEON(1);    % CHECK REMAINING POTENTIAL DIVISORS %    I := START;    LIMIT := N / START;    WHILE I <= LIMIT DO      BEGIN        IF MOD(N, I) = 0 THEN          BEGIN            IF DISPLAY <> 0 THEN WRITEON(I);            COUNT := COUNT + 1;          END;        I := I + DELTA;        IF COUNT = 1 THEN LIMIT := N / I;      END;   DIVISORS := COUNT;END; COMMENT MAIN PROGRAM BEGINS HERE;INTEGER I, NDIV, TRUE, FALSE, HIGHDIV, HIGHNUM;TRUE := -1;FALSE := 0; WRITE("PROPER DIVISORS OF FIRST TEN NUMBERS:");FOR I := 1 STEP 1 UNTIL 10 DO  BEGIN     WRITE(I, " : ");     NDIV := DIVISORS(I, TRUE);  END; WRITE("SEARCHING FOR NUMBER UP TO 10000 WITH MOST DIVISORS ...");HIGHDIV := 1;HIGHNUM := 1;FOR I := 1 STEP 1 UNTIL 10000 DO  BEGIN     NDIV := DIVISORS(I, FALSE);     IF NDIV > HIGHDIV THEN        BEGIN         HIGHDIV := NDIV;         HIGHNUM := I;       END;  END;WRITE("THE NUMBER IS:", HIGHNUM);WRITE("IT HAS", HIGHDIV, " DIVISORS"); END `
Output:
```PROPER DIVISORS OF FIRST TEN NUMBERS:
1 :
2 :      1
3 :      1
4 :      1     2
5 :      1
6 :      1     2     3
7 :      1
8 :      1     2     4
9 :      1     3
10 :      1     2     5
SEARCHING FOR NUMBER UP TO 10000 WITH MOST DIVISORS:
THE NUMBER IS:  7560
IT HAS    63 DIVISORS
```

## AppleScript

### Functional

Translation of: JavaScript
`-- PROPER DIVISORS ----------------------------------------------------------- -- properDivisors :: Int -> [Int]on properDivisors(n)    if n = 1 then        {1}    else        set realRoot to n ^ (1 / 2)        set intRoot to realRoot as integer        set blnPerfectSquare to intRoot = realRoot         -- isFactor :: Int -> Bool         script isFactor            on |λ|(x)                n mod x = 0            end |λ|        end script         -- Factors up to square root of n,        set lows to filter(isFactor, enumFromTo(1, intRoot))         -- and quotients of these factors beyond the square root,         -- integerQuotient :: Int -> Int        script integerQuotient            on |λ|(x)                (n / x) as integer            end |λ|        end script         -- excluding n itself (last item)        items 1 thru -2 of (lows & map(integerQuotient, ¬            items (1 + (blnPerfectSquare as integer)) thru -1 of reverse of lows))    end ifend properDivisors  -- TEST ----------------------------------------------------------------------on run    -- numberAndDivisors :: Int -> [Int]    script numberAndDivisors        on |λ|(n)            {num:n, divisors:properDivisors(n)}        end |λ|    end script     -- maxDivisorCount :: Record -> Int -> Record    script maxDivisorCount        on |λ|(a, n)            set intDivisors to length of properDivisors(n)             if intDivisors ≥ divisors of a then                {num:n, divisors:intDivisors}            else                a            end if        end |λ|    end script     {oneToTen:map(numberAndDivisors, ¬        enumFromTo(1, 10)), mostDivisors:foldl(maxDivisorCount, ¬        {num:0, divisors:0}, enumFromTo(1, 20000))} ¬ end run  -- GENERIC FUNCTIONS --------------------------------------------------------- -- enumFromTo :: Int -> Int -> [Int]on enumFromTo(m, n)    if m > n then        set d to -1    else        set d to 1    end if    set lst to {}    repeat with i from m to n by d        set end of lst to i    end repeat    return lstend enumFromTo -- filter :: (a -> Bool) -> [a] -> [a]on filter(f, xs)    tell mReturn(f)        set lst to {}        set lng to length of xs        repeat with i from 1 to lng            set v to item i of xs            if |λ|(v, i, xs) then set end of lst to v        end repeat        return lst    end tellend filter -- foldl :: (a -> b -> a) -> a -> [b] -> aon foldl(f, startValue, xs)    tell mReturn(f)        set v to startValue        set lng to length of xs        repeat with i from 1 to lng            set v to |λ|(v, item i of xs, i, xs)        end repeat        return v    end tellend foldl -- map :: (a -> b) -> [a] -> [b]on map(f, xs)    tell mReturn(f)        set lng to length of xs        set lst to {}        repeat with i from 1 to lng            set end of lst to |λ|(item i of xs, i, xs)        end repeat        return lst    end tellend map -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Scripton mReturn(f)    if class of f is script then        f    else        script            property |λ| : f        end script    end ifend mReturn`
Output:
`{oneToTen:{{num:1, divisors:{1}}, {num:2, divisors:{1}}, {num:3, divisors:{1}}, {num:4, divisors:{1, 2}}, {num:5, divisors:{1}}, {num:6, divisors:{1, 2, 3}}, {num:7, divisors:{1}}, {num:8, divisors:{1, 2, 4}}, {num:9, divisors:{1, 3}}, {num:10, divisors:{1, 2, 5}}}, mostDivisors:{num:18480, divisors:79}}`

### Idiomatic

`on properDivisors(n)    set output to {}     if (n > 1) then        set sqrt to n ^ 0.5        set limit to sqrt div 1        if (limit = sqrt) then            set end of output to limit            set limit to limit - 1        end if        repeat with i from limit to 2 by -1            if (n mod i is 0) then                set beginning of output to i                set end of output to n div i            end if        end repeat        set beginning of output to 1    end if     return outputend properDivisors -- Task code.local output, astid, i, maxPDs, maxPDNums, pdCountset output to {}set astid to AppleScript's text item delimitersset AppleScript's text item delimiters to ", "repeat with i from 1 to 10    set end of output to (i as text) & "'s proper divisors:  {" & properDivisors(i) & "}"end repeatset maxPDs to 0set maxPDNums to {}repeat with i from 1 to 20000    set pdCount to (count properDivisors(i))    if (pdCount > maxPDs) then        set maxPDs to pdCount        set maxPDNums to {i}    else if (pdCount = maxPDs) then        set end of maxPDNums to i    end ifend repeatset end of output to linefeed & "Largest number of proper divisors for any number from 1 to 20,000:  " & maxPDsset end of output to "Numbers with this many:  " & maxPDNumsset AppleScript's text item delimiters to linefeedset output to output as textset AppleScript's text item delimiters to astidreturn output`
Output:
`"1's proper divisors:  {}2's proper divisors:  {1}3's proper divisors:  {1}4's proper divisors:  {1, 2}5's proper divisors:  {1}6's proper divisors:  {1, 2, 3}7's proper divisors:  {1}8's proper divisors:  {1, 2, 4}9's proper divisors:  {1, 3}10's proper divisors:  {1, 2, 5} Largest number of proper divisors for any number from 1 to 20,000:  79Numbers with this many:  15120, 18480"`

## Arc

`  ;; Given num, return num and the list of its divisors(= divisor (fn (num)   (= dlist '())   (when (is 1 num) (= dlist '(1 0)))   (when (is 2 num) (= dlist '(2 1)))   (unless (or (is 1 num) (is 2 num))   (up i 1 (+ 1 (/ num 2))     (if (is 0 (mod num i))         (push i dlist)))   (= dlist (cons num dlist)))   dlist)) ;; Find out what number has the most divisors between 2 and 20,000.;; Print a list of the largest known number's divisors as it is found.(= div-lists (fn (cnt (o show 0))  (= tlist '()) (= clist tlist)  (when (> show 0) (prn tlist))  (up i 1 cnt    (divisor i)    (when (is 1 show) (prn dlist))    (when (>= (len dlist) (len tlist))        (= tlist dlist)        (when (is show 2) (prn tlist))        (let c (- (len dlist) 1)        (push (list i c) clist))))   (= many-divisors (list ((clist 0) 1)))  (for n 0 (is ((clist n) 1) ((clist 0) 1)) (= n (+ 1 n))    (push ((clist n) 0) many-divisors))  (= many-divisors (rev many-divisors))  (prn "The number with the most divisors under " cnt       " has " (many-divisors 0) " divisors.")  (prn "It is the number "  (if (> 2 (len many-divisors)) (cut (many-divisors) 1)      (many-divisors 1)) ".")  (prn "There are " (- (len many-divisors) 1) " numbers"       " with this trait, and they are "       (map [many-divisors _] (range 1 (- (len many-divisors) 1))))  (prn (map [divisor _] (cut many-divisors 1)))  many-divisors)) ;; Do the tasks (div-lists 10 1)(div-lists 20000) ;; This took about 10 minutes on my machine.  `
Output:
` (1 0)(2 1)(3 1)(4 2 1)(5 1)(6 3 2 1)(7 1)(8 4 2 1)(9 3 1)(10 5 2 1)The number with the most divisors under 10 has 3 divisors.It is the number 10.There are 3 numbers with this trait, and they are (10 8 6)((10 5 2 1) (8 4 2 1) (6 3 2 1))'(3 10 8 6)  The number with the most divisors under 20000 has 79 divisors.It is the number 18480.There are 2 numbers with this trait, and they are (18480 15120) `

## ARM Assembly

Works with: as version Raspberry Pi
` /* ARM assembly Raspberry PI  *//*  program proFactor.s   */ /* REMARK 1 : this program use routines in a include file    see task Include a file language arm assembly    for the routine affichageMess conversion10    see at end of this program the instruction include *//*******************************************//* Constantes                              *//*******************************************/.equ STDOUT, 1                           @ Linux output console.equ EXIT,   1                           @ Linux syscall.equ WRITE,  4                           @ Linux syscall /*******************************************//* Initialized data                        *//*******************************************/.dataszMessStartPgm:          .asciz "Program start \n"szMessEndPgm:            .asciz "Program normal end.\n"szMessError:             .asciz "\033[31mError Allocation !!!\n"szCarriageReturn:        .asciz "\n" /* datas message display */szMessEntete:            .ascii "Number :"sNumber:                 .space 12,' '                         .asciz " Divisors :"szMessResult:            .ascii " "sValue:                  .space 12,' '                         .asciz ""szMessDivNumber:         .ascii "\nnumber divisors :"sCounter:                .space 12,' '                         .asciz "\n"szMessNumberMax:         .ascii "Number :"sNumberMax:              .space 12,' '                         .ascii " has "sDivMax:                 .space 12, ' '                         .asciz " divisors\n"/*******************************************//* UnInitialized data                      *//*******************************************/.bss /*******************************************//*  code section                           *//*******************************************/.text.global main main:                               @ program start    ldr r0,iAdrszMessStartPgm       @ display start message    bl affichageMess    mov r2,#11:    mov r0,r2                       @  number    ldr r1,iAdrsNumber              @ and convert ascii string    bl conversion10    ldr r0,iAdrszMessEntete         @ display result message    bl affichageMess    mov r0,r2                       @  number    mov r1,#1                       @ display flag    bl divisors                     @ display divisors    ldr r1,iAdrsCounter              @ and convert ascii string    bl conversion10    ldr r0,iAdrszMessDivNumber      @ display result message    bl affichageMess    add r2,r2,#1    cmp r2,#10    ble 1b     mov r2,#2    mov r3,#0    mov r4,#0    ldr r5,iMaxi2:    mov r0,r2    mov r1,#0                       @ display flag    bl divisors                     @ display divisors    cmp r0,r3    movgt r3,r0    movgt r4,r2    add r2,r2,#1    cmp r2,r5    ble 2b    mov r0,r4    ldr r1,iAdrsNumberMax           @ and convert ascii string    bl conversion10    mov r0,r3    ldr r1,iAdrsDivMax              @ and convert ascii string    bl conversion10    ldr r0,iAdrszMessNumberMax    bl affichageMess      ldr r0,iAdrszMessEndPgm         @ display end message    bl affichageMess    b 100f99:                                 @ display error message     ldr r0,iAdrszMessError    bl affichageMess100:                                @ standard end of the program    mov r0, #0                      @ return code    mov r7, #EXIT                   @ request to exit program    svc 0                           @ perform system calliAdrszMessStartPgm:        .int szMessStartPgmiAdrszMessEndPgm:          .int szMessEndPgmiAdrszMessError:           .int szMessErroriAdrszCarriageReturn:      .int szCarriageReturniAdrszMessResult:          .int szMessResultiAdrsValue:                .int sValueiAdrszMessDivNumber:       .int szMessDivNumberiAdrsCounter:              .int sCounteriAdrszMessEntete:          .int szMessEnteteiAdrsNumber:               .int sNumberiAdrszMessNumberMax:       .int szMessNumberMaxiAdrsDivMax:               .int sDivMaxiAdrsNumberMax:            .int sNumberMaxiMaxi:                     .int 20000/******************************************************************//*     divisors function                         */ /******************************************************************//* r0 contains the number  *//* r1 contains display flag (<>0: display, 0: no display )/* r0 return divisors number */divisors:    push {r1-r8,lr}             @ save  registers     cmp r0,#1                   @ = 1 ?    movle r0,#0    ble 100f    mov r7,r0    mov r8,r1    cmp r8,#0    beq 1f    mov r0,#1                   @ first divisor = 1    ldr r1,iAdrsValue           @ and convert ascii string    bl conversion10    ldr r0,iAdrszMessResult     @ display result message    bl affichageMess1:                              @ begin loop    lsr r4,r7,#1                @ Maxi    mov r6,r4                   @ first divisor    mov r5,#1                   @ Counter divisors2:    mov r0,r7                   @ dividende = number    mov r1,r6                   @ divisor    bl division    cmp r3,#0                   @ remainder = 0 ?    bne 3f    add r5,r5,#1                @ increment counter    cmp r8,#0                   @ display divisor ?    beq 3f    mov r0,r2                   @ divisor    ldr r1,iAdrsValue           @ and convert ascii string    bl conversion10    ldr r0,iAdrszMessResult     @ display result message    bl affichageMess3:    sub r6,r6,#1                @ decrement divisor    cmp r6,#2                   @ End ?    bge 2b                      @ no loop    mov r0,r5                   @ return divisors number 100:    pop {r1-r8,lr}              @ restaur registers    bx lr                       @ return /***************************************************//*      ROUTINES INCLUDE                 *//***************************************************/.include "../affichage.inc" `
Output:
```Program start
Number :1            Divisors :
number divisors :0
Number :2            Divisors : 1            2
number divisors :2
Number :3            Divisors : 1            3
number divisors :2
Number :4            Divisors : 1            2
number divisors :2
Number :5            Divisors : 1
number divisors :1
Number :6            Divisors : 1            2            3
number divisors :3
Number :7            Divisors : 1
number divisors :1
Number :8            Divisors : 1            2            4
number divisors :3
Number :9            Divisors : 1            3
number divisors :2
Number :10           Divisors : 1            2            5
number divisors :3
Number :15120        has 79           divisors
Program normal end.

```

## Arturo

`properDivisors: function [x] ->    (factors x) -- x loop 1..10 'x ->    print ["proper divisors of" x "=>" properDivisors x] maxN: 0maxProperDivisors: 0 loop 1..20000 'x [    pd: size properDivisors x    if maxProperDivisors < pd [        maxN: x        maxProperDivisors: pd    ]] print ["The number with the most proper divisors (" maxProperDivisors ") is" maxN]`
Output:
```proper divisors of 1 => []
proper divisors of 2 => 
proper divisors of 3 => 
proper divisors of 4 => [1 2]
proper divisors of 5 => 
proper divisors of 6 => [1 2 3]
proper divisors of 7 => 
proper divisors of 8 => [1 2 4]
proper divisors of 9 => [1 3]
proper divisors of 10 => [1 2 5]
The number with the most proper divisors ( 79 ) is 15120```

## AutoHotkey

`proper_divisors(n) {	Array := []	if n = 1		return Array	Array := true	x := Floor(Sqrt(n))	loop, % x+1		if !Mod(n, i:=A_Index+1) && (floor(n/i) < n)			Array[floor(n/i)] := true	Loop % n/x		if !Mod(n, i:=A_Index+1) && (i < n)			Array[i] := true	return Array}`
Examples:
`output := "Number`tDivisors`tCount`n"loop, 10{	output .= A_Index "`t"	for n, bool in x := proper_divisors(A_Index)		output .= n " "	output .= "`t" x.count() "`n"}maxDiv := 0, numDiv := []loop, 20000{	Arr := proper_divisors(A_Index)	numDiv[Arr.count()] := (numDiv[Arr.count()] ? numDiv[Arr.count()] ", " : "") A_Index	maxDiv := maxDiv > Arr.count() ? maxDiv : Arr.count()}output .= "`nNumber(s) in the range 1 to 20,000 with the most proper divisors:`n" numDiv.Pop() " with " maxDiv " divisors"MsgBox % outputreturn`
Output:
```Number	Divisors	Count
1			0
2	1 		1
3	1 		1
4	1 2 		2
5	1 		1
6	1 2 3 		3
7	1 		1
8	1 2 4 		3
9	1 3 		2
10	1 2 5 		3

Number(s) in the range 1 to 20,000 with the most proper divisors:
15120, 18480 with 79 divisors```

## AWK

` # syntax: GAWK -f PROPER_DIVISORS.AWKBEGIN {    show = 0 # show divisors: 0=no, 1=yes    print("    N  cnt  DIVISORS")    for (i=1; i<=20000; i++) {      divisors(i)      if (i <= 10 || i == 100) { # including 100 as it was an example in task description        printf("%5d  %3d  %s\n",i,Dcnt,Dstr)      }      if (Dcnt < max_cnt) {        continue      }      if (Dcnt > max_cnt) {        rec = ""        max_cnt = Dcnt      }      rec = sprintf("%s%5d  %3d  %s\n",rec,i,Dcnt,show?Dstr:"divisors not shown")    }    printf("%s",rec)    exit(0)}function divisors(n,  i) {    if (n == 1) {      Dcnt = 0      Dstr = ""      return    }    Dcnt = Dstr = 1    for (i=2; i<n; i++) {      if (n % i == 0) {        Dcnt++        Dstr = sprintf("%s %s",Dstr,i)      }    }    return} `

output:

```    N  cnt  DIVISORS
1    0
2    1  1
3    1  1
4    2  1 2
5    1  1
6    3  1 2 3
7    1  1
8    3  1 2 4
9    2  1 3
10    3  1 2 5
100    8  1 2 4 5 10 20 25 50
15120   79  divisors not shown
18480   79  divisors not shown
```

## BASIC

Works with: QBasic version 1.1
Works with: QuickBasic version 4.5
`FUNCTION CountProperDivisors (number)    IF number < 2 THEN CountProperDivisors = 0    count = 0    FOR i = 1 TO number \ 2        IF number MOD i = 0 THEN count = count + 1    NEXT i    CountProperDivisors = countEND FUNCTION SUB ListProperDivisors (limit)    IF limit < 1 THEN EXIT SUB    FOR i = 1 TO limit        PRINT USING "## ->"; i;        IF i = 1 THEN PRINT " (None)";        FOR j = 1 TO i \ 2            IF i MOD j = 0 THEN PRINT " "; j;        NEXT j        PRINT    NEXT iEND SUB most = 1maxCount = 0 PRINT "The proper divisors of the following numbers are: "; CHR\$(10)ListProperDivisors (10) FOR n = 2 TO 20000'' It is extremely slow in this loop     count = CountProperDivisors(n)    IF count > maxCount THEN        maxCount = count        most = n    END IFNEXT n PRINTPRINT most; " has the most proper divisors, namely "; maxCountEND`
Output:
```Igual que la entrada de FreeBASIC o PureBasic.
```

### BASIC256

`subroutine ListProperDivisors(limit)	if limit < 1 then return	for i = 1 to limit		print i; " ->";		if i = 1 then			print " (None)"			continue for		end if		for j = 1 to i \ 2			if i mod j = 0 then print " "; j;		next j		print	next iend subroutine function CountProperDivisors(number)	if number < 2 then return 0	dim cont = 0	for i = 1 to number \ 2		if number mod i = 0 then cont += 1	next i	return contend function most = 1maxCount = 0 print "The proper divisors of the following numbers are:"printcall ListProperDivisors(10) for n = 2 to 20000	cont = CountProperDivisors(n)	if cont > maxCount then		maxCount = cont		most = n	end ifnext n printprint most; " has the most proper divisors, namely "; maxCountend`
Output:
```Igual que la entrada de FreeBASIC o PureBasic.
```

### True BASIC

`FUNCTION CountProperDivisors (number)    IF number < 2 THEN LET CountProperDivisors = 0    LET count = 0    FOR i = 1 TO INT(number / 2)        IF MOD(number, i) = 0 THEN LET count = count + 1    NEXT i    LET CountProperDivisors = countEND FUNCTION SUB ListProperDivisors (limit)    IF limit < 1 THEN EXIT SUB    FOR i = 1 TO limit        PRINT i; " ->";        IF i = 1 THEN PRINT " (None)";        FOR j = 1 TO INT(i / 2)            IF MOD(i, j) = 0 THEN PRINT " "; j;        NEXT j        PRINT    NEXT iEND SUB LET most = 1LET maxCount = 0 PRINT "The proper divisors of the following numbers are: "; CHR\$(10)CALL LISTPROPERDIVISORS (10) FOR n = 2 TO 20000    LET count = CountProperDivisors(n)    IF count > maxCount THEN       LET maxCount = count       LET most = n    END IFNEXT n PRINTPRINT most; "has the most proper divisors, namely"; maxCountEND`
Output:
```Igual que la entrada de FreeBASIC o PureBasic.
```

### Yabasic

`sub ListProperDivisors(limit)    if limit < 1 then return : fi    for i = 1 to limit        print i, " ->";        if i = 1 then             print " (None)"            continue //for        endif        for j = 1 to int(i / 2)            if mod(i, j) = 0 then print " ", j; : fi        next j        print    next iend sub sub CountProperDivisors(number)    if number < 2 then return 0 : fi    count = 0    for i = 1 to int(number / 2)        if mod(number, i) = 0 then count = count + 1 : fi    next i    return countend sub most = 1maxCount = 0 print "The proper divisors of the following numbers are: \n"ListProperDivisors(10) for n = 2 to 20000    count = CountProperDivisors(n)    if count > maxCount then        maxCount = count        most = n    endifnext n printprint most, " has the most proper divisors, namely ", maxCountend`
Output:
```Igual que la entrada de FreeBASIC o PureBasic.
```

## BaCon

` FUNCTION ProperDivisor(nr, show)     LOCAL probe, total     FOR probe = 1 TO nr-1        IF MOD(nr, probe) = 0 THEN            IF show THEN PRINT " ", probe;            INCR total        END IF    NEXT     RETURN total END FUNCTION FOR x = 1 TO 10    PRINT x, ":";    IF ProperDivisor(x, 1) = 0 THEN PRINT " 0";    PRINTNEXT FOR x = 1 TO 20000    DivisorCount = ProperDivisor(x, 0)    IF DivisorCount > MaxDivisors THEN        MaxDivisors = DivisorCount        MagicNumber = x    END IFNEXT PRINT "Most proper divisors for number in the range 1-20000: ", MagicNumber, " with ", MaxDivisors, " divisors." `
Output:
```1: 0
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
Most proper divisors for number in the range 1-20000: 15120 with 79 divisors.
```

## C

### Brute Force

C has tedious boilerplate related to allocating memory for dynamic arrays, so we just skip the problem of storing values altogether.

` #include <stdio.h>#include <stdbool.h> int proper_divisors(const int n, bool print_flag){    int count = 0;     for (int i = 1; i < n; ++i) {        if (n % i == 0) {            count++;            if (print_flag)                printf("%d ", i);        }    }     if (print_flag)        printf("\n");     return count;} int main(void){    for (int i = 1; i <= 10; ++i) {        printf("%d: ", i);        proper_divisors(i, true);    }     int max = 0;    int max_i = 1;     for (int i = 1; i <= 20000; ++i) {        int v = proper_divisors(i, false);        if (v >= max) {            max = v;            max_i = i;        }    }     printf("%d with %d divisors\n", max_i, max);    return 0;} `
Output:
```1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
18480 with 79 divisors
```

### Number Theoretic

There is no need to go through all the divisors if only the count is needed, this implementation refines the brute force approach by solving the second part of the task via a Number Theory formula. The running time is noticeably faster than the brute force method above. Output is same as the above.

` #include <stdio.h>#include <stdbool.h> int proper_divisors(const int n, bool print_flag){    int count = 0;     for (int i = 1; i < n; ++i) {        if (n % i == 0) {            count++;            if (print_flag)                printf("%d ", i);        }    }     if (print_flag)        printf("\n");     return count;} int countProperDivisors(int n){	int prod = 1,i,count=0; 	while(n%2==0){		count++;		n /= 2;	} 	prod *= (1+count); 	for(i=3;i*i<=n;i+=2){		count = 0; 		while(n%i==0){			count++;			n /= i;		} 		prod *= (1+count);	} 	if(n>2)		prod *= 2; 	return prod - 1;} int main(void){    for (int i = 1; i <= 10; ++i) {        printf("%d: ", i);        proper_divisors(i, true);    }     int max = 0;    int max_i = 1;     for (int i = 1; i <= 20000; ++i) {        int v = countProperDivisors(i);        if (v >= max) {            max = v;            max_i = i;        }    }     printf("%d with %d divisors\n", max_i, max);    return 0;} `

## C#

`namespace RosettaCode.ProperDivisors{    using System;    using System.Collections.Generic;    using System.Linq;     internal static class Program    {        private static IEnumerable<int> ProperDivisors(int number)        {            return                Enumerable.Range(1, number / 2)                    .Where(divisor => number % divisor == 0);        }         private static void Main()        {            foreach (var number in Enumerable.Range(1, 10))            {                Console.WriteLine("{0}: {{{1}}}", number,                    string.Join(", ", ProperDivisors(number)));            }             var record = Enumerable.Range(1, 20000).Select(number => new            {                Number = number,                Count = ProperDivisors(number).Count()            }).OrderByDescending(currentRecord => currentRecord.Count).First();            Console.WriteLine("{0}: {1}", record.Number, record.Count);        }    }}`
Output:
```1: {}
2: {1}
3: {1}
4: {1, 2}
5: {1}
6: {1, 2, 3}
7: {1}
8: {1, 2, 4}
9: {1, 3}
10: {1, 2, 5}
15120: 79```

## C++

`#include <vector>#include <iostream>#include <algorithm> std::vector<int> properDivisors ( int number ) {   std::vector<int> divisors ;   for ( int i = 1 ; i < number / 2 + 1 ; i++ )      if ( number % i == 0 )	 divisors.push_back( i ) ;   return divisors ;} int main( ) {   std::vector<int> divisors ;   unsigned int maxdivisors = 0 ;   int corresponding_number = 0 ;   for ( int i = 1 ; i < 11 ; i++ ) {      divisors =  properDivisors ( i ) ;      std::cout << "Proper divisors of " << i << ":\n" ;      for ( int number : divisors ) { 	 std::cout << number << " " ;      }      std::cout << std::endl ;      divisors.clear( ) ;   }   for ( int i = 11 ; i < 20001 ; i++ ) {      divisors =  properDivisors ( i ) ;      if ( divisors.size( ) > maxdivisors ) {	 maxdivisors = divisors.size( ) ;	 corresponding_number = i ;      }      divisors.clear( ) ;   }    std::cout << "Most divisors has " << corresponding_number <<      " , it has " << maxdivisors << " divisors!\n" ;    return 0 ;} `
Output:
```Proper divisors of 1:

Proper divisors of 2:
1
Proper divisors of 3:
1
Proper divisors of 4:
1 2
Proper divisors of 5:
1
Proper divisors of 6:
1 2 3
Proper divisors of 7:
1
Proper divisors of 8:
1 2 4
Proper divisors of 9:
1 3
Proper divisors of 10:
1 2 5
Most divisors has 15120 , it has 79 divisors!
```

## Ceylon

`shared void run() { 	function divisors(Integer int) => 			if(int <= 1) 			then {} 			else (1..int / 2).filter((Integer element) => element.divides(int)); 	for(i in 1..10) {		print("``i`` => ``divisors(i)``");	} 	value start = 1;	value end = 20k; 	value mostDivisors = 			map {for(i in start..end) i->divisors(i).size}			.inverse()			.max(byKey(byIncreasing(Integer.magnitude))); 	print("the number(s) with the most divisors between ``start`` and ``end`` is/are:	       ``mostDivisors?.item else "nothing"`` with ``mostDivisors?.key else "no"`` divisors");}`
Output:
```1 => []
2 => { 1 }
3 => { 1 }
4 => { 1, 2 }
5 => { 1 }
6 => { 1, 2, 3 }
7 => { 1 }
8 => { 1, 2, 4 }
9 => { 1, 3 }
10 => { 1, 2, 5 }
the number(s) with the most divisors between 1 and 20000 is/are:
[15120, 18480] with 79 divisors```

## Clojure

`(ns properdivisors  (:gen-class)) (defn proper-divisors [n]  " Proper divisors of n"  (if (= n 1)    []  (filter #(= 0 (rem n %)) (range 1 n)))) ;; Property divisors of numbers 1 to 20,000 inclusive(def data (for [n (range 1 (inc 20000))]            [n (proper-divisors n)])) ;; Find Max(defn maximal-key [k x & xs]  " Normal max-key only finds one key that produces maximum, while this function finds them all "  (reduce (fn [ys x]            (let [c (compare (k x) (k (peek ys)))]              (cond                (pos? c) [x]                (neg? c) ys                :else    (conj ys x))))          [x]          xs)) (println "n\tcnt\tPROPER DIVISORS")(doseq [n (range 1 11)]  (let [factors (proper-divisors n)]    (println n "\t" (count factors) "\t" factors))) (def max-data (apply maximal-key (fn [[i pd]] (count pd)) data)) (doseq [[n factors] max-data]  (println n " has " (count factors) " divisors"))  `
Output:
```n	cnt	PROPER DIVISORS
1 	 0 	 []
2 	 1 	 (1)
3 	 1 	 (1)
4 	 2 	 (1 2)
5 	 1 	 (1)
6 	 3 	 (1 2 3)
7 	 1 	 (1)
8 	 3 	 (1 2 4)
9 	 2 	 (1 3)
10 	 3 	 (1 2 5)
15120  has  79  divisors
18480  has  79  divisors
```

## Common Lisp

Ideally, the smallest-divisor function would only try prime numbers instead of odd numbers.

`(defun proper-divisors-recursive (product &optional (results '(1)))   "(int,list)->list::Function to find all proper divisors of a +ve integer."    (defun smallest-divisor (x)      "int->int::Find the smallest divisor of an integer > 1."      (if (evenp x) 2          (do ((lim (truncate (sqrt x)))               (sd 3 (+ sd 2)))              ((or (integerp (/ x sd)) (> sd lim)) (if (> sd lim) x sd)))))    (defun pd-rec (fac)      "(int,int)->nil::Recursive function to find proper divisors of a +ve integer"      (when (not (member fac results))         (push fac results)         (let ((hifac (/ fac (smallest-divisor fac))))            (pd-rec hifac)            (pd-rec (/ product hifac)))))    (pd-rec product)   (butlast (sort (copy-list results) #'<))) (defun task (method &optional (n 1) (most-pds '(0)))   (dotimes (i 19999)      (let ((npds (length (funcall method (incf n))))            (hiest (car most-pds)))         (when (>= npds hiest)            (if (> npds hiest)                (setf most-pds (list npds (list n)))                (setf most-pds (list npds (cons n (second most-pds))))))))   most-pds) (defun main ()   (format t "Task 1:Proper Divisors of [1,10]:~%")   (dotimes (i 10) (format t "~A:~A~%" (1+ i) (proper-divisors-recursive (1+ i))))   (format t "Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:~%~A~%"           (task #'proper-divisors-recursive)))`
Output:
```CL-USER(10): (main)
1:NIL
2:(1)
3:(1)
4:(1 2)
5:(1)
6:(1 2 3)
7:(1)
8:(1 2 4)
9:(1 3)
10:(1 2 5)
Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:
(79 (18480 15120))
NIL```

## Component Pascal

` MODULE RosettaProperDivisor;IMPORT StdLog; PROCEDURE Pd*(n: LONGINT;OUT r: ARRAY OF LONGINT):LONGINT;VAR	i,j: LONGINT;BEGIN	i := 1;j := 0;	IF n >  1 THEN		WHILE (i < n) DO			IF (n MOD i) = 0 THEN 				IF (j < LEN(r)) THEN r[j] := i END; INC(j)			END;			INC(i)		END;	END;	RETURN jEND Pd; PROCEDURE Do*;VAR	r: ARRAY 128 OF LONGINT;	i,j,found,max,idxMx: LONGINT;	mx: ARRAY 128 OF LONGINT;BEGIN	FOR i := 1 TO 10 DO		found := Pd(i,r);		IF found > LEN(r) THEN (* Error. more pd than r can admit *) HALT(1) END;		StdLog.Int(i);StdLog.String("[");StdLog.Int(found);StdLog.String("]:> ");		FOR j := 0 TO found - 1 DO			StdLog.Int(r[j]);StdLog.Char(' ');		END;		StdLog.Ln	END; 	max := 0;idxMx := 0;  FOR i := 1 TO 20000 DO  	found := Pd(i,r);  	IF found > max THEN    	idxMx:= 0;mx[idxMx] := i;max := found	  ELSIF found = max THEN    	INC(idxMx);mx[idxMx] := i  	END;  END;	StdLog.String("Found: ");StdLog.Int(idxMx + 1);  StdLog.String(" Numbers with the longest proper divisors [");	StdLog.Int(max);StdLog.String("]: ");StdLog.Ln;	FOR i := 0 TO idxMx DO  	StdLog.Int(mx[i]);StdLog.Ln	ENDEND Do; END RosettaProperDivisor. ^Q RosettaProperDivisor.Do~ `
Output:
``` 1[ 0]:>
2[ 1]:>  1
3[ 1]:>  1
4[ 2]:>  1  2
5[ 1]:>  1
6[ 3]:>  1  2  3
7[ 1]:>  1
8[ 3]:>  1  2  4
9[ 2]:>  1  3
10[ 3]:>  1  2  5
Found:  2 Numbers with the longest proper divisors [ 79]:
15120
18480
```

## D

Translation of: Python

Currently the lambda of the filter allocates a closure on the GC-managed heap.

`void main() /*@safe*/ {    import std.stdio, std.algorithm, std.range, std.typecons;     immutable properDivs = (in uint n) pure nothrow @safe /*@nogc*/ =>        iota(1, (n + 1) / 2 + 1).filter!(x => n % x == 0 && n != x);     iota(1, 11).map!properDivs.writeln;    iota(1, 20_001).map!(n => tuple(properDivs(n).count, n)).reduce!max.writeln;}`
Output:
```[[], , , [1, 2], , [1, 2, 3], , [1, 2, 4], [1, 3], [1, 2, 5]]
Tuple!(uint, int)(79, 18480)```

The Run-time is about 0.67 seconds with the ldc2 compiler.

## Delphi

Translation of: C#
` program ProperDivisors; {\$APPTYPE CONSOLE} {\$R *.res} uses  System.SysUtils,  System.Generics.Collections; type  TProperDivisors = TArray<Integer>; function GetProperDivisors(const value: Integer): TProperDivisors;var  i, count: Integer;begin  count := 0;   for i := 1 to value div 2 do  begin    if value mod i = 0 then    begin      inc(count);      SetLength(result, count);      Result[count - 1] := i;    end;  end;end; procedure Println(values: TProperDivisors);var  i: Integer;begin  Write('[');  if Length(values) > 0 then    for i := 0 to High(values) do      Write(Format('%2d', [values[i]]));  Writeln(']');end; var  number, max_count, count, max_number: Integer; begin  for number := 1 to 10 do  begin    write(number, ': ');    Println(GetProperDivisors(number));  end;   max_count := 0;  for number := 1 to 20000 do  begin    count := length(GetProperDivisors(number));    if count > max_count then    begin      max_count := count;      max_number := number;    end;  end;   Write(max_number, ': ', max_count);   readln;end. `
Output:
```1: []
2: [ 1]
3: [ 1]
4: [ 1 2]
5: [ 1]
6: [ 1 2 3]
7: [ 1]
8: [ 1 2 4]
9: [ 1 3]
10: [ 1 2 5]
15120: 79
```

Version with TParallel.For

` program ProperDivisors; {\$APPTYPE CONSOLE} {\$R *.res} uses  System.SysUtils,  System.Threading,  System.SyncObjs; type  TProperDivisors = array of Integer; function GetProperDivisors(const value: Integer): TProperDivisors;var  i, count: Integer;begin  count := 0;   for i := 1 to value div 2 do  begin    if value mod i = 0 then    begin      inc(count);      SetLength(result, count);      Result[count - 1] := i;    end;  end;end; procedure Println(values: TProperDivisors);var  i: Integer;begin  Write('[');  if Length(values) > 0 then    for i := 0 to High(values) do      Write(Format('%2d', [values[i]]));  Writeln(']');end; var  number, max_count, count, max_number: Integer; begin  for number := 1 to 10 do  begin    write(number, ': ');    Println(GetProperDivisors(number));  end;   max_count := 0;  TParallel.for (1, 20000,    procedure(I: Int64)    begin      count := length(GetProperDivisors(I));      if count > max_count then      begin        TInterlocked.Exchange(max_count, count);        TInterlocked.Exchange(max_number, I);      end;    end);   Writeln(max_number, ': ', max_count);  readln;end. `

## Dyalect

Translation of: Swift
`func properDivs(n) {    if n == 1 {        yield break    }    for x in 1..<n {        if n % x == 0 {            yield x        }    }} for i in 1..10 {    print("\(i): \(properDivs(i).ToArray())")} var (num, max) = (0,0) for i in 1..20000 {    let count = properDivs(i).Length()    if count > max {         set (num, max) = (i, count)     }} print("\(num): \(max)")`
Output:
```1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79```

## EchoLisp

` (lib 'list) ;; list-delete ;; let n = product p_i^a_i , p_i prime;; number of divisors = product (a_i + 1) - 1(define (numdivs n)    (1- (apply * (map (lambda(g) (1+ (length g))) (group (prime-factors n)))))) (remember 'numdivs) ;; prime powers;; input : a list g of grouped prime factors ( 3 3 3 ..);; returns (1 3 9 27 ...)(define (ppows g (mult 1))	(for/fold (ppows '(1)) ((a g))	(set! mult (* mult a))	(cons mult ppows))) ;; proper divisors;; decomp n into ((2 2 ..) ( 3 3 ..)  ) prime factors groups;; combines (1 2 4 8 ..) (1 3 9 ..) lists;; remove n from the list (define (divs n)   (if (<= n 1) null     (list-delete        (for/fold (divs'(1)) ((g (map  ppows (group (prime-factors n)))))		    (for*/list ((a divs) (b g)) (* a b)))    n ))) ;; find number(s) with max # of proper divisors;; returns list of (n . maxdivs)  for n in range 2..N (define (most-proper N)    (define maxdivs 1)    (define ndivs 0)    (for/fold (most-proper null) ((n (in-range 2 N)))       (set! ndivs (numdivs n))        #:continue (< ndivs maxdivs)        (when (> ndivs maxdivs)        (set!-values (most-proper maxdivs) (values null ndivs)))        (cons (cons n maxdivs) most-proper)))   `
Output:
` (for ((i (in-range 1 11))) (writeln i (divs i)))1     null    2     (1)    3     (1)    4     (2 1)    5     (1)    6     (2 3 1)    7     (1)    8     (4 2 1)    9     (3 1)    10     (2 5 1)    (most-proper 20000)    → ((18480 . 79) (15120 . 79))(most-proper 1_000_000)    → ((997920 . 239) (982800 . 239) (942480 . 239) (831600 . 239) (720720 . 239))   (lib 'bigint)(numdivs 95952222101012742144)  → 666 ;; 🎩 `

## Eiffel

` class	APPLICATION create	make feature 	make			-- Test the feature proper_divisors.		local			list: LINKED_LIST [INTEGER]			count, number: INTEGER		do			across				1 |..| 10 as c			loop				list := proper_divisors (c.item)				io.put_string (c.item.out + ": ")				across					list as l				loop					io.put_string (l.item.out + " ")				end				io.new_line			end			across				1 |..| 20000 as c			loop				list := proper_divisors (c.item)				if list.count > count then					count := list.count					number := c.item				end			end			io.put_string (number.out + " has with " + count.out + " divisors the highest number of proper divisors.")		end 	proper_divisors (n: INTEGER): LINKED_LIST [INTEGER]			-- Proper divisors of 'n'.		do			create Result.make			across				1 |..| (n - 1) as c			loop				if n \\ c.item = 0 then					Result.extend (c.item)				end			end		end end `
Output:
```1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
15120 has with 79 divisors the highest number of proper divisors.
```

## Elixir

Translation of: Erlang
`defmodule Proper do  def divisors(1), do: []  def divisors(n), do: [1 | divisors(2,n,:math.sqrt(n))] |> Enum.sort   defp divisors(k,_n,q) when k>q, do: []  defp divisors(k,n,q) when rem(n,k)>0, do: divisors(k+1,n,q)  defp divisors(k,n,q) when k * k == n, do: [k | divisors(k+1,n,q)]  defp divisors(k,n,q)                , do: [k,div(n,k) | divisors(k+1,n,q)]   def most_divisors(limit) do    {length,nums} = Enum.group_by(1..limit, fn n -> length(divisors(n)) end)                    |> Enum.max_by(fn {length,_nums} -> length end)    IO.puts "With #{length}, Number #{inspect nums} has the most divisors"  endend Enum.each(1..10, fn n ->  IO.puts "#{n}: #{inspect Proper.divisors(n)}"end)Proper.most_divisors(20000)`
Output:
```1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
With 79, Number [18480, 15120] has the most divisors
```

## Erlang

`-module(properdivs).-export([divs/1,sumdivs/1,longest/1]). divs(0) -> [];divs(1) -> [];divs(N) -> lists:sort( ++ divisors(2,N,math:sqrt(N))). divisors(K,_N,Q) when K > Q -> [];divisors(K,N,Q) when N rem K =/= 0 ->     divisors(K+1,N,Q);divisors(K,N,Q) when K * K  == N ->     [K] ++ divisors(K+1,N,Q);divisors(K,N,Q) ->    [K, N div K] ++ divisors(K+1,N,Q). sumdivs(N) -> lists:sum(divs(N)). longest(Limit) -> longest(Limit,0,0,1). longest(L,Current,CurLeng,Acc) when Acc >= L ->     io:format("With ~w, Number ~w has the most divisors~n", [CurLeng,Current]);longest(L,Current,CurLeng,Acc) ->         A = length(divs(Acc)),    if A > CurLeng ->        longest(L,Acc,A,Acc+1);        true -> longest(L,Current,CurLeng,Acc+1)    end.`
Output:
```1> [io:format("X: ~w, N: ~w~n", [N,properdivs:divs(N)]) ||  N <- lists:seq(1,10)].
X: 1, N: []
X: 2, N: 
X: 3, N: 
X: 4, N: [1,2]
X: 5, N: 
X: 6, N: [1,2,3]
X: 7, N: 
X: 8, N: [1,2,4]
X: 9, N: [1,3]
X: 10, N: [1,2,5]
[ok,ok,ok,ok,ok,ok,ok,ok,ok,ok]

2> properdivs:longest(20000).
With 79, Number 15120 has the most divisors
```

## F#

` // the simple function with the answerlet propDivs n = [1..n/2] |> List.filter (fun x->n % x = 0) // to cache the result length; helpful for a long searchlet propDivDat n = propDivs n |> fun xs -> n, xs.Length, xs // UI: always the longest and messiestlet show (n,count,divs) =  let showCount = count |> function | 0-> "no proper divisors" | 1->"1 proper divisor" | _-> sprintf "%d proper divisors" count  let showDiv = divs |> function | []->"" | x::[]->sprintf ": %d" x | _->divs |> Seq.map string |> String.concat "," |> sprintf ": %s"  printfn "%d has %s%s" n showCount showDiv // generate output[1..10] |> List.iter (propDivDat >> show) // use a sequence: we don't really need to hold this data, just iterate over itSeq.init 20000 ( ((+) 1) >> propDivDat)|> Seq.fold (fun a b ->match a,b with | (_,c1,_),(_,c2,_) when c2 > c1 -> b | _-> a) (0,0,[]) |> fun (n,count,_) -> (n,count,[]) |> show `
Output:
```1 has no proper divisors
2 has 1 proper divisor: 1
3 has 1 proper divisor: 1
4 has 2 proper divisors: 1,2
5 has 1 proper divisor: 1
6 has 3 proper divisors: 1,2,3
7 has 1 proper divisor: 1
8 has 3 proper divisors: 1,2,4
9 has 2 proper divisors: 1,3
10 has 3 proper divisors: 1,2,5
15120 has 79 proper divisors
```

## Factor

`USING: formatting io kernel math math.functionsmath.primes.factors math.ranges prettyprint sequences ; : #divisors ( m -- n )    dup sqrt >integer 1 + [1,b] [ divisor? ] with count dup +    1 - ; 10 [1,b] [ dup pprint bl divisors but-last . ] each20000 [1,b] [ #divisors ] supremum-by dup #divisors"%d with %d divisors.\n" printf`
Output:
```1 { }
2 { 1 }
3 { 1 }
4 { 1 2 }
5 { 1 }
6 { 1 2 3 }
7 { 1 }
8 { 1 2 4 }
9 { 1 3 }
10 { 1 2 5 }
15120 with 79 divisors.
```

## Fermat

`Func Divisors(n) =    [d]:=[(1)];                                            {start divisor list with just 1, which is a divisor of everything}    for i = 2 to n\2 do                                    {loop through possible divisors of n}        if Divides(i, n) then [d]:=[d]_[(i)] fi               od;    .; for n = 1 to 10 do    Divisors(n);    !!(n,'    ',[d);od; record:=0;champ:=1;for n=2 to 20000 do    Divisors(n);    m:=Cols[d];                                            {this gets the length of the array}    if m > record then        champ:=n;        record:=m;    fi;od; !!('The number up to 20,000 with the most divisors was ',champ,' with ',record,' divisors.');`
Output:
```
1    1
2    1
3    1
4    1,  2
5    1
6    1,  2,  3
7    1
8    1,  2,  4
9    1,  3
10    1,  2,  5

The number up to 20,000 with the most divisors was  15120 with  79 divisors.```

## Forth

Works with: gforth version 0.7.3
`: .proper-divisors  dup 1 ?do    dup i mod 0= if i . then  loop cr drop; : proper-divisors-count  0 swap  dup 1 ?do    dup i mod 0= if swap 1 + swap then  loop drop; : rosetta-proper-divisors  cr  11 1 do    i . ." : " i .proper-divisors  loop   1 0  20000 2 do    i proper-divisors-count    2dup < if nip nip i swap else drop then  loop  swap cr . ." has " . ." divisors" cr; rosetta-proper-divisors`
Output:
```1 :
2 : 1
3 : 1
4 : 1 2
5 : 1
6 : 1 2 3
7 : 1
8 : 1 2 4
9 : 1 3
10 : 1 2 5

15120 has 79 divisors
ok```

## Fortran

Compiled using G95 compiler, run on x86 system under Puppy Linux

`        function icntprop(num  )      icnt=0      do i=1 , num-1          if (mod(num , i)  .eq. 0)  then          icnt = icnt + 1          if (num .lt. 11) print *,'    ',i          end if          end do      icntprop =  icnt      end function       limit = 20000      maxcnt = 0      print *,'N   divisors'      do j=1,limit,1      if (j .lt. 11) print *,j      icnt = icntprop(j)       if (icnt .gt. maxcnt) then      maxcnt = icnt      maxj = j      end if       end do       print *,' '      print *,' from 1 to ',limit      print *,maxj,' has max proper divisors: ',maxcnt      end `
Output:
``` N   divisors
1
2
1
3
1
4
1
2
5
1
6
1
2
3
7
1
8
1
2
4
9
1
3
10
1
2
5

from 1 to  20000
15120  has max proper divisors:  79
```

## FreeBASIC

` ' FreeBASIC v1.05.0 win64 Sub ListProperDivisors(limit As Integer)  If limit < 1 Then Return  For i As Integer = 1 To limit     Print Using "##"; i;      Print " ->";     If i = 1 Then        Print " (None)"       Continue For     End if     For j As Integer = 1 To i \ 2       If i Mod j = 0 Then Print " "; j;     Next j     Print  Next iEnd Sub Function CountProperDivisors(number As Integer) As Integer  If number < 2 Then Return 0  Dim count As Integer = 0  For i As Integer = 1 To number \ 2    If number Mod i = 0 Then count += 1  Next  Return countEnd Function Dim As Integer n, count, most = 1, maxCount = 0 Print "The proper divisors of the following numbers are :"PrintListProperDivisors(10) For n As Integer = 2 To 20000  count = CountProperDivisors(n)  If count > maxCount Then    maxCount = count    most = n  EndIfNext PrintPrint Str(most); " has the most proper divisors, namely"; maxCountPrintPrint "Press any key to exit the program"SleepEnd `
Output:
```The proper divisors of the following numbers are :

1 -> (None)
2 ->  1
3 ->  1
4 ->  1  2
5 ->  1
6 ->  1  2  3
7 ->  1
8 ->  1  2  4
9 ->  1  3
10 ->  1  2  5

15120 has the most proper divisors, namely 79
```

## Free Pascal

` Program ProperDivisors; Uses fgl; Type   TIntegerList = Specialize TfpgList<longint>; Var list : TintegerList; Function GetProperDivisors(x : longint): longint;{this function will return the number of proper divisors and put them in the list} Var i : longint;Begin  list.clear;  If x = 1 Then {by default 1 has no proper divisors}    GetProperDivisors := 0  Else    Begin      list.add(1); //add 1 as a proper divisor;      i := 2;      While i * i < x Do        Begin          If (x Mod i) = 0 Then //found a proper divisor            Begin              list.add(i); // add divisor              list.add(x Div i); // add result            End;          inc(i);        End;      If i*i=x Then list.add(i); //make sure to capture the sqrt only once      GetProperDivisors := list.count;    End;End; Var i,j,count,most : longint;Begin   list := TIntegerList.Create;  For i := 1 To 10 Do    Begin      write(i:4,' has ', GetProperDivisors(i),' proper divisors:');      For j := 0 To pred(list.count) Do        write(list[j]:3);      writeln();    End;  count := 0; //store highest number of proper divisors  most := 0;  //store number with highest number of proper divisors  For i := 1 To 20000 Do    If GetProperDivisors(i) > count Then      Begin        count := list.count;        most := i;      End;  writeln(most,' has ',count,' proper divisors');  list.free;End. `
Output:
```   1 has 0 proper divisors:
2 has 1 proper divisors:  1
3 has 1 proper divisors:  1
4 has 2 proper divisors:  1  2
5 has 1 proper divisors:  1
6 has 3 proper divisors:  1  2  3
7 has 1 proper divisors:  1
8 has 3 proper divisors:  1  2  4
9 has 2 proper divisors:  1  3
10 has 3 proper divisors:  1  2  5
15120 has 79 proper divisors
```

## Frink

Frink's built-in factorization routines efficiently find factors of arbitrary-sized integers.

` for n = 1 to 10   println["\$n\t" + join[" ", properDivisors[n]]] println[] d = new dictfor n = 1 to 20000{   c = length[properDivisors[n]]   d.addToList[c, n]} most = max[keys[d]]println[[email protected] + " have \$most factors"] properDivisors[n] := allFactors[n, true, false, true] `
Output:
```1
2       1
3       1
4       1 2
5       1
6       1 2 3
7       1
8       1 2 4
9       1 3
10      1 2 5

[15120, 18480] have 79 factors
```

## GFA Basic

` OPENW 1CLEARW 1'' Array f% is used to hold the divisorsDIM f%(SQR(20000)) ! cannot redim arrays, so set size to largest needed'' 1. Show proper divisors of 1 to 10, inclusive'FOR i%=1 TO 10  num%[email protected]_divisors(i%)  PRINT "Divisors for ";i%;":";  FOR j%=1 TO num%    PRINT " ";f%(j%);  NEXT j%  PRINTNEXT i%'' 2. Find (smallest) number <= 20000 with largest number of proper divisors'result%=1 ! largest so farnumber%=0 ! its number of divisorsFOR i%=1 TO 20000  num%[email protected]_divisors(i%)  IF num%>number%    result%=i%    number%=num%  ENDIFNEXT i%PRINT "Largest number of divisors is ";number%;" for ";result%'~INP(2)CLOSEW 1'' find the proper divisors of n%, placing results in f%' and return the number found'FUNCTION proper_divisors(n%)  LOCAL i%,root%,count%  '  ARRAYFILL f%(),0  count%=1 ! index of next slot in f% to fill  '  IF n%>1    f%(count%)=1    count%=count%+1    root%=SQR(n%)    FOR i%=2 TO root%      IF n% MOD i%=0        f%(count%)=i%        count%=count%+1        IF i%*i%<>n% ! root% is an integer, so check if i% is actual squa- lists:seq(1,10)].                                      X: 1, N: []X: 2, N: X: 3, N: X: 4, N: [1,2]X: 5, N: X: 6, N: [1,2,3]X: 7, N: X: 8, N: [1,2,4]X: 9, N: [1,3]X: 10, N: [1,2,5][ok,ok,ok,ok,ok,ok,ok,ok,ok,ok] 2> properdivs:longest(20000).With 79, Number 15120 has the most divisorsre root of n%          f%(count%)=n%/i%          count%=count%+1        ENDIF      ENDIF    NEXT i%  ENDIF  '  RETURN count%-1ENDFUNC `

Output is:

```Divisors for 1:
Divisors for 2: 1
Divisors for 3: 1
Divisors for 4: 1 2
Divisors for 5: 1
Divisors for 6: 1 2 3
Divisors for 7: 1
Divisors for 8: 1 2 4
Divisors for 9: 1 3
Divisors for 10: 1 2 5
Largest number of divisors is 79 for 15120
```

## Go

Translation of: Kotlin
`package main import (    "fmt"    "strconv") func listProperDivisors(limit int) {    if limit < 1 {        return    }    width := len(strconv.Itoa(limit))    for i := 1; i <= limit; i++ {        fmt.Printf("%*d -> ", width, i)        if i == 1 {            fmt.Println("(None)")            continue        }        for j := 1; j <= i/2; j++ {            if i%j == 0 {                fmt.Printf(" %d", j)            }        }        fmt.Println()    }} func countProperDivisors(n int) int {    if n < 2 {        return 0    }    count := 0    for i := 1; i <= n/2; i++ {        if n%i == 0 {            count++        }    }    return count} func main() {    fmt.Println("The proper divisors of the following numbers are :\n")    listProperDivisors(10)    fmt.Println()    maxCount := 0    most := []int{1}    for n := 2; n <= 20000; n++ {        count := countProperDivisors(n)        if count == maxCount {            most = append(most, n)        } else if count > maxCount {            maxCount = count            most = most[0:1]            most = n        }    }    fmt.Print("The following number(s) <= 20000 have the most proper divisors, ")    fmt.Println("namely", maxCount, "\b\n")    for _, n := range most {        fmt.Println(n)    }}`
Output:
```The proper divisors of the following numbers are :

1 -> (None)
2 ->  1
3 ->  1
4 ->  1 2
5 ->  1
6 ->  1 2 3
7 ->  1
8 ->  1 2 4
9 ->  1 3
10 ->  1 2 5

The following number(s) <= 20000 have the most proper divisors, namely 79

15120
18480
```

`import Data.Ordimport Data.List divisors :: (Integral a) => a -> [a]divisors n = filter ((0 ==) . (n `mod`)) [1 .. (n `div` 2)] main :: IO ()main = do  putStrLn "divisors of 1 to 10:"  mapM_ (print . divisors) [1 .. 10]  putStrLn "a number with the most divisors within 1 to 20000 (number, count):"  print \$ maximumBy (comparing snd)    [(n, length \$ divisors n) | n <- [1 .. 20000]]`
Output:
```divisors of 1 to 10:
[]


[1,2]

[1,2,3]

[1,2,4]
[1,3]
[1,2,5]
a number with the most divisors within 1 to 20000 (number, count):
(18480,79)```

For a little more efficiency, we can filter only up to the root – deriving the higher proper divisors from the lower ones, as quotients:

`import Data.List (maximumBy)import Data.Ord (comparing)import Data.Bool (bool) properDivisors  :: Integral a  => a -> [a]properDivisors n =  let root = (floor . sqrt . fromIntegral) n      lows = filter ((0 ==) . rem n) [1 .. root]  in init (lows ++ bool id tail (n == root * root) (reverse (quot n <\$> lows))) main :: IO ()main = do  putStrLn "Proper divisors of 1 to 10:"  mapM_ (print . properDivisors) [1 .. 10]  mapM_    putStrLn    [ ""    , "A number in the range 1 to 20,000 with the most proper divisors,"    , "as (number, count of proper divisors):"    , ""    ]  print \$    maximumBy (comparing snd) \$    (,) <*> (length . properDivisors) <\$> [1 .. 20000]`
Output:
```Proper divisors of 1 to 10:
[]


[1,2]

[1,2,3]

[1,2,4]
[1,3]
[1,2,5]

A number in the range 1 to 20,000 with the most proper divisors,
as (number, count of proper divisors):

(18480,79)```

and we can also define properDivisors in terms of primeFactors:

`import Data.Numbers.Primes (primeFactors)import Data.List (group, maximumBy, sort)import Data.Ord (comparing) properDivisors :: Int -> [Int]properDivisors =  init . sort . foldr (    flip ((<*>) . fmap (*)) . scanl (*) 1  )  . group . primeFactors ---------------------------TEST----------------------------main :: IO ()main = do  putStrLn \$    fTable "Proper divisors of [1..10]:" show show properDivisors [1 .. 10]  mapM_    putStrLn    [ ""    , "A number in the range 1 to 20,000 with the most proper divisors,"    , "as (number, count of proper divisors):"    , ""    ]  print \$    maximumBy (comparing snd) \$    (,) <*> (length . properDivisors) <\$> [1 .. 20000] --------------------------DISPLAY--------------------------fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> StringfTable s xShow fxShow f xs =  let rjust n c = (drop . length) <*> (replicate n c ++)      w = maximum (length . xShow <\$> xs)  in unlines \$     s : fmap (((++) . rjust w ' ' . xShow) <*> ((" -> " ++) . fxShow . f)) xs`
Output:
```Proper divisors of [1..10]:
1 -> []
2 -> 
3 -> 
4 -> [1,2]
5 -> 
6 -> [1,2,3]
7 -> 
8 -> [1,2,4]
9 -> [1,3]
10 -> [1,2,5]

A number in the range 1 to 20,000 with the most proper divisors,
as (number, count of proper divisors):

(18480,79)```

## J

The proper divisors of an integer are the Factors of an integer without the integer itself.

So, borrowing from the J implementation of that related task:

`factors=: [: /:[email protected], */&>@{@((^ [email protected]>:)&.>/)@q:~&__properDivisors=: factors -. ]`

Proper divisors of numbers 1 through 10:

`   (,&": ' -- ' ,&": properDivisors)&>1+i.101 --       2 -- 1     3 -- 1     4 -- 1 2   5 -- 1     6 -- 1 2 3 7 -- 1     8 -- 1 2 4 9 -- 1 3   10 -- 1 2 5`

Number(s) not exceeding 20000 with largest number of proper divisors (and the count of those divisors):

`   (, #@properDivisors)&> 1+I.(= >./) #@[email protected]> 1+i.2000015120 7918480 79`

Note that it's a bit more efficient to simply count factors here, when selecting the candidate numbers.

`      (, #@properDivisors)&> 1+I.(= >./) #@[email protected]> 1+i.2000015120 7918480 79`

We could also arbitrarily toss either 15120 or 18480 (keeping the other number), if it were important that we produce only one result.

## Java

Works with: Java version 1.5+
`import java.util.Collections;import java.util.LinkedList;import java.util.List; public class Proper{    public static List<Integer> properDivs(int n){        List<Integer> divs = new LinkedList<Integer>();        if(n == 1) return divs;        divs.add(1);        for(int x = 2; x < n; x++){            if(n % x == 0) divs.add(x);        }         Collections.sort(divs);         return divs;    }     public static void main(String[] args){        for(int x = 1; x <= 10; x++){            System.out.println(x + ": " + properDivs(x));        }         int x = 0, count = 0;        for(int n = 1; n <= 20000; n++){            if(properDivs(n).size() > count){                x = n;                count = properDivs(n).size();            }        }        System.out.println(x + ": " + count);    }}`
Output:
```1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79```

## JavaScript

### ES5

`(function () {     // Proper divisors    function properDivisors(n) {        if (n < 2) return [];        else {            var rRoot = Math.sqrt(n),                intRoot = Math.floor(rRoot),                 lows = range(1, intRoot).filter(function (x) {                    return (n % x) === 0;                });             return lows.concat(lows.slice(1).map(function (x) {                return n / x;            }).reverse().slice((rRoot === intRoot) | 0));        }    }     // [m..n]    function range(m, n) {        var a = Array(n - m + 1),            i = n + 1;        while (i--) a[i - 1] = i;        return a;    }     var tblOneToTen = [            ['Number', 'Proper Divisors', 'Count']        ].concat(range(1, 10).map(function (x) {            var ds = properDivisors(x);             return [x, ds.join(', '), ds.length];        })),         dctMostBelow20k = range(1, 20000).reduce(function (a, x) {            var lng = properDivisors(x).length;             return lng > a.divisorCount ? {                n: x,                divisorCount: lng            } : a;        }, {            n: 0,            divisorCount: 0        });      // [[a]] -> bool -> s -> s    function wikiTable(lstRows, blnHeaderRow, strStyle) {        return '{| class="wikitable" ' + (            strStyle ? 'style="' + strStyle + '"' : ''        ) + lstRows.map(function (lstRow, iRow) {            var strDelim = ((blnHeaderRow && !iRow) ? '!' : '|');             return '\n|-\n' + strDelim + ' ' + lstRow.map(function (v) {                return typeof v === 'undefined' ? ' ' : v;            }).join(' ' + strDelim + strDelim + ' ');        }).join('') + '\n|}';    }     return wikiTable(        tblOneToTen,        true    ) + '\n\nMost proper divisors below 20,000:\n\n  ' + JSON.stringify(        dctMostBelow20k    ); })();`
Output:
Number Proper Divisors Count
1 0
2 1 1
3 1 1
4 1, 2 2
5 1 1
6 1, 2, 3 3
7 1 1
8 1, 2, 4 3
9 1, 3 2
10 1, 2, 5 3

Most proper divisors below 20,000:

``` {"n":15120,"divisorCount":79}
```

### ES6

`(() => {    'use strict';     // properDivisors :: Int -> [Int]    const properDivisors = n => {        // The integer divisors of n, excluding n itself.        const            rRoot = Math.sqrt(n),            intRoot = Math.floor(rRoot),            blnPerfectSquare = rRoot === intRoot,            lows = enumFromTo(1)(intRoot)            .filter(x => 0 === (n % x));         // For perfect squares, we can drop        // the head of the 'highs' list        return lows.concat(lows                .map(x => n / x)                .reverse()                .slice(blnPerfectSquare | 0)            )            .slice(0, -1); // except n itself    };      // ------------------------TESTS-----------------------    // main :: IO ()    const main = () =>        console.log([            fTable('Proper divisors of [1..10]:')(str)(                JSON.stringify            )(properDivisors)(enumFromTo(1)(10)),            '',            'Example of maximum divisor count in the range [1..20000]:',            '    ' + maximumBy(comparing(snd))(                enumFromTo(1)(20000).map(                    n => [n, properDivisors(n).length]                )            ).join(' has ') + ' proper divisors.'        ].join('\n'));      // -----------------GENERIC FUNCTIONS------------------     // comparing :: (a -> b) -> (a -> a -> Ordering)    const comparing = f =>        x => y => {            const                a = f(x),                b = f(y);            return a < b ? -1 : (a > b ? 1 : 0);        };     // enumFromTo :: Int -> Int -> [Int]    const enumFromTo = m => n =>        Array.from({            length: 1 + n - m        }, (_, i) => m + i);     // fTable :: String -> (a -> String) -> (b -> String)    //                      -> (a -> b) -> [a] -> String    const fTable = s => xShow => fxShow => f => xs => {        // Heading -> x display function ->        //           fx display function ->        //    f -> values -> tabular string        const            ys = xs.map(xShow),            w = Math.max(...ys.map(x => x.length));        return s + '\n' + zipWith(            a => b => a.padStart(w, ' ') + ' -> ' + b        )(ys)(            xs.map(x => fxShow(f(x)))        ).join('\n');    };     // maximumBy :: (a -> a -> Ordering) -> [a] -> a    const maximumBy = f => xs =>        0 < xs.length ? (            xs.slice(1)            .reduce((a, x) => 0 < f(x)(a) ? x : a, xs)        ) : undefined;     // snd :: (a, b) -> b    const snd = tpl => tpl;     // str :: a -> String    const str = x => x.toString();     // until :: (a -> Bool) -> (a -> a) -> a -> a    const until = p => f => x => {        let v = x;        while (!p(v)) v = f(v);        return v;    };     // zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]    const zipWith = f => xs => ys => {        const            lng = Math.min(xs.length, xs.length),            as = xs.slice(0, lng),            bs = ys.slice(0, lng);        return Array.from({            length: lng        }, (_, i) => f(as[i])(            bs[i]        ));    };     // MAIN ---    return main();})();`
Output:
```Proper divisors of [1..10]:
1 -> []
2 -> 
3 -> 
4 -> [1,2]
5 -> 
6 -> [1,2,3]
7 -> 
8 -> [1,2,4]
9 -> [1,3]
10 -> [1,2,5]

Example of maximum divisor count in the range [1..20000]:
15120 has 79 proper divisors.```

## jq

Works with: jq version 1.4

In the following, proper_divisors returns a stream. In order to count the number of items in the stream economically, we first define "count(stream)":

`def count(stream): reduce stream as \$i (0; . + 1); # unordereddef proper_divisors:  . as \$n  | if \$n > 1 then 1,      ( range(2; 1 + (sqrt|floor)) as \$i        | if (\$n % \$i) == 0 then \$i,            ((\$n / \$i) | if . == \$i then empty else . end)         else empty	 end)    else empty    end; # The first integer in 1 .. n inclusive# with the maximal number of proper divisors in that range:def most_proper_divisors(n):  reduce range(1; n+1) as \$i    ( [null, 0];      count( \$i | proper_divisors ) as \$count      | if \$count > . then [\$i, \$count] else . end);`

`"The proper divisors of the numbers 1 to 10 inclusive are:",(range(1;11) as \$i | "\(\$i): \( [ \$i | proper_divisors] )"),"","The pair consisting of the least number in the range 1 to 20,000 with","the maximal number proper divisors together with the corresponding","count of proper divisors is:",most_proper_divisors(20000) `
Output:
`\$ jq -n -c -r -f /Users/peter/jq/proper_divisors.jqThe proper divisors of the numbers 1 to 10 inclusive are:1: []2: 3: 4: [1,2]5: 6: [1,2,3]7: 8: [1,2,4]9: [1,3]10: [1,2,5] The pair consisting of the least number in the range 1 to 20,000 withthe maximal number proper divisors together with the correspondingcount of proper divisors is:[15120,79]`

## Julia

Use `factor` to obtain the prime factorization of the target number. I adopted the argument handling style of `factor` in my `properdivisors` function.

` function properdivisors{T<:Integer}(n::T)    0 < n || throw(ArgumentError("number to be factored must be ≥ 0, got \$n"))    1 < n || return T[]    !isprime(n) || return T[one(T), n]    f = factor(n)    d = T[one(T)]    for (k, v) in f        c = T[k^i for i in 0:v]        d = d*c'        d = reshape(d, length(d))    end    sort!(d)    return d[1:end-1]end lo = 1hi = 10println("List the proper divisors for ", lo, " through ", hi, ".")for i in lo:hi    println(@sprintf("%4d", i), " ", properdivisors(i))end hi = 2*10^4println("\nFind the numbers within [", lo, ",", hi, "] having the most divisors.") maxdiv = 0nlst = Int[] for i in lo:hi    ndiv = length(properdivisors(i))    if ndiv > maxdiv        maxdiv = ndiv        nlst = [i]    elseif ndiv == maxdiv        push!(nlst, i)    endend println(nlst, " have the maximum proper divisor count of ", maxdiv, ".") `
Output:
```List the proper divisors for 1 through 10.
1 []
2 [1,2]
3 [1,3]
4 [1,2]
5 [1,5]
6 [1,2,3]
7 [1,7]
8 [1,2,4]
9 [1,3]
10 [1,2,5]

Find the numbers within [1,20000] having the most divisors.
[15120,18480] have the maximum proper divisor count of 79.
```

## Kotlin

`// version 1.0.5-2 fun listProperDivisors(limit: Int) {    if (limit < 1) return    for(i in 1..limit) {        print(i.toString().padStart(2) + " -> ")        if (i == 1) {            println("(None)")            continue        }        (1..i/2).filter{ i % it == 0 }.forEach { print(" \$it") }        println()    }} fun countProperDivisors(n: Int): Int {    if (n < 2) return 0    return (1..n/2).count { (n % it) == 0 }} fun main(args: Array<String>) {     println("The proper divisors of the following numbers are :\n")    listProperDivisors(10)    println()    var count: Int    var maxCount = 0    val most: MutableList<Int> = mutableListOf(1)    for (n in 2..20000) {        count = countProperDivisors(n)        if (count == maxCount)            most.add(n)        else if (count > maxCount) {            maxCount = count            most.clear()            most.add(n)        }    }    println("The following number(s) have the most proper divisors, namely " + maxCount + "\n")    for (n in most) println(n)}`
Output:
```The proper divisors of the following numbers are :

1 -> (None)
2 ->  1
3 ->  1
4 ->  1 2
5 ->  1
6 ->  1 2 3
7 ->  1
8 ->  1 2 4
9 ->  1 3
10 ->  1 2 5

The following number(s) have the most proper divisors, namely 79

15120
18480
```

## Lua

`-- Return a table of the proper divisors of nfunction propDivs (n)    if n < 2 then return {} end    local divs, sqr = {1}, math.sqrt(n)    for d = 2, sqr do        if n % d == 0 then            table.insert(divs, d)            if d ~= sqr then table.insert(divs, n/d) end        end    end    table.sort(divs)    return divsend -- Show n followed by all values in tfunction show (n, t)    io.write(n .. ":\t")    for _, v in pairs(t) do io.write(v .. " ") end    print()end -- Main procedurelocal mostDivs, numDivs, answer = 0for i = 1, 10 do show(i, propDivs(i)) endfor i = 1, 20000 do    numDivs = #propDivs(i)    if numDivs > mostDivs then        mostDivs = numDivs        answer = i    endendprint(answer .. " has " .. mostDivs .. " proper divisors.")`
Output:
```1:
2:      1
3:      1
4:      1 2
5:      1
6:      1 2 3
7:      1
8:      1 2 4
9:      1 3
10:     1 2 5
15120 has 79 proper divisors.```

## Mathematica / Wolfram Language

A Function that yields the proper divisors of an integer n:

`ProperDivisors[n_Integer /; n > 0] := [email protected]@n;`

Proper divisors of n from 1 to 10:

`[email protected][{n, ProperDivisors[n]}, {n, 1, 10}]`
Output:
```1	{}
2	{1}
3	{1}
4	{1,2}
5	{1}
6	{1,2,3}
7	{1}
8	{1,2,4}
9	{1,3}
10	{1,2,5}```

The number with the most divisors between 1 and 20,000:

`Fold[ Last[SortBy[{#1, {#2, [email protected][#2]}}, Last]] &, {0, 0}, Range]`
Output:
`{18480, 79}`

An alternate way to find the number with the most divisors between 1 and 20,000:

`[email protected][  Table[    {n, [email protected][n]},    {n, 1, 20000}],  Last]`
Output:
`{15120, 79}`

## MATLAB

`function D=pd(N)K=1:ceil(N/2);D=K(~(rem(N, K)));`
Output:
```for I=1:10
disp([num2str(I) ' : ' num2str(pd(I))])
end
1 : 1
2 : 1
3 : 1
4 : 1  2
5 : 1
6 : 1  2  3
7 : 1
8 : 1  2  4
9 : 1  3
10 : 1  2  5

maxL=0; maxI=0;
for I=1:20000
L=length(pd(I));
if L>maxL
maxL=L; maxI=I;
end
end
maxI

maxI =

15120

maxL

maxL =

79
```

## Modula-2

`MODULE ProperDivisors;FROM FormatString IMPORT FormatString;FROM Terminal IMPORT WriteString,WriteLn,ReadChar; PROCEDURE WriteInt(n : INTEGER);VAR buf : ARRAY[0..15] OF CHAR;BEGIN    FormatString("%i", buf, n);    WriteString(buf)END WriteInt; PROCEDURE proper_divisors(n : INTEGER; print_flag : BOOLEAN) : INTEGER;VAR count,i : INTEGER;BEGIN    count := 0;    FOR i:=1 TO n-1 DO        IF n MOD i = 0 THEN            INC(count);            IF print_flag THEN                WriteInt(i);                WriteString(" ")            END        END    END;    IF print_flag THEN WriteLn END;    RETURN count;END proper_divisors; VAR    buf : ARRAY[0..63] OF CHAR;    i,max,max_i,v : INTEGER;BEGIN    FOR i:=1 TO 10 DO        WriteInt(i);        WriteString(": ");        proper_divisors(i, TRUE)    END;     max := 0;    max_i := 1;     FOR i:=1 TO 20000 DO        v := proper_divisors(i, FALSE);        IF v>= max THEN            max := v;            max_i := i        END    END;     FormatString("%i with %i divisors\n", buf, max_i, max);    WriteString(buf);     ReadCharEND ProperDivisors.`

## Nim

Translation of: C
`import strformat proc properDivisors(n: int) =  var count = 0  for i in 1..<n:    if n mod i == 0:      inc count      write(stdout, fmt"{i} ")  write(stdout, "\n") proc countProperDivisors(n: int): int =  var nn = n  var prod = 1  var count = 0  while nn mod 2 == 0:    inc count    nn = nn div 2  prod *= (1 + count)  for i in countup(3, n, 2):    count = 0    while nn mod i == 0:      inc count      nn = nn div i    prod *= (1 + count)  if nn > 2:    prod *= 2  prod - 1 for i in 1..10:  write(stdout, fmt"{i:2}: ")  properDivisors(i) var max = 0var maxI = 1 for i in 1..20000:  var v = countProperDivisors(i)  if v >= max:    max = v    maxI = i echo fmt"{maxI} with {max} divisors"`
Output:
``` 1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
18480 with 79 divisors
```

## Oberon-2

` MODULE ProperDivisors;IMPORT  Out; CONST     initialSize = 128;TYPE  Result* = POINTER TO ResultDesc;  ResultDesc = RECORD     found-: LONGINT; (* number of slots in pd *)    pd-: POINTER TO ARRAY OF LONGINT;    cap: LONGINT;   (* Capacity *)  END; VAR  i,found,max,idxMx: LONGINT;  mx: ARRAY 32 OF LONGINT;  rs: Result;   PROCEDURE (r: Result) Init(size: LONGINT);  BEGIN     r.found := 0;    r.cap := size;    NEW(r.pd,r.cap);  END Init;   PROCEDURE (r: Result) Add(n: LONGINT);  BEGIN    (* Out.String("--->");Out.LongInt(n,0);Out.String(" At: ");Out.LongInt(r.found,0);Out.Ln; *)    IF (r.found < LEN(r.pd^) - 1) THEN       r.pd[r.found] := n;    ELSE      (* expand pd for more room *)    END;    INC(r.found);  END Add;   PROCEDURE (r:Result) Show();  VAR    i: LONGINT;  BEGIN      Out.String("(Result:");Out.LongInt(r.found + 1,0);(* Out.String("/");Out.LongInt(r.cap,0);*)      Out.String("-");      IF r.found > 0 THEN        FOR i:= 0 TO r.found - 1 DO          Out.LongInt(r.pd[i],0);          IF i = r.found - 1 THEN Out.Char(')') ELSE Out.Char(',') END        END      END;      Out.Ln  END Show;   PROCEDURE (r:Result) Reset();  BEGIN     r.found := 0;  END Reset;   PROCEDURE GetFor(n: LONGINT;VAR rs: Result);  VAR    i: LONGINT;  BEGIN    IF n > 1 THEN       rs.Add(1);i := 2;      WHILE (i < n) DO        IF (n MOD i) = 0 THEN rs.Add(i) END;        INC(i)      END    END;  END GetFor; BEGIN  NEW(rs);rs.Init(initialSize);  FOR i := 1 TO 10 DO     Out.LongInt(i,4);Out.Char(':');    GetFor(i,rs);    rs.Show();    rs.Reset();  END;  Out.LongInt(100,4);Out.Char(':');GetFor(100,rs);rs.Show();rs.Reset();  max := 0;idxMx := 0;found := 0;  FOR i := 1 TO 20000 DO    GetFor(i,rs);    IF rs.found > max THEN      idxMx:= 0;mx[idxMx] := i;max := rs.found    ELSIF rs.found = max THEN      INC(idxMx);mx[idxMx] := i    END;    rs.Reset()  END;  Out.String("Found: ");Out.LongInt(idxMx + 1,0);  Out.String(" Numbers with most proper divisors ");  Out.LongInt(max,0);Out.String(": ");Out.Ln;  FOR i := 0 TO idxMx DO    Out.LongInt(mx[i],0);Out.Ln  ENDEND ProperDivisors. `
Output:
```   1:(Result:1-
2:(Result:2-1)
3:(Result:2-1)
4:(Result:3-1,2)
5:(Result:2-1)
6:(Result:4-1,2,3)
7:(Result:2-1)
8:(Result:4-1,2,4)
9:(Result:3-1,3)
10:(Result:4-1,2,5)
100:(Result:9-1,2,4,5,10,20,25,50)
Found: 2 Numbers with most proper divisors 79:
15120
18480
```

## Objeck

`use Collection; class Proper{  function : Main(args : String[]) ~ Nil {    for(x := 1; x <= 10; x++;) {      Print(x, ProperDivs(x));    };     x := 0;    count := 0;     for(n := 1; n <= 20000; n++;) {      if(ProperDivs(n)->Size() > count) {        x := n;        count := ProperDivs(n)->Size();      };    };    "{\$x}: {\$count}"->PrintLine();  }   function : ProperDivs(n : Int) ~ IntVector {    divs := IntVector->New();     if(n = 1) {      return divs;    };    divs->AddBack(1);     for(x := 2; x < n; x++;) {      if(n % x = 0) {         divs->AddBack(x);      };    };    divs->Sort();     return divs;  }   function : Print(x : Int, result : IntVector) ~ Nil {    "{\$x}: "->Print();    result->ToArray()->ToString()->PrintLine();  }} `

Output:

```1: []
2: 
3: 
4: [1,2]
5: 
6: [1,2,3]
7: 
8: [1,2,4]
9: [1,3]
10: [1,2,5]
15120: 79
```

## Oforth

`Integer method: properDivs  self 2 / seq filter(#[ self swap mod 0 == ]) } 10 seq apply(#[ dup print " : " print properDivs println ])20000 seq map(#[ dup properDivs size Pair new ]) reduce(#maxKey) println`
Output:
```1 : []
2 : 
3 : 
4 : [1, 2]
5 : 
6 : [1, 2, 3]
7 : 
8 : [1, 2, 4]
9 : [1, 3]
10 : [1, 2, 5]
[79, 15120]
```

## PARI/GP

`proper(n)=if(n==1, [], my(d=divisors(n)); d[2..#d]);apply(proper, [1..10])r=at=0; for(n=1,20000, t=numdiv(n); if(t>r, r=t; at=n)); [at, numdiv(t)-1]`
Output:
```%1 = [[], , , [2, 4], , [2, 3, 6], , [2, 4, 8], [3, 9], [2, 5, 10]]
%2 = [15120, 7]```

## Pascal

Works with: Free Pascal

Using prime factorisation

`{\$IFDEF FPC}{\$MODE DELPHI}{\$ELSE}{\$APPTYPE CONSOLE}{\$ENDIF}uses  sysutils;const  MAXPROPERDIVS = 1920;type  tRes = array[0..MAXPROPERDIVS] of LongWord;  tPot = record           potPrim,           potMax :LongWord;         end;   tprimeFac = record                 pfPrims : array[1..10] of tPot;                 pfCnt,                 pfNum   : LongWord;               end;  tSmallPrimes = array[0..6541] of longWord; var  SmallPrimes: tSmallPrimes; procedure InitSmallPrimes;var  pr,testPr,j,maxprimidx: Longword;  isPrime : boolean;Begin  maxprimidx := 0;  SmallPrimes := 2;  pr := 3;  repeat    isprime := true;    j := 0;    repeat      testPr := SmallPrimes[j];      IF testPr*testPr > pr then        break;      If pr mod testPr = 0 then      Begin        isprime := false;        break;      end;      inc(j);    until false;     if isprime then    Begin      inc(maxprimidx);      SmallPrimes[maxprimidx]:= pr;    end;    inc(pr,2);  until pr > 1 shl 16 -1;end; procedure PrimeFacOut(primeDecomp:tprimeFac);var  i : LongWord;begin  with primeDecomp do  Begin    write(pfNum,' = ');    For i := 1 to pfCnt-1 do      with pfPrims[i] do        If potMax = 1 then          write(potPrim,'*')        else          write(potPrim,'^',potMax,'*');    with pfPrims[pfCnt] do      If potMax = 1 then        write(potPrim)      else        write(potPrim,'^',potMax);  end;end; procedure PrimeDecomposition(n:LongWord;var res:tprimeFac);var  i,pr,cnt,quot{to minimize divisions} : LongWord;Begin  res.pfNum := n;  res.pfCnt:= 0;  i := 0;  cnt := 0;  repeat    pr := SmallPrimes[i];    IF pr*pr>n then      Break;     quot := n div pr;    IF pr*quot = n then      with res do      Begin        inc(pfCnt);        with pfPrims[pfCnt] do        Begin          potPrim := pr;          potMax := 0;          repeat            n := quot;            quot := quot div pr;            inc(potMax);          until pr*quot <> n;        end;      end;     inc(i);  until false;  //a big prime left over?  IF n <> 1 then    with res do    Begin      inc(pfCnt);      with pfPrims[pfCnt] do      Begin        potPrim := n;        potMax := 1;      end;    end;end; function CntProperDivs(const primeDecomp:tprimeFac):LongWord;//count of proper divisorsvar   i: LongWord;begin  result := 1;  with primeDecomp do    For i := 1 to pfCnt do      result := result*(pfPrims[i].potMax+1);  //remove  dec(result);end; function findProperdivs(n:LongWord;var res:TRes):LongWord;//simple trial division to get a sorted list of all proper divisorsvar  i,j: LongWord;Begin  result := 0;  i := 1;  j := n;  while j>i do  begin    j := n DIV i;    IF i*j = n then    Begin      //smaller factor part at the beginning upwards      res[result]:= i;      IF i <> j then        //bigger factor at the end downwards        res[MAXPROPERDIVS-result]:= j      else        //n is square number        res[MAXPROPERDIVS-result]:= 0;      inc(result);    end;    inc(i);  end;   If result>0 then  Begin    //move close together    i := result;    j := MAXPROPERDIVS-result+1;    result := 2*result-1;    repeat      res[i] := res[j];      inc(j);      inc(i);    until i > result;     if res[result-1] = 0 then      dec(result);  end;end; procedure AllFacsOut(n: Longword);var  res:TRes;  i,k,j:LongInt;Begin   j := findProperdivs(n,res);   write(n:5,' : ');   For k := 0 to j-2 do write(res[k],',');   IF j>=1 then     write(res[j-1]);   writeln;end; var  primeDecomp: tprimeFac;  rs : tRes;  i,j,max,maxcnt: LongWord;BEGIN  InitSmallPrimes;  For i := 1 to 10 do    AllFacsOut(i);  writeln;  max    := 0;  maxCnt := 0;  For i := 1 to 20*1000 do  Begin    PrimeDecomposition(i,primeDecomp);    j := CntProperDivs(primeDecomp);    IF j> maxCnt then    Begin      maxcnt := j;      max := i;    end;  end;  PrimeDecomposition(max,primeDecomp);  j := CntProperDivs(primeDecomp);   PrimeFacOut(primeDecomp);writeln('  ',j:10,' factors'); writeln;  //https://en.wikipedia.org/wiki/Highly_composite_number <= HCN  //http://wwwhomes.uni-bielefeld.de/achim/highly.txt the first 1200 HCN  max := 3491888400;  PrimeDecomposition(max,primeDecomp);  j := CntProperDivs(primeDecomp);  PrimeFacOut(primeDecomp);writeln('  ',j:10,' factors'); writeln;END.`
Output:
```
1 :
2 : 1
3 : 1
4 : 1,2
5 : 1
6 : 1,2,3
7 : 1
8 : 1,2,4
9 : 1,3
10 : 1,2,5

15120 = 2^4*3^3*5*7          79 factors
3491888400 = 2^4*3^3*5^2*7*11*13*17*19        1919 factors

real    0m0.004s```

## Perl

### Using a module for divisors

Library: ntheory
`use ntheory qw/divisors/;sub proper_divisors {  my \$n = shift;  # Like Pari/GP, divisors(0) = (0,1) and divisors(1) = ()  return 1 if \$n == 0;  my @d = divisors(\$n);  pop @d;  # divisors are in sorted order, so last entry is the input  @d;}say "\$_: ", join " ", proper_divisors(\$_) for 1..10;# 1. For the max, we can do a traditional loop.my(\$max,\$ind) = (0,0);for (1..20000) {  my \$nd = scalar proper_divisors(\$_); (\$max,\$ind) = (\$nd,\$_) if \$nd > \$max;}say "\$max \$ind";# 2. Or we can use List::Util's max with decoration (this exploits its implementation){  use List::Util qw/max/;  no warnings 'numeric';  say max(map { scalar(proper_divisors(\$_)) . " \$_" } 1..20000);}`
Output:
```1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
79 15120
79 18480```

Note that the first code will choose the first max, while the second chooses the last.

## Phix

The factors routine is an auto-include. The actual implementation of it, from builtins\pfactors.e is

```global function factors(atom n, integer include1=0)
--
-- returns a list of all integer factors of n
--  if include1 is 0 (the default), result does not contain either 1 or n
--  if include1 is 1 the result contains 1 and n
--  if include1 is -1 the result contains 1 but not n
--
if n=0 then return {} end if
check_limits(n,"factors")
sequence lfactors = {}, hfactors = {}
atom hfactor
integer p = 2,
lim = floor(sqrt(n))

if include1!=0 then
lfactors = {1}
if n!=1 and include1=1 then
hfactors = {n}
end if
end if
while p<=lim do
if remainder(n,p)=0 then
lfactors = append(lfactors,p)
hfactor = n/p
if hfactor=p then exit end if
hfactors = prepend(hfactors,hfactor)
end if
p += 1
end while
return lfactors & hfactors
end function
```

The compiler knows where to find that, so the main program is just:

```for i=0 to 10 do
printf(1,"%d: %v\n",{i,factors(i,-1)})
end for

sequence candidates = {}
integer maxd = 0
for i=1 to 20000 do
integer k = length(factors(i,-1))
if k>=maxd then
if k=maxd then
candidates &= i
else
candidates = {i}
maxd = k
end if
end if
end for
printf(1,"%d divisors: %v\n", {maxd,candidates})
```
Output:
```0: {}
1: {1}
2: {1}
3: {1}
4: {1,2}
5: {1}
6: {1,2,3}
7: {1}
8: {1,2,4}
9: {1,3}
10: {1,2,5}
79 divisors: {15120,18480}
```

## PHP

`<?phpfunction ProperDivisors(\$n) {  yield 1;  \$large_divisors = [];  for (\$i = 2; \$i <= sqrt(\$n); \$i++) {    if (\$n % \$i == 0) {      yield \$i;      if (\$i*\$i != \$n) {        \$large_divisors[] = \$n / \$i;      }    }  }  foreach (array_reverse(\$large_divisors) as \$i) {    yield \$i;  }} assert([1, 2, 4, 5, 10, 20, 25, 50] ==        iterator_to_array(ProperDivisors(100))); foreach (range(1, 10) as \$n) {  echo "\$n =>";  foreach (ProperDivisors(\$n) as \$divisor) {    echo " \$divisor";  }  echo "\n";} \$divisorsCount = [];for (\$i = 1; \$i < 20000; \$i++) {  \$divisorsCount[sizeof(iterator_to_array(ProperDivisors(\$i)))][] = \$i;}ksort(\$divisorsCount); echo "Numbers with most divisors: ", implode(", ", end(\$divisorsCount)), ".\n";echo "They have ", key(\$divisorsCount), " divisors.\n";  `

Outputs:

```1 => 1
2 => 1
3 => 1
4 => 1 2
5 => 1
6 => 1 2 3
7 => 1
8 => 1 2 4
9 => 1 3
10 => 1 2 5
Numbers with most divisors: 15120, 18480.
They have 79 divisors.```

## Picat

`go =>  println(11111=proper_divisors(11111)),  nl,  foreach(N in 1..10)    println(N=proper_divisors(N))  end,  nl,   find_most_divisors(20_000),  nl. % Proper divisors of number Nproper_divisors(N) = Divisors =>   Div1 = [ I : I in 1..ceiling(sqrt(N)), N mod I == 0],  Divisors = (Div1 ++ [N div I : I in Div1]).sort_remove_dups().delete(N).  % Find the number(s) with the most proper divisors below Limitfind_most_divisors(Limit) =>  MaxN = [],  MaxNumDivisors = [],  MaxLen = 1,   foreach(N in 1..Limit, not prime(N))    D = proper_divisors(N),    Len = D.len,    % Get all numbers with most proper divisors    if Len = MaxLen then      MaxN := MaxN ++ [N],      MaxNumDivisors := MaxNumDivisors ++ [[N=D]]    elseif Len > MaxLen then      MaxLen := Len,      MaxN := [N],      MaxNumDivisors := [N=D]    end  end,   println(maxN=MaxN),  println(maxLen=MaxLen),  nl.`
Output:
```11111 = [1,41,271]

1 = []
2 = 
3 = 
4 = [1,2]
5 = 
6 = [1,2,3]
7 = 
8 = [1,2,4]
9 = [1,3]
10 = [1,2,5]

maxN = [15120,18480]
maxLen = 79```

### Larger tests

Some larger tests of most number of divisors:

`go2 =>  time(find_most_divisors(100_000)),  nl,  time(find_most_divisors(1_000_000)),  nl.`
Output:
```maxN = [83160,98280]
maxLen = 127

maxN = [720720,831600,942480,982800,997920]
maxLen = 239```

## PicoLisp

`# Generate all proper divisors.(de propdiv (N)   (head -1 (filter      '((X) (=0 (% N X)))      (range 1 N) )) ) # Obtaining the values from 1 to 10 inclusive.(mapcar propdiv (range 1 10))# Output:# (NIL (1) (1) (1 2) (1) (1 2 3) (1) (1 2 4) (1 3) (1 2 5))`

### Brute-force

`(de propdiv (N)   (cdr      (rot         (make            (for I N               (and (=0 (% N I)) (link I)) ) ) ) ) )(de countdiv (N)    (let C -1       (for I N         (and (=0 (% N I)) (inc 'C)) )      C ) )(let F (-5 -8)   (tab F "N" "LIST")   (for I 10      (tab F         I         (glue " + " (propdiv I)) ) ) )(println   (maxi      countdiv      (range 1 20000) ) )`

### Factorization

`(de accu1 (Var Key)   (if (assoc Key (val Var))      (con @ (inc (cdr @)))      (push Var (cons Key 2)) )   Key )(de factor (N)   (let      (R NIL         D 2         L (1 2 2 . (4 2 4 2 4 6 2 6 .))         M (sqrt N) )      (while (>= M D)         (if (=0 (% N D))            (setq M               (sqrt (setq N (/ N (accu1 'R D)))) )            (inc 'D (pop 'L)) ) )      (accu1 'R N)      (dec (apply * (mapcar cdr R))) ) )(bench   (println      (maxi         factor         (range 1 20000) )       @@ ) )`

Output:

```15120 79
0.081 sec
```

## PL/I

`*process source xref; (subrg): cpd: Proc Options(main); p9a=time(); Dcl (p9a,p9b) Pic'(9)9'; Dcl cnt(3) Bin Fixed(31) Init((3)0); Dcl x Bin Fixed(31); Dcl pd(300) Bin Fixed(31); Dcl sumpd   Bin Fixed(31); Dcl npd     Bin Fixed(31); Dcl hi      Bin Fixed(31) Init(0); Dcl (xl(10),xi) Bin Fixed(31); Dcl i       Bin Fixed(31); Do x=1 To 10;   Call proper_divisors(x,pd,npd);   Put Edit(x,' -> ',(pd(i) Do i=1 To npd))(Skip,f(2),a,10(f(2)));   End; xi=0; Do x=1 To 20000;   Call proper_divisors(x,pd,npd);   Select;     When(npd>hi) Do;       xi=1;       xl(1)=x;       hi=npd;       End;     When(npd=hi) Do;       xi+=1;       xl(xi)=x;       End;     Otherwise;     End;   End; Put Edit(hi,' -> ',(xl(i) Do i=1 To xi))(Skip,f(3),a,10(f(6)));  x= 166320; Call proper_divisors(x,pd,npd); Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4)); x=1441440; Call proper_divisors(x,pd,npd); Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4));   p9b=time(); Put Edit((p9b-p9a)/1000,' seconds elapsed')(Skip,f(6,3),a); Return;  proper_divisors: Proc(n,pd,npd); Dcl (n,pd(300),npd) Bin Fixed(31); Dcl (d,delta)       Bin Fixed(31); npd=0; If n>1 Then Do;   If mod(n,2)=1 Then  /* odd number  */     delta=2;   Else                /* even number */     delta=1;   Do d=1 To n/2 By delta;     If mod(n,d)=0 Then Do;       npd+=1;       pd(npd)=d;       End;     End;   End; End;  End;`
Output:
``` 1 ->
2 ->  1
3 ->  1
4 ->  1 2
5 ->  1
6 ->  1 2 3
7 ->  1
8 ->  1 2 4
9 ->  1 3
10 ->  1 2 5
79 ->  15120 18480
166320 ->  159
1441440 ->  287
0.530 seconds elapsed```

## PowerShell

### version 1

` function proper-divisor (\$n) {    if(\$n -ge 2) {        \$lim = [Math]::Floor([Math]::Sqrt(\$n))        \$less, \$greater = @(1), @()        for(\$i = 2; \$i -lt \$lim; \$i++){            if(\$n%\$i -eq 0) {                \$less += @(\$i)                \$greater = @(\$n/\$i) + \$greater            }        }        if((\$lim -ne 1) -and (\$n%\$lim -eq 0)) {\$less += @(\$lim)}        \$(\$less + \$greater)    } else {@()}}"\$(proper-divisor 100)""\$(proper-divisor 496)""\$(proper-divisor 2048)" `

### version 2

` function proper-divisor (\$n) {    if(\$n -ge 2) {        \$lim = [Math]::Floor(\$n/2)+1        \$proper = @(1)        for(\$i = 2; \$i -lt \$lim; \$i++){            if(\$n%\$i -eq 0) {                \$proper += @(\$i)            }        }        \$proper    } else {@()}}"\$(proper-divisor 100)""\$(proper-divisor 496)""\$(proper-divisor 2048)" `

### version 3

` function eratosthenes (\$n) {    if(\$n -gt 1){        \$prime = @(0..\$n| foreach{\$true})        \$m = [Math]::Floor([Math]::Sqrt(\$n))        function multiple(\$i) {            for(\$j = \$i*\$i; \$j -le \$n; \$j += \$i) {                \$prime[\$j] = \$false            }        }        multiple 2        for(\$i = 3; \$i -le \$m; \$i += 2) {            if(\$prime[\$i]) {multiple \$i}        }        2        for(\$i = 3; \$i -le \$n; \$i += 2) {            if(\$prime[\$i]) {\$i}        }     } else {        Write-Error "\$n is not greater than 1"    }}function prime-decomposition (\$n) {    \$array = eratosthenes \$n    \$prime = @()    foreach(\$p in \$array) {        while(\$n%\$p -eq 0) {            \$n /= \$p            \$prime += @(\$p)        }    }    \$prime}function proper-divisor (\$n) {    if(\$n -ge 2) {        \$array = prime-decomposition \$n        \$lim = \$array.Count        function state(\$res, \$i){              if(\$i -lt \$lim) {                state (\$res) (\$i + 1)                state (\$res*\$array[\$i]) (\$i + 1)               } elseif(\$res -lt \$n) {\$res}        }        state 1 0 | sort -Unique    } else {@()}}"\$(proper-divisor 100)""\$(proper-divisor 496)""\$(proper-divisor 2048)" `

Output:

```1 2 4 5 10 20 25 50
1 2 4 8 16 31 62 124 248
1 2 4 8 16 32 64 128 256 512 1024
```

## Prolog

Works with: SWI-Prolog 7

Taking a cue from an SO answer:

`divisor(N, Divisor) :-    UpperBound is round(sqrt(N)),    between(1, UpperBound, D),    0 is N mod D,    (        Divisor = D     ;        LargerDivisor is N/D,        LargerDivisor =\= D,        Divisor = LargerDivisor    ). proper_divisor(N, D) :-    divisor(N, D),    D =\= N.  %% Task 1% proper_divisors(N, Ds) :-    setof(D, proper_divisor(N, D), Ds).  %% Task 2% show_proper_divisors_of_range(Low, High) :-    findall( N:Ds,             ( between(Low, High, N),               proper_divisors(N, Ds) ),             Results ),    maplist(writeln, Results).  %% Task 3% proper_divisor_count(N, Count) :-    proper_divisors(N, Ds),    length(Ds, Count). find_most_proper_divisors_in_range(Low, High, Result) :-    aggregate_all( max(Count, N),                   ( between(Low, High, N),                     proper_divisor_count(N, Count) ),                   max(MaxCount, Num) ),    Result = (num(Num)-divisor_count(MaxCount)).`

Output:

`?- show_proper_divisors_of_range(1,10).2:3:4:[1,2]5:6:[1,2,3]7:8:[1,2,4]9:[1,3]10:[1,2,5]true. ?- find_most_proper_divisors_in_range(1,20000,Result).Result = num(15120)-divisor_count(79). `

## PureBasic

` EnableExplicit Procedure ListProperDivisors(Number, List Lst())  If Number < 2 : ProcedureReturn : EndIf  Protected i  For i = 1 To Number / 2    If Number % i = 0      AddElement(Lst())      Lst() = i    EndIf  NextEndProcedure Procedure.i CountProperDivisors(Number)  If Number < 2 : ProcedureReturn 0 : EndIf  Protected i, count = 0  For i = 1 To Number / 2    If Number % i = 0      count + 1    EndIf  Next  ProcedureReturn countEndProcedure Define n, count, most = 1, maxCount = 0If OpenConsole()  PrintN("The proper divisors of the following numbers are : ")  PrintN("")  NewList lst()  For n = 1 To 10    ListProperDivisors(n, lst())    Print(RSet(Str(n), 3) + " -> ")    If ListSize(lst()) = 0      Print("(None)")    Else        ForEach lst()        Print(Str(lst()) + " ")      Next    EndIf    ClearList(lst())    PrintN("")  Next  For n = 2 To 20000    count = CountProperDivisors(n)    If count > maxCount      maxCount = count      most = n    EndIf  Next  PrintN("")  PrintN(Str(most) + " has the most proper divisors, namely " + Str(maxCount))  PrintN("")  PrintN("Press any key to close the console")  Repeat: Delay(10) : Until Inkey() <> ""  CloseConsole()EndIf `
Output:
```The proper divisors of the following numbers are :

1 -> (None)
2 -> 1
3 -> 1
4 -> 1 2
5 -> 1
6 -> 1 2 3
7 -> 1
8 -> 1 2 4
9 -> 1 3
10 -> 1 2 5

15120 has the most proper divisors, namely 79
```

## Python

### Python: Literal

A very literal interpretation

`>>> def proper_divs2(n):...     return {x for x in range(1, (n + 1) // 2 + 1) if n % x == 0 and n != x}... >>> [proper_divs2(n) for n in range(1, 11)][set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}]>>> >>> n, length = max(((n, len(proper_divs2(n))) for n in range(1, 20001)), key=lambda pd: pd)>>> n15120>>> length79>>> `

### Python: From prime factors

I found a reference on how to generate factors from all the prime factors and the number of times each prime factor goes into N - its multiplicity.

For example, given N having prime factors P(i) with associated multiplicity M(i}) then the factors are given by:

```for m in range(M(0) + 1):
for m in range(M + 1):
...
for m[i - 1] in range(M[i - 1] + 1):
mult = 1
for j in range(i):
mult *= P[j] ** m[j]
yield mult```

This version is over an order of magnitude faster for generating the proper divisors of the first 20,000 integers; at the expense of simplicity.

`from math import sqrtfrom functools import lru_cache, reducefrom collections import Counterfrom itertools import product  MUL = int.__mul__  def prime_factors(n):    'Map prime factors to their multiplicity for n'    d = _divs(n)    d = [] if d == [n] else (d[:-1] if d[-1] == d else d)    pf = Counter(d)    return dict(pf) @lru_cache(maxsize=None)def _divs(n):    'Memoized recursive function returning prime factors of n as a list'    for i in range(2, int(sqrt(n)+1)):        d, m  = divmod(n, i)        if not m:            return [i] + _divs(d)    return [n]  def proper_divs(n):    '''Return the set of proper divisors of n.'''    pf = prime_factors(n)    pfactors, occurrences = pf.keys(), pf.values()    multiplicities = product(*(range(oc + 1) for oc in occurrences))    divs = {reduce(MUL, (pf**m for pf, m in zip(pfactors, multis)), 1)            for multis in multiplicities}    try:        divs.remove(n)    except KeyError:        pass    return divs or ({1} if n != 1 else set())  if __name__ == '__main__':    rangemax = 20000     print([proper_divs(n) for n in range(1, 11)])    print(*max(((n, len(proper_divs(n))) for n in range(1, 20001)), key=lambda pd: pd))`
Output:
```[set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}]
15120 79```

### Python: Functional

Defining a list of proper divisors in terms of the prime factorization:

Works with: Python version 3.7
`'''Proper divisors''' from itertools import accumulate, chain, groupby, productfrom functools import reducefrom math import floor, sqrtfrom operator import mul  # properDivisors :: Int -> [Int]def properDivisors(n):    '''The ordered divisors of n, excluding n itself.    '''    def go(a, group):        return [x * y for x, y in product(            a,            accumulate(chain(, group), mul)        )]    return sorted(        reduce(go, [            list(g) for _, g in groupby(primeFactors(n))        ], )    )[:-1] if 1 < n else []  # --------------------------TEST---------------------------# main :: IO ()def main():    '''Tests'''     print(        fTable('Proper divisors of [1..10]:')(str)(str)(            properDivisors        )(range(1, 1 + 10))    )     print('\nExample of maximum divisor count in the range [1..20000]:')    print(        max(            [(n, len(properDivisors(n))) for n in range(1, 1 + 20000)],            key=snd        )    )  # -------------------------DISPLAY------------------------- # fTable :: String -> (a -> String) -># (b -> String) -> (a -> b) -> [a] -> Stringdef fTable(s):    '''Heading -> x display function -> fx display function ->       f -> xs -> tabular string.    '''    def go(xShow, fxShow, f, xs):        ys = [xShow(x) for x in xs]        w = max(map(len, ys))        return s + '\n' + '\n'.join(map(            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),            xs, ys        ))    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(        xShow, fxShow, f, xs    )  # -------------------------GENERIC------------------------- # primeFactors :: Int -> [Int]def primeFactors(n):    '''A list of the prime factors of n.    '''    def f(qr):        r = qr        return step(r), 1 + r     def step(x):        return 1 + (x << 2) - ((x >> 1) << 1)     def go(x):        root = floor(sqrt(x))         def p(qr):            q = qr            return root < q or 0 == (x % q)         q = until(p)(f)(            (2 if 0 == x % 2 else 3, 1)        )        return [x] if q > root else [q] + go(x // q)     return go(n)  # snd :: (a, b) -> bdef snd(tpl):    '''Second member of a pair.'''    return tpl  # until :: (a -> Bool) -> (a -> a) -> a -> adef until(p):    '''The result of repeatedly applying f until p holds.       The initial seed value is x.    '''    def go(f, x):        v = x        while not p(v):            v = f(v)        return v    return lambda f: lambda x: go(f, x)  # MAIN ---if __name__ == '__main__':    main()`
Output:
```Proper divisors of [1..10]:
1 -> []
2 -> 
3 -> 
4 -> [1, 2]
5 -> 
6 -> [1, 2, 3]
7 -> 
8 -> [1, 2, 4]
9 -> [1, 3]
10 -> [1, 2, 5]

Example of maximum divisor count in the range [1..20000]:
(15120, 79)```

### Python: The Simple Way

Not all the code submitters realized that it's a tie for the largest number of factors inside the limit. The task description clearly indicates only one answer is needed. But both numbers are provided for the curious. Also shown is the result for 25000, as there is no tie for that, just to show the program can handle either scenario.

`def pd(num):	factors = []	for divisor in range(1,1+num//2):		if num % divisor == 0: factors.append(divisor)	return factors def pdc(num):	count = 0	for divisor in range(1,1+num//2):		if num % divisor == 0: count += 1	return count def fmtres(title, lmt, best, bestc):	return "The " + title + " number up to and including " + str(lmt) + " with the highest number of proper divisors is " + str(best) + ", which has " + str(bestc) def showcount(limit):	best, bestc, bh, bhc = 0, 0, 0, 0	for i in range(limit+1):		divc = pdc(i)		if divc > bestc: bestc, best = divc, i		if divc >= bhc: bhc, bh = divc, i	if best == bh:		print(fmtres("only", limit, best, bestc))	else:		print(fmtres("lowest", limit, best, bestc))		print(fmtres("highest", limit, bh, bhc))	print() lmt = 10for i in range(1, lmt + 1):	divs = pd(i)	if len(divs) == 0:		print("There are no proper divisors of", i)	elif len(divs) == 1:		print(divs, "is the only proper divisor of", i)	else:		print(divs, "are the proper divisors of", i)print()showcount(20000)showcount(25000)`
Output:
```There are no proper divisors of 1
1 is the only proper divisor of 2
1 is the only proper divisor of 3
[1, 2] are the proper divisors of 4
1 is the only proper divisor of 5
[1, 2, 3] are the proper divisors of 6
1 is the only proper divisor of 7
[1, 2, 4] are the proper divisors of 8
[1, 3] are the proper divisors of 9
[1, 2, 5] are the proper divisors of 10

The lowest number up to and including 20000 with the highest number of proper divisors is 15120, which has 79
The highest number up to and including 20000 with the highest number of proper divisors is 18480, which has 79

The only number up to and including 25000 with the highest number of proper divisors is 20160, which has 83```

## Quackery

`factors` is defined at Factors of an integer#Quackery.

`  [ factors -1 split drop ]      is properdivisors ( n --> [ )   10 times [ i^ 1+ properdivisors echo cr ]    0 0  20000 times     [ i^ 1+ properdivisors size      2dup < iff        [ dip [ 2drop i^ 1+ ] ]      else drop ]  swap echo say " has "  echo say " proper divisors." cr`
Output:
```[ ]
[ 1 ]
[ 1 ]
[ 1 2 ]
[ 1 ]
[ 1 2 3 ]
[ 1 ]
[ 1 2 4 ]
[ 1 3 ]
[ 1 2 5 ]
15120 has 79 proper divisors.
```

## R

### Package solution

Works with: R version 3.3.2 and above
`# Proper divisors. 12/10/16 aevrequire(numbers);V <- sapply(1:20000, Sigma, k = 0, proper = TRUE); ind <- which(V==max(V));cat("  *** max number of divisors:", max(V), "\n"," *** for the following indices:",ind, "\n");`
Output:
```Loading required package: numbers
*** max number of divisors: 79
*** for the following indices: 15120 18480 ```

### Filter solution

`#Task 1properDivisors <- function(N) Filter(function(x) N %% x == 0, seq_len(N %/% 2)) #Task 2Vectorize(properDivisors)(1:10) #Task 3#Although there are two, the task only asks for one suitable number so that is all we give.#Similarly, we have seen no need to make sure that "divisors" is only a plural when it should be.#Be aware that this solution uses both length and lengths. It would not work if the index of the#desired number was not also the number itself. However, this is always the case.mostProperDivisors <- function(N){  divisorList <- Vectorize(properDivisors)(seq_len(N))  numberWithMostDivisors <- which.max(lengths(divisorList))  paste0("The number with the most proper divisors between 1 and ", N,        " is ", numberWithMostDivisors,        ". It has ", length(divisorList[[numberWithMostDivisors]]),        " proper divisors.")}mostProperDivisors(20000)`
Output:
```#Task 2
> Vectorize(properDivisors)(1:10)
[]
integer(0)

[]
 1

[]
 1

[]
 1 2

[]
 1

[]
 1 2 3

[]
 1

[]
 1 2 4

[]
 1 3

[]
 1 2 5

> mostProperDivisors(20000)
 "The number with the most proper divisors between 1 and 20000 is 15120. It has 79 proper divisors."```

## Racket

### Short version

`#lang racket(require math)(define (proper-divisors n) (drop-right (divisors n) 1))(for ([n (in-range 1 (add1 10))])  (printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))(define most-under-20000  (for/fold ([best '(1)]) ([n (in-range 2 (add1 20000))])    (define divs (proper-divisors n))    (if (< (length (cdr best)) (length divs)) (cons n divs) best)))(printf "~a has ~a proper divisors\n"        (car most-under-20000) (length (cdr most-under-20000)))`
Output:
```proper divisors of: 1	()
proper divisors of: 2	(1)
proper divisors of: 3	(1)
proper divisors of: 4	(1 2)
proper divisors of: 5	(1)
proper divisors of: 6	(1 2 3)
proper divisors of: 7	(1)
proper divisors of: 8	(1 2 4)
proper divisors of: 9	(1 3)
proper divisors of: 10	(1 2 5)
15120 has 79 proper divisors```

### Long version

The main module will only be executed when this file is executed. When used as a library, it will not be used.

`#lang racket/base(provide fold-divisors ; name as per "Abundant..."         proper-divisors) (define (fold-divisors v n k0 kons)  (define n1 (add1 n))  (cond    [(>= n1 (vector-length v))     (define rv (make-vector n1 k0))     (for* ([n (in-range 1 n1)] [m (in-range (* 2 n) n1 n)])       (vector-set! rv m (kons n (vector-ref rv m))))     rv]    [else v])) (define proper-divisors  (let ([p.d-v (vector)])    (λ (n)      (set! p.d-v (reverse (fold-divisors p.d-v n null cons)))      (vector-ref p.d-v n)))) (module+ main  (for ([n (in-range 1 (add1 10))])    (printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))   (define count-proper-divisors    (let ([p.d-v (vector)])      (λ(n) (set! p.d-v (fold-divisors p.d-v n 0 (λ (d n) (add1 n))))            (vector-ref p.d-v n))))   (void (count-proper-divisors 20000))   (define-values [C I]    (for*/fold ([C 0] [I 1])               ([i (in-range 1 (add1 20000))]                [c (in-value (count-proper-divisors i))]                #:when [> c C])      (values c i)))  (printf "~a has ~a proper divisors\n" I C))`

The output is the same as the short version above.

## Raku

(formerly Perl 6)

Works with: rakudo version 2018.10

Once your threshold is over 1000, the maximum proper divisors will always include 2, 3 and 5 as divisors, so only bother to check multiples of 2, 3 and 5.

There really isn't any point in using concurrency for a limit of 20_000. The setup and bookkeeping drowns out any benefit. Really doesn't start to pay off until the limit is 50_000 and higher. Try swapping in the commented out race map iterator line below for comparison.

`sub propdiv (\x) {    my @l = 1 if x > 1;    (2 .. x.sqrt.floor).map: -> \d {        unless x % d { @l.push: d; my \y = x div d; @l.push: y if y != d }    }    @l} put "\$_ [{propdiv(\$_)}]" for 1..10; my @candidates;loop (my int \$c = 30; \$c <= 20_000; \$c += 30) {#(30, *+30 …^ * > 500_000).race.map: -> \$c {    my \mx = +propdiv(\$c);    next if mx < @candidates - 1;    @candidates[mx].push: \$c} say "max = {@candidates - 1}, candidates = {@candidates.tail}";`
Output:
```1 []
2 
3 
4 [1 2]
5 
6 [1 2 3]
7 
8 [1 2 4]
9 [1 3]
10 [1 2 5]
max = 79, candidates = 15120 18480```

## REXX

### version 1

` /*REXX*/ Call time 'R'Do x=1 To 10  Say x '->' proper_divisors(x)  End hi=1Do x=1 To 20000  /* If x//1000=0 Then Say x */  npd=count_proper_divisors(x)  Select    When npd>hi Then Do      list.npd=x      hi=npd      End    When npd=hi Then      list.hi=list.hi x    Otherwise      Nop    End  End Say hi '->' list.hi Say ' 166320 ->' count_proper_divisors(166320)Say '1441440 ->' count_proper_divisors(1441440) Say time('E') 'seconds elapsed'Exit proper_divisors: ProcedureParse Arg nIf n=1 Then Return ''pd=''/* Optimization reduces 37 seconds to 28 seconds */If n//2=1 Then  /* odd number  */  delta=2Else            /* even number */  delta=1Do d=1 To n%2 By delta  If n//d=0 Then    pd=pd d  EndReturn space(pd) count_proper_divisors: ProcedureParse Arg nReturn words(proper_divisors(n))`
Output:
```1 ->
2 -> 1
3 -> 1
4 -> 1 2
5 -> 1
6 -> 1 2 3
7 -> 1
8 -> 1 2 4
9 -> 1 3
10 -> 1 2 5
79 -> 15120 18480
166320 -> 159
1441440 -> 287
28.342000 seconds elapsed```

### version 2

The following REXX version is an adaptation of the   optimized   version for the REXX language example for the Rosetta
code task:       Factors of an integer.

This REXX version handles all integers   (negative, zero, positive)   and automatically adjusts the precision (decimal digits).
It also allows the specification of the ranges (for display and for finding the maximum),   and allows for extra numbers to be
specified.

With the (function) optimization, it's over   20   times faster.

`/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/parse arg bot top inc range xtra                 /*obtain optional arguments from the CL*/if   bot=='' |   bot==","  then    bot=     1    /*Not specified?  Then use the default.*/if   top=='' |   top==","  then    top=    10    /* "      "         "   "   "     "    */if   inc=='' |   inc==","  then    inc=     1    /* "      "         "   "   "     "    */if range=='' | range==","  then  range= 20000    /* "      "         "   "   "     "    */w= max( length(top), length(bot), length(range)) /*determine the biggest number of these*/numeric digits max(9, w + 1)                     /*have enough digits for  //  operator.*/@.= 'and'                                        /*a literal used to separate #s in list*/      do n=bot  to top  by inc                   /*process the first range specified.   */      q= Pdivs(n);    #= words(q)                /*get proper divs; get number of Pdivs.*/      if q=='∞'  then #= q                       /*adjust number of Pdivisors for zero. */      say right(n, max(20, w) )   'has'   center(#, 4)     "proper divisors: "    q      end   /*n*/m= 0                                             /*M ≡ maximum number of Pdivs (so far).*/      do r=1  for range;    q= Pdivs(r)          /*process the second range specified.  */      #= words(q);          if #<m  then iterate /*get proper divs; get number of Pdivs.*/      if #<m  then iterate                       /*Less then max?   Then ignore this #. */      @.#= @.#  @.  r;      m=#                  /*add this Pdiv to max list; set new M.*/      end   /*r*/                                /* [↑]   process 2nd range of integers.*/saysay m  ' is the highest number of proper divisors in range 1──►'range,       ", and it's for: "       subword(@.m, 3)say                                              /* [↓]  handle any given extra numbers.*/      do i=1  for words(xtra);  n= word(xtra, i) /*obtain an extra number from XTRA list*/      w= max(w, 1 + length(n) )                  /*use maximum width for aligned output.*/      numeric digits max(9, 1 + length(n) )      /*have enough digits for  //  operator.*/      q= Pdivs(n);              #= words(q)      /*get proper divs; get number of Pdivs.*/      say  right(n, max(20, w) )     'has'     center(#, 4)      "proper divisors."      end   /*i*/                                /* [↑] support extra specified integers*/exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/Pdivs: procedure; parse arg x,b;  x= abs(x);   if x==1  then return ''          /*unity?*/       odd= x // 2;                            if x==0  then return '∞'         /*zero ?*/       a= 1                                      /* [↓]  use all, or only odd #s.    ___*/           do j=2+odd  by 1+odd  while j*j < x   /*divide by some integers up to    √ X */           if x//j==0  then do;  a=a j;  b=x%j b /*if ÷, add both divisors to α & ß.    */                            end           end   /*j*/                           /* [↑]  %  is the REXX integer division*/                                                 /* [↓]  adjust for a square.        ___*/       if j*j==x  then  return  a j b            /*Was  X  a square?    If so, add  √ X */                        return  a   b            /*return the divisors  (both lists).   */`
output   when using the following input:   0   10   1       20000       166320   1441440   11796480000
```                   0 has  ∞   proper divisors:  ∞
1 has  0   proper divisors:
2 has  1   proper divisors:  1
3 has  1   proper divisors:  1
4 has  2   proper divisors:  1 2
5 has  1   proper divisors:  1
6 has  3   proper divisors:  1 2 3
7 has  1   proper divisors:  1
8 has  3   proper divisors:  1 2 4
9 has  2   proper divisors:  1 3
10 has  3   proper divisors:  1 2 5

79  is the highest number of proper divisors in range 1──►20000, and it's for:  15120 and 18480

166320 has 159  proper divisors.
1441440 has 287  proper divisors.
11796480000 has 329  proper divisors.
```

### version 3

When factoring          20,000 integers,   this REXX version is about   10%   faster than the REXX version 2.
When factoring        200,000 integers,   this REXX version is about   30%   faster.
When factoring     2,000,000 integers,   this REXX version is about   40%   faster.
When factoring   20,000,000 integers,   this REXX version is about   38%   faster.

(The apparent slowdown for the last example above is probably due to a shortage of real storage, causing more page faults.)

It accomplishes a faster speed by incorporating the calculation of an   integer square root   of an integer   (without using any floating point arithmetic).

`/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/parse arg bot top inc range xtra                 /*obtain optional arguments from the CL*/if   bot=='' |   bot==","  then    bot=     1    /*Not specified?  Then use the default.*/if   top=='' |   top==","  then    top=    10    /* "      "         "   "   "     "    */if   inc=='' |   inc==","  then    inc=     1    /* "      "         "   "   "     "    */if range=='' | range==","  then  range= 20000    /* "      "         "   "   "     "    */w= max( length(top), length(bot), length(range)) /*determine the biggest number of these*/numeric digits max(9, w + 1)                     /*have enough digits for  //  operator.*/@.= 'and'                                        /*a literal used to separate #s in list*/      do n=bot  to top  by inc                   /*process the first range specified.   */      q= Pdivs(n);    #= words(q)                /*get proper divs; get number of Pdivs.*/      if q=='∞'  then #= q                       /*adjust number of Pdivisors for zero. */      say right(n, max(20, w) )   'has'   center(#, 4)     "proper divisors: "    q      end   /*n*/m= 0                                             /*M ≡ maximum number of Pdivs (so far).*/      do r=1  for range;    q= Pdivs(r)          /*process the second range specified.  */      #= words(q);          if #<m  then iterate /*get proper divs; get number of Pdivs.*/      if #<m  then iterate                       /*Less then max?   Then ignore this #. */      @.#= @.#  @.  r;      m=#                  /*add this Pdiv to max list; set new M.*/      end   /*r*/                                /* [↑]   process 2nd range of integers.*/saysay m  ' is the highest number of proper divisors in range 1──►'range,       ", and it's for: "       subword(@.m, 3)say                                              /* [↓]  handle any given extra numbers.*/      do i=1  for words(xtra);  n= word(xtra, i) /*obtain an extra number from XTRA list*/      w= max(w, 1 + length(n) )                  /*use maximum width for aligned output.*/      numeric digits max(9, 1 + length(n) )      /*have enough digits for  //  operator.*/      q= Pdivs(n);              #= words(q)      /*get proper divs; get number of Pdivs.*/      say  right(n, max(20, w) )     'has'     center(#, 4)      "proper divisors."      end   /*i*/                                /* [↑] support extra specified integers*/exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/Pdivs: procedure; parse arg x 1 z,b;  x= abs(x);   if x==1  then return ''      /*unity?*/       odd= x // 2;                                if x==0  then return '∞'     /*zero ?*/       r= 0;         q= 1                        /* [↓] ══integer square root══     ___ */            do while q<=z; q=q*4; end            /*R:  an integer which will be    √ X  */            do while q>1;  q=q%4; _= z-r-q;  r=r%2;  if _>=0  then  do;  z=_;  r=r+q;  end            end   /*while q>1*/                  /* [↑]  compute the integer sqrt of  X.*/       a=1                                       /* [↓]  use all, or only odd #s.   ___ */           do j=2 +odd  by 1 +odd to r -(r*r==x) /*divide by some integers up to   √ X  */           if x//j==0  then do;  a=a j;  b=x%j b /*if ÷, add both divisors to α & ß.    */                            end           end   /*j*/                           /* [↑]  %  is the REXX integer division*/                                                 /* [↓]  adjust for a square.        ___*/       if j*j==x  then  return  a j b            /*Was  X  a square?    If so, add  √ X */                        return  a   b            /*return the divisors  (both lists).   */`
output   is identical to the 2nd REXX version when using the same inputs.

### version 4

This REXX version uses an integer square root function to find the upper limit for the divisions.

For larger numbers,   it is about   7%   faster.

`/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/parse arg bot top inc range xtra                 /*obtain optional arguments from the CL*/if   bot=='' |   bot==","  then    bot=     1    /*Not specified?  Then use the default.*/if   top=='' |   top==","  then    top=    10    /* "      "         "   "   "     "    */if   inc=='' |   inc==","  then    inc=     1    /* "      "         "   "   "     "    */if range=='' | range==","  then  range= 20000    /* "      "         "   "   "     "    */w= max( length(top), length(bot), length(range)) /*determine the biggest number of these*/numeric digits max(9, w + 1)                     /*have enough digits for  //  operator.*/@.= 'and'                                        /*a literal used to separate #s in list*/      do n=bot  to top  by inc                   /*process the first range specified.   */      q= Pdivs(n);    #= words(q)                /*get proper divs; get number of Pdivs.*/      if q=='∞'  then #= q                       /*adjust number of Pdivisors for zero. */      say right(n, max(20, w) )   'has'   center(#, 4)     "proper divisors: "    q      end   /*n*/m= 0                                             /*M ≡ maximum number of Pdivs (so far).*/      do r=1  for range;    q= Pdivs(r)          /*process the second range specified.  */      #= words(q);          if #<m  then iterate /*get proper divs; get number of Pdivs.*/      if #<m  then iterate                       /*Less then max?   Then ignore this #. */      @.#= @.#  @.  r;      m=#                  /*add this Pdiv to max list; set new M.*/      end   /*r*/                                /* [↑]   process 2nd range of integers.*/saysay m  ' is the highest number of proper divisors in range 1──►'range,       ", and it's for: "       subword(@.m, 3)say                                              /* [↓]  handle any given extra numbers.*/      do i=1  for words(xtra);  n= word(xtra, i) /*obtain an extra number from XTRA list*/      w= max(w, 1 + length(n) )                  /*use maximum width for aligned output.*/      numeric digits max(9, 1 + length(n) )      /*have enough digits for  //  operator.*/      q= Pdivs(n);              #= words(q)      /*get proper divs; get number of Pdivs.*/      say  right(n, max(20, w) )     'has'     center(#, 4)      "proper divisors."      end   /*i*/                                /* [↑] support extra specified integers*/exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/iSqrt: procedure; parse arg x;  r=0;  q=1;             do while q<=x;  q=q*4;  end                  do while q>1; q=q%4; _=x-r-q; r=r%2; if _>=0 then do;x=_;r=r+q; end; end       return r/*──────────────────────────────────────────────────────────────────────────────────────*/Pdivs: procedure; parse arg x,b;  x= abs(x)                                  if x==1  then return ''                    /*null set.*/                                  if x==0  then return '∞'                   /*infinity.*/       odd= x // 2                               /*oddness of  X.        ___            */       r= iSqrt(x)                               /* obtain the integer  √ X             */       a= 1                                      /* [↓]  use all,  or only odd numbers. */                                                 /*                                  ___*/       if odd then do j=3  by 2 for r%2-(r*r==x) /*divide by some integers up to    √ X */                   if x//j==0  then do;  a=a j; b=x%j b /*÷?  Add both divisors to A & B*/                                    end                   end   /*j*/                   /*                                  ___*/              else do j=2       for r-1-(r*r==x) /*divide by some integers up to    √ X */                   if x//j==0  then do;  a=a j; b=x%j b /*÷?  Add both divisors to A & B*/                                    end                   end   /*j*/       if r*r==x  then  return  a j b            /*Was  X  a square?    If so, add  √ X */                        return  a   b            /*return proper divisors  (both lists).*/`
output   is identical to the 2nd REXX version when using the same inputs.

## Ring

` # Project : Proper divisors limit = 10for n=1 to limit    if n=1       see "" + 1 + " -> (None)" + nl       loop    ok    see "" + n + " -> "    for m=1 to n-1        if n%m = 0           see " " + m         ok    next    see nlnext `

Output:

```1 -> (None)
2 ->  1
3 ->  1
4 ->  1 2
5 ->  1
6 ->  1 2 3
7 ->  1
8 ->  1 2 4
9 ->  1 3
10 ->  1 2 5
```

## Ruby

`require "prime" class Integer  def proper_divisors    return [] if self == 1    primes = prime_division.flat_map{|prime, freq| [prime] * freq}    (1...primes.size).each_with_object() do |n, res|      primes.combination(n).map{|combi| res << combi.inject(:*)}    end.flatten.uniq  endend (1..10).map{|n| puts "#{n}: #{n.proper_divisors}"} size, select = (1..20_000).group_by{|n| n.proper_divisors.size}.maxselect.each do |n|  puts "#{n} has #{size} divisors"end`
Output:
```1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120 has 79 divisors
18480 has 79 divisors
```

### An Alternative Approach

`#Determine the integer within a range of integers that has the most proper divisors#Nigel Galloway: December 23rd., 2014require "prime"n, g = 0(1..20000).each{|i| e = i.prime_division.inject(1){|n,g| n * (g+1)}                    n, g = e, i if e > n}puts "#{g} has #{n-1} proper divisors"`
Output:

In the range 1..200000

```15120 has 79 proper divisors
```

and in the ranges 1..2000000 & 1..20000000

```166320 has 159 proper divisors
1441440 has 287 proper divisors
```

## Rust

`trait ProperDivisors {    fn proper_divisors(&self) -> Option<Vec<u64>>;} impl ProperDivisors for u64 {    fn proper_divisors(&self) -> Option<Vec<u64>> {        if self.le(&1) {            return None;        }        let mut divisors: Vec<u64> = Vec::new();         for i in 1..*self {            if *self % i == 0 {                divisors.push(i);            }        }        Option::from(divisors)    }} fn main() {    for i in 1..11 {        println!("Proper divisors of {:2}: {:?}", i,                 i.proper_divisors().unwrap_or(vec![]));    }     let mut most_idx: u64 = 0;    let mut most_divisors: Vec<u64> = Vec::new();    for i in 1..20_001 {        let divs = i.proper_divisors().unwrap_or(vec![]);        if divs.len() > most_divisors.len() {            most_divisors = divs;            most_idx = i;        }    }    println!("In 1 to 20000, {} has the most proper divisors at {}", most_idx,             most_divisors.len());} `
Output:
```Proper divisors of  1: []
Proper divisors of  2: 
Proper divisors of  3: 
Proper divisors of  4: [1, 2]
Proper divisors of  5: 
Proper divisors of  6: [1, 2, 3]
Proper divisors of  7: 
Proper divisors of  8: [1, 2, 4]
Proper divisors of  9: [1, 3]
Proper divisors of 10: [1, 2, 5]
In 1 to 20000, 15120 has the most proper divisors at 79
```

## S-BASIC

` \$lines \$constant false = 0\$constant true = FFFFH rem - compute p mod qfunction mod(p, q = integer) = integerend = p - q * (p/q) rem - count, and optionally display, proper divisors of nfunction divisors(n, display = integer) = integer  var i, limit, count, start, delta = integer  if mod(n, 2) = 0 then    begin      start = 2      delta = 1    end  else    begin      start = 3      delta = 2    end  if n < 2 then count = 0 else count = 1  if display and (count = 1) then print using "#####"; 1;  i = start  limit = n / start  while i <= limit do    begin      if mod(n, i) = 0 then        begin          if display then print using "#####"; i;          count = count + 1        end      i = i + delta      if count = 1 then limit = n / i    end  if display then printend = count rem - main program begins herevar i, ndiv, highdiv, highnum = integer print "Proper divisors of first 10 numbers:" for i = 1 to 10   print using "### : "; i;  ndiv = divisors(i, true)next i print "Searching for number with most divisors ..."highdiv = 1highnum = 1for i = 1 to 20000   ndiv = divisors(i, false)  if ndiv > highdiv then    begin      highdiv = ndiv      highnum = i    endnext iprint "Searched up to"; iprint highnum; " has the most divisors: "; highdiv end `
Output:
```Proper divisors of first 10 numbers:
1 :
2 :     1
3 :     1
4 :     1     2
5 :     1
6 :     1     2    3
7 :     1
8 :     1     2    4
9 :     1     3
10 :     1     2    5
Searching for number with most divisors ...
Searched up to 20000
15120 has the most divisors:  79
```

## Scala

### Simple proper divisors

`def properDivisors(n: Int) = (1 to n/2).filter(i => n % i == 0)def format(i: Int, divisors: Seq[Int]) = f"\$i%5d    \${divisors.length}%2d   \${divisors mkString " "}" println(f"    n   cnt   PROPER DIVISORS")val (count, list) = (1 to 20000).foldLeft( (0, List[Int]()) ) { (max, i) =>    val divisors = properDivisors(i)    if (i <= 10 || i == 100) println( format(i, divisors) )    if (max._1 < divisors.length) (divisors.length, List(i))    else if (max._1 == divisors.length) (divisors.length, max._2 ::: List(i))    else max} list.foreach( number => println(f"\$number%5d    \${properDivisors(number).length}") )`
Output:
```    n   cnt   PROPER DIVISORS
1     0
2     1   1
3     1   1
4     2   1 2
5     1   1
6     3   1 2 3
7     1   1
8     3   1 2 4
9     2   1 3
10     3   1 2 5
100     8   1 2 4 5 10 20 25 50
15120    79
18480    79```

### Proper divisors for integers for big integers

If Longs are enough to you you can replace every BigInt with Long and the one BigInt(1) with 1L

`import scala.annotation.tailrec def factorize(x: BigInt): List[BigInt] = {  @tailrec  def foo(x: BigInt, a: BigInt = 2, list: List[BigInt] = Nil): List[BigInt] = a * a > x match {    case false if x % a == 0 => foo(x / a, a, a :: list)    case false => foo(x, a + 1, list)    case true => x :: list  }   foo(x)} def properDivisors(n: BigInt): List[BigInt] = {  val factors = factorize(n)  val products = (1 until factors.length).flatMap(i => factors.combinations(i).map(_.product).toList).toList  (BigInt(1) :: products).filter(_ < n)}`

## Seed7

`\$ include "seed7_05.s7i"; const proc: writeProperDivisors (in integer: n) is func  local    var integer: i is 0;  begin    for i range 1 to n div 2 do      if n rem i = 0 then        write(i <& " ");      end if;    end for;    writeln;  end func; const func integer: countProperDivisors (in integer: n) is func  result    var integer: count is 0;  local    var integer: i is 0;  begin    for i range 1 to n div 2 step succ(n rem 2) do      if n rem i = 0 then        incr(count);      end if;    end for;  end func; const proc: main is func  local    var integer: i is 0;    var integer: v is 0;    var integer: max is 0;    var integer: max_i is 1;  begin    for i range 1 to 10 do      write(i <& ": ");      writeProperDivisors(i);    end for;    for i range 1 to 20000 do      v := countProperDivisors(i);      if v > max then        max := v;        max_i := i;      end if;    end for;    writeln(max_i <& " with " <& max <& " divisors");  end func;`
Output:
```1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
15120 with 79 divisors
```

## Sidef

Translation of: Raku
`func propdiv (n) {    n.divisors.slice(0, -2)} {|i| printf("%2d: %s\n", i, propdiv(i)) } << 1..10 var max = 0var candidates = [] for i in (1..20_000) {    var divs = propdiv(i).len    if (divs > max) {        candidates = []        max = divs    }    candidates << i if (divs == max)} say "max = #{max}, candidates = #{candidates}"`
Output:
``` 1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
max = 79, candidates = [15120, 18480]
```

## Swift

Simple function:

`func properDivs1(n: Int) -> [Int] {     return filter (1 ..< n) { n % \$0 == 0 }}`

More efficient function:

`import func Darwin.sqrt func sqrt(x:Int) -> Int { return Int(sqrt(Double(x))) } func properDivs(n: Int) -> [Int] {     if n == 1 { return [] }     var result = [Int]()     for div in filter (1 ... sqrt(n), { n % \$0 == 0 }) {         result.append(div)         if n/div != div && n/div != n { result.append(n/div) }    }     return sorted(result) }`

`for i in 1...10 {    println("\(i): \(properDivs(i))")} var (num, max) = (0,0) for i in 1...20_000 {     let count = properDivs(i).count    if (count > max) { (num, max) = (i, count) }} println("\(num): \(max)")`
Output:
```1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79
```

## tbas

` dim _proper_divisors(100) sub proper_divisors(n)	dim i	dim _proper_divisors_count = 0	if n <> 1 then		for i = 1 to (n \ 2)			if n %% i = 0 then				_proper_divisors_count = _proper_divisors_count + 1				_proper_divisors(_proper_divisors_count) = i 			end if		next	end if	return _proper_divisors_countend sub sub show_proper_divisors(n, tabbed)	dim cnt = proper_divisors(n)	print str\$(n) + ":"; tab(4);"(" + str\$(cnt) + " items) ";	dim j 	for j = 1 to cnt		if tabbed then			print str\$(_proper_divisors(j)),		else 			print str\$(_proper_divisors(j));		end if		if (j < cnt) then print ",";	next	printend sub dim ifor i = 1 to 10    show_proper_divisors(i, false)next dim c dim maxindex = 0dim maxlength = 0for t = 1 to 20000	c = proper_divisors(t)	if c > maxlength then		maxindex = t		maxlength = c	end ifnext print "A maximum at ";show_proper_divisors(maxindex, false) `
```>tbas proper_divisors.bas
1:  (0 items)
2:  (1 items) 1
3:  (1 items) 1
4:  (2 items) 1,2
5:  (1 items) 1
6:  (3 items) 1,2,3
7:  (1 items) 1
8:  (3 items) 1,2,4
9:  (2 items) 1,3
10: (3 items) 1,2,5
A maximum at 15120:(79 items) 1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,21,24,27,28,30,
35,36,40,42,45,48,54,56,60,63,70,72,80,84,90,105,108,112,120,126,135,
140,144,168,180,189,210,216,240,252,270,280,315,336,360,378,420,432,
504,540,560,630,720,756,840,945,1008,1080,1260,1512,1680,1890,2160,
2520,3024,3780,5040,7560
```

## Tcl

Note that if a number, $k$, greater than 1 divides $n$ exactly, both $k$ and $n/k$ are proper divisors. (The raw answers are not sorted; the pretty-printer code sorts.)

`proc properDivisors {n} {    if {\$n == 1} return    set divs 1    for {set i 2} {\$i*\$i <= \$n} {incr i} {	if {!(\$n % \$i)} {	    lappend divs \$i	    if {\$i*\$i < \$n} {		lappend divs [expr {\$n / \$i}]	    }	}    }    return \$divs} for {set i 1} {\$i <= 10} {incr i} {    puts "\$i => {[join [lsort -int [properDivisors \$i]] ,]}"}set maxI [set maxC 0]for {set i 1} {\$i <= 20000} {incr i} {    set c [llength [properDivisors \$i]]    if {\$c > \$maxC} {	set maxI \$i	set maxC \$c    }}puts "max: \$maxI => (...\$maxC…)"`
Output:
```1 => {}
2 => {1}
3 => {1}
4 => {1,2}
5 => {1}
6 => {1,2,3}
7 => {1}
8 => {1,2,4}
9 => {1,3}
10 => {1,2,5}
max: 15120 => (...79...)
```

## uBasic/4tH

Translation of: True BASIC
`LET m = 1LET c = 0 PRINT "The proper divisors of the following numbers are:\n"PROC _ListProperDivisors (10) FOR n = 2 TO 20000    LET d = FUNC(_CountProperDivisors(n))    IF d > c THEN       LET c = d       LET m = n    ENDIFNEXT PRINTPRINT m; " has the most proper divisors, namely "; cEND _CountProperDivisors    PARAM (1)    LOCAL (2)     IF [email protected] < 2 THEN RETURN (0)    LET [email protected] = 0    FOR [email protected] = 1 TO [email protected] / 2        IF [email protected] % [email protected] = 0 THEN LET [email protected] = [email protected] + 1    NEXTRETURN ([email protected]) _ListProperDivisors    PARAM (1)    LOCAL (2)     IF [email protected] < 1 THEN RETURN    FOR [email protected] = 1 TO [email protected]        PRINT [email protected]; " ->";        IF [email protected] = 1 THEN PRINT " (None)";        FOR [email protected] = 1 TO [email protected] / 2            IF [email protected] % [email protected] = 0 THEN PRINT " "; [email protected];        NEXT        PRINT    NEXTRETURN`
Output:
```The proper divisors of the following numbers are:

1 -> (None)
2 -> 1
3 -> 1
4 -> 1 2
5 -> 1
6 -> 1 2 3
7 -> 1
8 -> 1 2 4
9 -> 1 3
10 -> 1 2 5

15120 has the most proper divisors, namely 79

0 OK, 0:415```

## VBA

`Public Sub Proper_Divisor()Dim t() As Long, i As Long, l As Long, j As Long, c As Long    For i = 1 To 10        Debug.Print "Proper divisor of " & i & " : " & Join(S(i), ", ")    Next    For i = 2 To 20000        l = UBound(S(i)) + 1        If l > c Then c = l: j = i    Next    Debug.Print "Number in the range 1 to 20,000 with the most proper divisors is : " & j    Debug.Print j & " count " & c & " proper divisors"End Sub Private Function S(n As Long) As String()'returns the proper divisors of nDim j As Long, t() As String, c As Long    't = list of proper divisor of n    If n > 1 Then        For j = 1 To n \ 2            If n Mod j = 0 Then                ReDim Preserve t(c)                t(c) = j                c = c + 1            End If        Next    End If    S = tEnd Function`
Output:
```Proper divisor of 1 :
Proper divisor of 2 : 1
Proper divisor of 3 : 1
Proper divisor of 4 : 1, 2
Proper divisor of 5 : 1
Proper divisor of 6 : 1, 2, 3
Proper divisor of 7 : 1
Proper divisor of 8 : 1, 2, 4
Proper divisor of 9 : 1, 3
Proper divisor of 10 : 1, 2, 5
Number in the range 1 to 20,000 with the most proper divisors is : 15120
15120 count 79 proper divisors```

## Visual Basic .NET

Translation of: C#
`Module Module1     Function ProperDivisors(number As Integer) As IEnumerable(Of Integer)        Return Enumerable.Range(1, number / 2).Where(Function(divisor As Integer) number Mod divisor = 0)    End Function     Sub Main()        For Each number In Enumerable.Range(1, 10)            Console.WriteLine("{0}: {{{1}}}", number, String.Join(", ", ProperDivisors(number)))        Next         Dim record = Enumerable.Range(1, 20000).Select(Function(number) New With {.Number = number, .Count = ProperDivisors(number).Count()}).OrderByDescending(Function(currentRecord) currentRecord.Count).First()        Console.WriteLine("{0}: {1}", record.Number, record.Count)    End Sub End Module`
Output:
```1: {}
2: {1}
3: {1}
4: {1, 2}
5: {1}
6: {1, 2, 3}
7: {1}
8: {1, 2, 4}
9: {1, 3}
10: {1, 2, 5}
15120: 79```

## Wren

Library: Wren-fmt
Library: Wren-math
`import "/fmt" for Fmtimport "/math" for Int for (i in 1..10) System.print("%(Fmt.d(2, i)) -> %(Int.properDivisors(i))") System.print("\nThe number in the range [1, 20000] with the most proper divisors is:")var number = 1var maxDivs = 0for (i in 2..20000) {    var divs = Int.properDivisors(i).count    if (divs > maxDivs) {        number = i        maxDivs = divs    }}System.print("%(number) which has %(maxDivs) proper divisors.")`
Output:
``` 1 -> []
2 -> 
3 -> 
4 -> [1, 2]
5 -> 
6 -> [1, 2, 3]
7 -> 
8 -> [1, 2, 4]
9 -> [1, 3]
10 -> [1, 2, 5]

The number in the range [1, 20000] with the most proper divisors is:
15120 which has 79 proper divisors.
```

## XPL0

`func PropDiv(N, Show);  \Count and optionally show proper divisors of Nint  N, Show, D, C;[C:= 0;if N > 1 then    [D:= 1;    repeat  if rem(N/D) = 0 then                [C:= C+1;                if Show then                    [if D > 1 then ChOut(0, ^ );                    IntOut(0, D);                    ];                ];            D:= D+1;    until   D >= N;    ];return C;]; int N, SN, Cnt, Max;[for N:= 1 to 10 do    [ChOut(0, ^[);  PropDiv(N, true);  ChOut(0, ^]);    ChOut(0, ^ );    ];CrLf(0); Max:= 0;for N:= 1 to 20000 do    [Cnt:= PropDiv(N, false);    if Cnt > Max then        [Max:= Cnt;  SN:= N];    ];IntOut(0, SN);  ChOut(0, ^ ); IntOut(0, Max);  CrLf(0);]`
Output:
```[]   [1 2]  [1 2 3]  [1 2 4] [1 3] [1 2 5]
15120 79
```

## zkl

Translation of: D

This is the simple version :

`fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) }`

This version is MUCH faster (the output isn't ordered however):

`fcn properDivs(n){    if(n==1) return(T);   ( pd:=[1..(n).toFloat().sqrt()].filter('wrap(x){ n%x==0 }) )   .pump(pd,'wrap(pd){ if(pd!=1 and (y:=n/pd)!=pd ) y else Void.Skip })}`
`[1..10].apply(properDivs).println();[1..20_001].apply('wrap(n){ T(properDivs(n).len(),n) })   .reduce(fcn([(a,_)]ab, [(c,_)]cd){ a>c and ab or cd },T(0,0))   .println();`
Output:
```L(L(),L(1),L(1),L(1,2),L(1),L(1,2,3),L(1),L(1,2,4),L(1,3),L(1,2,5))
L(79,18480)
```