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.
- Task
- Create a routine to generate all the proper divisors of a number.
- use it to show the proper divisors of the numbers 1 to 10 inclusive.
- 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.
- Related tasks
- Amicable pairs
- Abundant, deficient and perfect number classifications
- Aliquot sequence classifications
- Factors of an integer
- Prime decomposition
11l
<lang 11l>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[1]) print(n‘ ’leng)</lang>
- Output:
[[], [1], [1], [1, 2], [1], [1, 2, 3], [1], [1, 2, 4], [1, 3], [1, 2, 5]] 15120 79
360 Assembly
This program uses two ASSIST macros (XDECO, XPRNT) to keep the code as short as possible. <lang 360asm>* Proper divisors 14/06/2016 PROPDIV 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=1
LOOPN1 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=1
LOOPNI 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 LOOPNI
ELOOPNI XPRNT PG,80 print buffer
LA R10,1(R10) n=n+1
B LOOPN1
ELOOPN1 SR R0,R0 0
ST R0,M m=0
LA R10,1 n=1
LOOPN2 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=nn
NNNHM LA R10,1(R10) n=n+1
B LOOPN2
ELOOPN2 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 R14
NOTONE 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+1
LOOPJ 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/j
ITERJ A R7,ODD j=j+odd
B LOOPJ
ELOOPJ 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)=j
JTJNEX 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 ELOOPI
LOOPI 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-1
ELOOPI LR R0,R8 return(ia)
BR R14 return to caller
- ---- ----------------------------------------
TDIV DS 80F TDIVB DS 40F M DS F NN DS F II DS F X DS F ODD DS F PGT DC CL80'... has .. proper divisors:' PGR DC CL80'..... has ... proper divisors.' PG DC CL80' ' XDEC DS CL12
YREGS
END PROPDIV</lang>
- 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.
Ada
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 [[1]], Abundant Odd Number [[2]], and Amicable Pairs [[3]], we define this routine as a function of a generic package:
<lang Ada>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;</lang>
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:
<lang Ada>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; </lang>
- 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
<lang algol68># 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
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
) )</lang>
- 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, this uses an array that is too large for ALGOL 68G under Windows.
<lang algol68>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
OD
END</lang>
- 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. <lang algol> 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 </lang>
- 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
<lang AppleScript>-- 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 if
end 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 lst
end 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 tell
end filter
-- foldl :: (a -> b -> a) -> a -> [b] -> a on 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 tell
end 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 tell
end map
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn</lang>
- Output:
<lang AppleScript>{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}}</lang>
Idiomatic
<lang applescript>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 output
end properDivisors
-- Task code. local output, astid, i, maxPDs, maxPDNums, pdCount set output to {} set astid to AppleScript's text item delimiters set 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 repeat set maxPDs to 0 set 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 if
end repeat set end of output to linefeed & "Largest number of proper divisors for any number from 1 to 20,000: " & maxPDs set end of output to "Numbers with this many: " & maxPDNums set AppleScript's text item delimiters to linefeed set output to output as text set AppleScript's text item delimiters to astid return output</lang>
- Output:
<lang applescript>"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: 79 Numbers with this many: 15120, 18480"</lang>
Arc
<lang 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.
</lang>
- Output:
<lang Arc> (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)
</lang>
ARM Assembly
<lang ARM Assembly> /* 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 */ /*******************************************/ .data szMessStartPgm: .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,#1
1:
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,iMaxi
2:
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 100f
99: @ display error message
ldr r0,iAdrszMessError bl affichageMess
100: @ standard end of the program
mov r0, #0 @ return code mov r7, #EXIT @ request to exit program svc 0 @ perform system call
iAdrszMessStartPgm: .int szMessStartPgm iAdrszMessEndPgm: .int szMessEndPgm iAdrszMessError: .int szMessError iAdrszCarriageReturn: .int szCarriageReturn iAdrszMessResult: .int szMessResult iAdrsValue: .int sValue iAdrszMessDivNumber: .int szMessDivNumber iAdrsCounter: .int sCounter iAdrszMessEntete: .int szMessEntete iAdrsNumber: .int sNumber iAdrszMessNumberMax: .int szMessNumberMax iAdrsDivMax: .int sDivMax iAdrsNumberMax: .int sNumberMax iMaxi: .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 affichageMess
1: @ begin loop
lsr r4,r7,#1 @ Maxi mov r6,r4 @ first divisor mov r5,#1 @ Counter divisors
2:
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 affichageMess
3:
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" </lang>
- 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
<lang rebol>properDivisors: function [x] ->
(factors x) -- x
loop 1..10 'x ->
print ["proper divisors of" x "=>" properDivisors x]
maxN: 0 maxProperDivisors: 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]</lang>
- 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] The number with the most proper divisors ( 79 ) is 15120
AutoHotkey
<lang AutoHotkey>proper_divisors(n) { Array := [] if n = 1 return Array Array[1] := 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 }</lang> Examples:<lang AutoHotkey>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 % output return</lang>
- 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
<lang AWK>
- syntax: GAWK -f PROPER_DIVISORS.AWK
BEGIN {
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
} </lang>
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
BaCon
<lang qbasic> 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"; PRINT
NEXT
FOR x = 1 TO 20000
DivisorCount = ProperDivisor(x, 0)
IF DivisorCount > MaxDivisors THEN
MaxDivisors = DivisorCount
MagicNumber = x
END IF
NEXT
PRINT "Most proper divisors for number in the range 1-20000: ", MagicNumber, " with ", MaxDivisors, " divisors." </lang>
- 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. <lang c>
- 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;
} </lang>
- 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. <lang C>
- 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;
} </lang>
C#
<lang csharp>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);
}
}
}</lang>
- 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++
<lang cpp>#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 ;
} </lang>
- 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
<lang 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"); }</lang>
- 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
<lang lisp>(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"))
</lang>
- 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. <lang lisp>(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)))</lang>
- Output:
CL-USER(10): (main) Task 1:Proper Divisors of [1,10]: 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
<lang oberon2> 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 j END 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
END END Do;
END RosettaProperDivisor.
^Q RosettaProperDivisor.Do~ </lang>
- 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
Currently the lambda of the filter allocates a closure on the GC-managed heap. <lang d>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;
}</lang>
- Output:
[[], [1], [1], [1, 2], [1], [1, 2, 3], [1], [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
<lang Delphi> 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. </lang>
- 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 <lang Delphi> 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. </lang>
Dyalect
<lang dyalect>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).len()
if count > max {
set (num, max) = (i, count)
}
}
print("\(num): \(max)")</lang>
- 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
EchoLisp
<lang scheme> (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)))
</lang>
- Output:
<lang scheme> (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 ;; 🎩 </lang>
Eiffel
<lang 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 </lang>
- 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
<lang elixir>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"
end
end
Enum.each(1..10, fn n ->
IO.puts "#{n}: #{inspect Proper.divisors(n)}"
end) Proper.most_divisors(20000)</lang>
- 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] With 79, Number [18480, 15120] has the most divisors
Erlang
<lang erlang>-module(properdivs). -export([divs/1,sumdivs/1,longest/1]).
divs(0) -> []; divs(1) -> []; divs(N) -> lists:sort([1] ++ 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.</lang>
- Output:
1> [io:format("X: ~w, N: ~w~n", [N,properdivs:divs(N)]) || N <- lists:seq(1,10)].
X: 1, N: []
X: 2, N: [1]
X: 3, N: [1]
X: 4, N: [1,2]
X: 5, N: [1]
X: 6, N: [1,2,3]
X: 7, N: [1]
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#
<lang fsharp> // the simple function with the answer let propDivs n = [1..n/2] |> List.filter (fun x->n % x = 0)
// to cache the result length; helpful for a long search let propDivDat n = propDivs n |> fun xs -> n, xs.Length, xs
// UI: always the longest and messiest let 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 it Seq.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 </lang>
- 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
<lang factor>USING: formatting io kernel math math.functions math.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 . ] each 20000 [1,b] [ #divisors ] supremum-by dup #divisors "%d with %d divisors.\n" printf</lang>
- 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.
Fortran
Compiled using G95 compiler, run on x86 system under Puppy Linux <lang Fortran>
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
</lang>
- 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
<lang 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 i
End 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 count
End Function
Dim As Integer n, count, most = 1, maxCount = 0
Print "The proper divisors of the following numbers are :" Print ListProperDivisors(10)
For n As Integer = 2 To 20000
count = CountProperDivisors(n) If count > maxCount Then maxCount = count most = n EndIf
Next
Print Print Str(most); " has the most proper divisors, namely"; maxCount Print Print "Press any key to exit the program" Sleep End </lang>
- 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
Frink
Frink's built-in factorization routines efficiently find factors of arbitrary-sized integers.
<lang frink> for n = 1 to 10
println["$n\t" + join[" ", properDivisors[n]]]
println[]
d = new dict for n = 1 to 20000 {
c = length[properDivisors[n]] d.addToList[c, n]
}
most = max[keys[d]] println[d@most + " have $most factors"]
properDivisors[n] := allFactors[n, true, false, true] </lang>
- 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
<lang> OPENW 1 CLEARW 1 ' ' Array f% is used to hold the divisors DIM 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%=@proper_divisors(i%) PRINT "Divisors for ";i%;":"; FOR j%=1 TO num% PRINT " ";f%(j%); NEXT j% PRINT
NEXT i% ' ' 2. Find (smallest) number <= 20000 with largest number of proper divisors ' result%=1 ! largest so far number%=0 ! its number of divisors FOR i%=1 TO 20000
num%=@proper_divisors(i%) IF num%>number% result%=i% number%=num% ENDIF
NEXT 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: [1] X: 3, N: [1] X: 4, N: [1,2] X: 5, N: [1] X: 6, N: [1,2,3] X: 7, N: [1] 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 re root of n%
f%(count%)=n%/i%
count%=count%+1
ENDIF
ENDIF
NEXT i%
ENDIF
'
RETURN count%-1
ENDFUNC </lang>
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
<lang go>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[0] = 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)
}
}</lang>
- 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
Haskell
<lang Haskell>import Data.Ord import 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]]</lang>
- Output:
divisors of 1 to 10: [] [1] [1] [1,2] [1] [1,2,3] [1] [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:
<lang haskell>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]</lang>
- Output:
Proper divisors of 1 to 10: [] [1] [1] [1,2] [1] [1,2,3] [1] [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:
<lang haskell>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 ) [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) <$> [1 .. 20000]
DISPLAY--------------------------
fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String fTable s xShow fxShow f xs =
let rjust n c = (drop . length) <*> (replicate n c ++)
w = maximum (length . xShow <$> xs)
in unlines $
s : fmap (((++) . rjust w ' ' . xShow) <*> ((" -> " ++) . fxShow . f)) xs</lang>
- Output:
Proper divisors of [1..10]: 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] 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:
<lang J>factors=: [: /:~@, */&>@{@((^ i.@>:)&.>/)@q:~&__ properDivisors=: factors -. ]</lang>
Proper divisors of numbers 1 through 10:
<lang J> (,&": ' -- ' ,&": properDivisors)&>1+i.10 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</lang>
Number(s) not exceeding 20000 with largest number of proper divisors (and the count of those divisors):
<lang J> (, #@properDivisors)&> 1+I.(= >./) #@properDivisors@> 1+i.20000 15120 79 18480 79</lang>
Note that it's a bit more efficient to simply count factors here, when selecting the candidate numbers.
<lang J> (, #@properDivisors)&> 1+I.(= >./) #@factors@> 1+i.20000 15120 79 18480 79</lang>
We could also arbitrarily toss either 15120 or 18480 (keeping the other number), if it were important that we produce only one result.
Java
<lang java5>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);
}
}</lang>
- 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
JavaScript
ES5
<lang JavaScript>(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
);
})();</lang>
- 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
<lang JavaScript>(() => {
'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[0])
) : undefined;
// snd :: (a, b) -> b const snd = tpl => tpl[1];
// 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();
})();</lang>
- Output:
Proper divisors of [1..10]:
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]
Example of maximum divisor count in the range [1..20000]:
15120 has 79 proper divisors.
jq
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)": <lang jq>def count(stream): reduce stream as $i (0; . + 1);
- unordered
def 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 $count > .[1] then [$i, $count] else . end);</lang>
The tasks: <lang jq>"The proper divisors of the numbers 1 to 10 inclusive are:", (range(1;11) as $i | "\($i): \( [ $i | proper_divisors] )"), "", "The pair consisting of the least number in the range 1 to 20,000 with", "the maximal number proper divisors together with the corresponding", "count of proper divisors is:", most_proper_divisors(20000) </lang>
- Output:
<lang sh>$ jq -n -c -r -f /Users/peter/jq/proper_divisors.jq The proper divisors of the numbers 1 to 10 inclusive are: 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 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: [15120,79]</lang>
Julia
Use factor to obtain the prime factorization of the target number. I adopted the argument handling style of factor in my properdivisors function.
<lang Julia>
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 = 1 hi = 10 println("List the proper divisors for ", lo, " through ", hi, ".") for i in lo:hi
println(@sprintf("%4d", i), " ", properdivisors(i))
end
hi = 2*10^4 println("\nFind the numbers within [", lo, ",", hi, "] having the most divisors.")
maxdiv = 0 nlst = Int[]
for i in lo:hi
ndiv = length(properdivisors(i))
if ndiv > maxdiv
maxdiv = ndiv
nlst = [i]
elseif ndiv == maxdiv
push!(nlst, i)
end
end
println(nlst, " have the maximum proper divisor count of ", maxdiv, ".") </lang>
- 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
<lang scala>// 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)
}</lang>
- 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
<lang Lua>-- Return a table of the proper divisors of n function 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 divs
end
-- Show n followed by all values in t function show (n, t)
io.write(n .. ":\t") for _, v in pairs(t) do io.write(v .. " ") end print()
end
-- Main procedure local mostDivs, numDivs, answer = 0 for i = 1, 10 do show(i, propDivs(i)) end for i = 1, 20000 do
numDivs = #propDivs(i)
if numDivs > mostDivs then
mostDivs = numDivs
answer = i
end
end print(answer .. " has " .. mostDivs .. " proper divisors.")</lang>
- 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: <lang Mathematica>ProperDivisors[n_Integer /; n > 0] := Most@Divisors@n;</lang>
Proper divisors of n from 1 to 10: <lang Mathematica>Grid@Table[{n, ProperDivisors[n]}, {n, 1, 10}]</lang>
- 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: <lang Mathematica>Fold[
Last[SortBy[{#1, {#2, Length@ProperDivisors[#2]}}, Last]] &,
{0, 0},
Range[20000]]</lang>
- Output:
{18480, 79}
An alternate way to find the number with the most divisors between 1 and 20,000: <lang Mathematica>Last@SortBy[
Table[
{n, Length@ProperDivisors[n]},
{n, 1, 20000}],
Last]</lang>
- Output:
{15120, 79}
MATLAB
<lang matlab>function D=pd(N) K=1:ceil(N/2); D=K(~(rem(N, K)));</lang>
- 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
<lang modula2>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);
ReadChar
END ProperDivisors.</lang>
Nim
<lang nim>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 = 0 var maxI = 1
for i in 1..20000:
var v = countProperDivisors(i) if v >= max: max = v maxI = i
echo fmt"{maxI} with {max} divisors"</lang>
- 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
<lang oberon2> 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
END
END ProperDivisors. </lang>
- 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
<lang 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();
}
} </lang>
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
Oforth
<lang 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</lang>
- 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]
PARI/GP
<lang parigp>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]</lang>
- Output:
%1 = [[], [2], [3], [2, 4], [5], [2, 3, 6], [7], [2, 4, 8], [3, 9], [2, 5, 10]] %2 = [15120, 7]
Pascal
Using prime factorisation <lang pascal>{$IFDEF FPC}{$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[0] := 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 divisors var
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 divisors var
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.</lang>
- 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 : 1,2,515120 = 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
<lang perl>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 $ind";
- 2. Or we can use List::Util's max with decoration (this exploits its implementation)
{
use List::Util qw/max/;
no warnings 'numeric';
say max(map { scalar(proper_divisors($_)) . " $_" } 1..20000);
}</lang>
- Output:
1: 2: 1 3: 1 4: 1 2 5: 1 6: 1 2 3 7: 1 8: 1 2 4 9: 1 3 10: 1 2 5 79 15120 79 18480
Note that the first code will choose the first max, while the second chooses the last.
Phix
The factors routine is an auto-include. The actual implementation of it, from builtins\pfactors.e is
global function factors(atom n, integer include1=0) -- -- returns a list of all integer factors of n -- if include1 is 0 (the default), result does not contain either 1 or n -- if include1 is 1 the result contains 1 and n -- if include1 is -1 the result contains 1 but not n -- if n=0 then return {} end if check_limits(n,"factors") sequence lfactors = {}, hfactors = {} atom hfactor integer p = 2, lim = floor(sqrt(n)) if include1!=0 then lfactors = {1} if n!=1 and include1=1 then hfactors = {n} end if end if while p<=lim do if remainder(n,p)=0 then lfactors = append(lfactors,p) hfactor = n/p if hfactor=p then exit end if hfactors = prepend(hfactors,hfactor) end if p += 1 end while return lfactors & hfactors end function
The compiler knows where to find that, so the main program is just:
for i=0 to 10 do printf(1,"%d: %v\n",{i,factors(i,-1)}) end for sequence candidates = {} integer maxd = 0 for i=1 to 20000 do integer k = length(factors(i,-1)) if k>=maxd then if k=maxd then candidates &= i else candidates = {i} maxd = k end if end if end for printf(1,"%d divisors: %v\n", {maxd,candidates})
- Output:
0: {}
1: {1}
2: {1}
3: {1}
4: {1,2}
5: {1}
6: {1,2,3}
7: {1}
8: {1,2,4}
9: {1,3}
10: {1,2,5}
79 divisors: {15120,18480}
PHP
<lang php><?php function ProperDivisors($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";
</lang>
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.
PicoLisp
<lang 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))</lang>
Brute-force
<lang PicoLisp>(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) ) )</lang>
Factorization
<lang PicoLisp>(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) )
@@ ) )</lang>
Output:
15120 79 0.081 sec
PL/I
<lang pli>*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;</lang>
- 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
<lang PowerShell> 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)" </lang>
version 2
<lang PowerShell> 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)" </lang>
version 3
<lang PowerShell> 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)" </lang> 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
Taking a cue from an SO answer:
<lang prolog>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)).</lang>
Output:
<lang prolog>?- show_proper_divisors_of_range(1,10). 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] true.
?- find_most_proper_divisors_in_range(1,20000,Result). Result = num(15120)-divisor_count(79). </lang>
PureBasic
<lang 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
Next
EndProcedure
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 count
EndProcedure
Define n, count, most = 1, maxCount = 0 If 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 </lang>
- 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 <lang python>>>> 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[1]) >>> n 15120 >>> length 79 >>> </lang>
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[0] in range(M(0) + 1):
for m[1] in range(M[1] + 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. <lang python>from math import sqrt from functools import lru_cache, reduce from collections import Counter from 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[1]))</lang>
- 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:
<lang python>Proper divisors
from itertools import accumulate, chain, groupby, product from functools import reduce from math import floor, sqrt from 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([1], group), mul)
)]
return sorted(
reduce(go, [
list(g) for _, g in groupby(primeFactors(n))
], [1])
)[:-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] -> String
def 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[1]
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[0]
return root < q or 0 == (x % q)
q = until(p)(f)(
(2 if 0 == x % 2 else 3, 1)
)[0]
return [x] if q > root else [q] + go(x // q)
return go(n)
- snd :: (a, b) -> b
def snd(tpl):
Second member of a pair. return tpl[1]
- until :: (a -> Bool) -> (a -> a) -> a -> a
def 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()</lang>
- Output:
Proper divisors of [1..10]: 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] Example of maximum divisor count in the range [1..20000]: (15120, 79)
Python: The Simple Way
<lang python>import random propdiv = [] num = random.randint(1,200) for n in range(1, num+1): propdiv = [] for x in range(1,n+1): n = int(n) if n%x == 0: propdiv.append(x) propdiv.pop(len(propdiv)-1) print(str(propdiv) + " are the Proper Divisors of " + str(num))
print("\n")
Ften = 10 propdten = [] for toten in range(1,11): for n1 in range(1, toten+1): propdten = [] for x in range(1,n1+1): n1 = int(n1) if n1%x == 0: propdten.append(x) propdten.pop(len(propdten)-1) print(str(propdten) + " are the Proper Divisors of " + str(toten))
print("\n")
findiv = 1 normdiv = [] greatestdiv = [] twentykrng = 200 for nnn in range(1, twentykrng+1): if len(normdiv) > len(greatestdiv): greatestdiv = [] greatestdiv = normdiv findiv = nnn - 1 normdiv = [] for xx in range(1,nnn+1): if nnn%xx == 0: normdiv.append(xx) normdiv.pop(len(normdiv)-1)
print(findiv) print(len(greatestdiv))</lang>
- Output:
[1, 2, 5, 10, 17, 34, 85] are the Proper Divisors of 170 [] are the Proper Divisors of 1 [1] are the Proper Divisors of 2 [1] are the Proper Divisors of 3 [1, 2] are the Proper Divisors of 4 [1] are the Proper Divisors of 5 [1, 2, 3] are the Proper Divisors of 6 [1] are the Proper Divisors 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 15120 79
Quackery
factors is defined at Factors of an integer#Quackery.
<lang 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</lang>
- 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
<lang r># Proper divisors. 12/10/16 aev require(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");</lang>
- Output:
Loading required package: numbers *** max number of divisors: 79 *** for the following indices: 15120 18480
Filter solution
<lang r>#Task 1
- Has no input error checking.
properDivisors<-function(n) {
#Returning NULL seems like bad code, but task 2 demands some output for n=1, which has no proper divisors. if(n==1) NULL else Filter(function(x) n %% x == 0, 1:(n%/%2))
}
- Task 2
- The output could be put in to a cleaner form than a list, but this is the idiomatic way.
Vectorize(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)(1:N)
numberWithMostDivisors<-which.max(lengths(divisorList))
return(paste0("The number with the most proper divisors between 1 and ",N,
" is ",numberWithMostDivisors,
". It has ",length(divisorListnumberWithMostDivisors)," proper divisors."))
} mostProperDivisors(20000)</lang>
- Output:
#Task 2 > Vectorize(properDivisors)(1:10) [[1]] NULL [[2]] [1] 1 [[3]] [1] 1 [[4]] [1] 1 2 [[5]] [1] 1 [[6]] [1] 1 2 3 [[7]] [1] 1 [[8]] [1] 1 2 4 [[9]] [1] 1 3 [[10]] [1] 1 2 5 #Task 3 > mostProperDivisors(20000) [1] "The number with the most proper divisors between 1 and 20000 is 15120. It has 79 proper divisors."
Racket
Short version
<lang racket>#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)))</lang>
- 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>#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))</lang>
The output is the same as the short version above.
Raku
(formerly Perl 6)
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. <lang perl6>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}";</lang>
- 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 = 79, candidates = 15120 18480
REXX
version 1
<lang rexx>Call time 'R' Do x=1 To 10
Say x '->' proper_divisors(x) End
hi=1 Do 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: Procedure Parse Arg n If n=1 Then Return pd= /* Optimization reduces 37 seconds to 28 seconds */ If n//2=1 Then /* odd number */
delta=2
Else /* even number */
delta=1
Do d=1 To n%2 By delta
If n//d=0 Then pd=pd d End
Return space(pd)
count_proper_divisors: Procedure Parse Arg n Return words(proper_divisors(n))</lang>
- 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. <lang rexx>/*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.*/
say say 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 /*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). */</lang>
- 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). <lang rexx>/*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.*/
say say 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 /*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). */</lang>
- 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. <lang rexx>/*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.*/
say say 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 /*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).*/</lang>
- output is identical to the 2nd REXX version when using the same inputs.
Ring
<lang ring>
- Project : Proper divisors
limit = 10 for 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 nl
next </lang> 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
<lang 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([1]) do |n, res|
primes.combination(n).map{|combi| res << combi.inject(:*)}
end.flatten.uniq
end
end
(1..10).map{|n| puts "#{n}: #{n.proper_divisors}"}
size, select = (1..20_000).group_by{|n| n.proper_divisors.size}.max select.each do |n|
puts "#{n} has #{size} divisors"
end</lang>
- 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 18480 has 79 divisors
An Alternative Approach
<lang ruby>#Determine the integer within a range of integers that has the most proper divisors
- Nigel Galloway: December 23rd., 2014
require "prime" n, g = 0 (1..20000).each{|i| e = i.prime_division.inject(1){|n,g| n * (g[1]+1)}
n, g = e, i if e > n}
puts "#{g} has #{n-1} proper divisors"</lang>
- 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
<lang 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());
} </lang>
- 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] In 1 to 20000, 15120 has the most proper divisors at 79
S-BASIC
<lang Basic> $constant false = 0 $constant true = FFFFH
rem - compute p mod q function mod(p, q = integer) = integer end = p - q * (p/q)
rem - count, and optionally display, proper divisors of n function divisors(n, display = integer) = integer
var i, limit, count, start, delta = integer
if mod(n, 2) = 0 then
begin
start = 2
delta = 1
end
else
begin
start = 3
delta = 2
end
if n < 2 then count = 0 else count = 1
if display and (count = 1) then print using "#####"; 1;
i = start
limit = n / start
while i <= limit do
begin
if mod(n, i) = 0 then
begin
if display then print using "#####"; i;
count = count + 1
end
i = i + delta
if count = 1 then limit = n / i
end
if display then print
end = count
rem - main program begins here var i, ndiv, highdiv, highnum = integer
print "Proper divisors of first 10 numbers:" for i = 1 to 10
print using "### : "; i; ndiv = divisors(i, true)
next i
print "Searching for number with most divisors ..." highdiv = 1 highnum = 1 for i = 1 to 20000
ndiv = divisors(i, false)
if ndiv > highdiv then
begin
highdiv = ndiv
highnum = i
end
next i print "Searched up to"; i print highnum; " has the most divisors: "; highdiv
end </lang>
- Output:
Proper divisors of first 10 numbers: 1 : 2 : 1 3 : 1 4 : 1 2 5 : 1 6 : 1 2 3 7 : 1 8 : 1 2 4 9 : 1 3 10 : 1 2 5 Searching for number with most divisors ... Searched up to 20000 15120 has the most divisors: 79
Scala
Simple proper divisors
<lang Scala>def properDivisors(n: Int) = (1 to n/2).filter(i => n % i == 0) def format(i: Int, divisors: Seq[Int]) = f"$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 ${properDivisors(number).length}") )</lang>
- Output:
n cnt PROPER DIVISORS
1 0
2 1 1
3 1 1
4 2 1 2
5 1 1
6 3 1 2 3
7 1 1
8 3 1 2 4
9 2 1 3
10 3 1 2 5
100 8 1 2 4 5 10 20 25 50
15120 79
18480 79
Proper divisors for integers for big integers
If Longs are enough to you you can replace every BigInt with Long and the one BigInt(1) with 1L
<lang Scala>import scala.annotation.tailrec
def factorize(x: BigInt): List[BigInt] = {
@tailrec
def foo(x: BigInt, a: BigInt = 2, list: List[BigInt] = Nil): List[BigInt] = a * a > x match {
case false if x % a == 0 => foo(x / a, a, a :: list)
case false => foo(x, a + 1, list)
case true => x :: list
}
foo(x)
}
def properDivisors(n: BigInt): List[BigInt] = {
val factors = factorize(n) val products = (1 until factors.length).flatMap(i => factors.combinations(i).map(_.product).toList).toList (BigInt(1) :: products).filter(_ < n)
}</lang>
Seed7
<lang seed7>$ 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;</lang>
- 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
<lang ruby>func propdiv (n) {
n.divisors.slice(0, -2)
}
var max = 0 var 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}"</lang>- 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 = 79, candidates = [15120, 18480]
Swift
Simple function: <lang Swift>func properDivs1(n: Int) -> [Int] {
return filter (1 ..< n) { n % $0 == 0 }
}</lang> More efficient function: <lang Swift>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)
}</lang> Rest of the task: <lang Swift>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)")</lang>
- 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
tbas
<lang vb> 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_count end 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 print end sub
dim i for i = 1 to 10
show_proper_divisors(i, false)
next
dim c dim maxindex = 0 dim maxlength = 0 for t = 1 to 20000 c = proper_divisors(t) if c > maxlength then maxindex = t maxlength = c end if next
print "A maximum at "; show_proper_divisors(maxindex, false) </lang>
>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, , greater than 1 divides exactly, both and are proper divisors. (The raw answers are not sorted; the pretty-printer code sorts.) <lang tcl>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…)"</lang>
- 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...)
VBA
<lang vb>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 n Dim 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 = t
End Function</lang>
- 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
<lang vbnet>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</lang>
- 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: 79Wren
<lang ecmascript>import "/fmt" for Fmt import "/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 = 1 var maxDivs = 0 for (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.")</lang>
- 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 in the range [1, 20000] with the most proper divisors is: 15120 which has 79 proper divisors.
zkl
This is the simple version : <lang zkl>fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) }</lang> This version is MUCH faster (the output isn't ordered however): <lang zkl>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 })
}</lang>
<lang zkl>[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();</lang>
- 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)
- Programming Tasks
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