# Abundant, deficient and perfect number classifications

Abundant, deficient and perfect number classifications
You are encouraged to solve this task according to the task description, using any language you may know.

These define three classifications of positive integers based on their   proper divisors.

Let   P(n)   be the sum of the proper divisors of   n   where the proper divisors are all positive divisors of   n   other than   n   itself.

   if    P(n) <  n    then  n  is classed as  deficient  (OEIS A005100).
if    P(n) == n    then  n  is classed as  perfect    (OEIS A000396).
if    P(n) >  n    then  n  is classed as  abundant   (OEIS A005101).


Example

6   has proper divisors of   1,   2,   and   3.

1 + 2 + 3 = 6,   so   6   is classed as a perfect number.

Calculate how many of the integers   1   to   20,000   (inclusive) are in each of the three classes.

Show the results here.

## 11l

Translation of: Kotlin
F sum_proper_divisors(n)   R I n < 2 {0} E sum((1 .. n I/ 2).filter(it -> (@n % it) == 0)) V deficient = 0V perfect = 0V abundant = 0 L(n) 1..20000   V sp = sum_proper_divisors(n)   I sp < n      deficient++   E I sp == n      perfect++   E I sp > n      abundant++ print(‘Deficient = ’deficient)print(‘Perfect   = ’perfect)print(‘Abundant  = ’abundant)
Output:
Deficient = 15043
Perfect   = 4
Abundant  = 4953


## 360 Assembly

Translation of: VBScript

For maximum compatibility, this program uses only the basic instruction set (S/360) with 2 ASSIST macros (XDECO,XPRNT).

*        Abundant, deficient and perfect number  08/05/2016ABUNDEFI CSECT         USING  ABUNDEFI,R13       set base registerSAVEAR   B      STM-SAVEAR(R15)    skip savearea         DC     17F'0'             saveareaSTM      STM    R14,R12,12(R13)    save registers         ST     R13,4(R15)         link backward SA         ST     R15,8(R13)         link forward SA         LR     R13,R15            establish addressability         SR     R10,R10            deficient=0         SR     R11,R11            perfect  =0         SR     R12,R12            abundant =0         LA     R6,1               i=1LOOPI    C      R6,NN              do i=1 to nn         BH     ELOOPI         SR     R8,R8              sum=0         LR     R9,R6              i         SRA    R9,1               i/2         LA     R7,1               j=1LOOPJ    CR     R7,R9              do j=1 to i/2         BH     ELOOPJ         LR     R2,R6              i         SRDA   R2,32         DR     R2,R7              i//j=0         LTR    R2,R2              if i//j=0         BNZ    NOTMOD         AR     R8,R7              sum=sum+jNOTMOD   LA     R7,1(R7)           j=j+1         B      LOOPJELOOPJ   CR     R8,R6              if sum?i         BL     SLI                      <          BE     SEI                      =         BH     SHI                      >SLI      LA     R10,1(R10)         deficient+=1         B      EIFSEI      LA     R11,1(R11)         perfect  +=1         B      EIFSHI      LA     R12,1(R12)         abundant +=1EIF      LA     R6,1(R6)           i=i+1         B      LOOPIELOOPI   XDECO  R10,XDEC           edit deficient         MVC    PG+10(5),XDEC+7         XDECO  R11,XDEC           edit perfect         MVC    PG+24(5),XDEC+7         XDECO  R12,XDEC           edit abundant         MVC    PG+39(5),XDEC+7         XPRNT  PG,80              print buffer         L      R13,4(0,R13)       restore savearea pointer         LM     R14,R12,12(R13)    restore registers         XR     R15,R15            return code = 0         BR     R14                return to callerNN       DC     F'20000'PG       DC     CL80'deficient=xxxxx perfect=xxxxx abundant=xxxxx'XDEC     DS     CL12         REGEQU         END    ABUNDEFI
Output:
deficient=15043 perfect=    4 abundant= 4953


## ALGOL 68

# resturns the sum of the proper divisors of n                    ## if n = 1, 0 or -1, we return 0                                  #PROC sum proper divisors = ( INT n )INT:     BEGIN         INT result := 0;         INT abs n = ABS n;         IF abs n > 1 THEN             FOR d FROM ENTIER sqrt( abs n ) BY -1 TO 2 DO                 IF abs n MOD d = 0 THEN                     # found another divisor                      #                     result +:= d;                     IF d * d /= n THEN                         # include the other divisor              #                         result +:= n OVER d                     FI                 FI             OD;             # 1 is always a proper divisor of numbers > 1        #             result +:= 1         FI;         result     END # sum proper divisors # ; # classify the numbers 1 : 20 000 as abudant, deficient or perfect #INT abundant count    := 0;INT deficient count   := 0;INT perfect count     := 0;INT abundant example  := 0;INT deficient example := 0;INT perfect example   := 0;INT max number         = 20 000;FOR n TO max number DO    IF     INT pd sum = sum proper divisors( n );           pd sum < n    THEN        # have a deficient number                                  #        deficient count    +:= 1;        deficient example   := n    ELIF   pd sum = n    THEN        # have a perfect number                                    #        perfect count      +:= 1;        perfect example     := n    ELSE # pd sum > n #        # have an abundant number                                  #        abundant count     +:= 1;        abundant example    := n    FIOD; # show how many of each type of number there are and an example    # # displays the classification, count and example                   #PROC show result = ( STRING classification, INT count, example )VOID:     print( ( "There are "            , whole( count, -8 )            , " "            , classification            , " numbers up to "            , whole( max number, 0 )            , " e.g.: "            , whole( example, 0 )            , newline            )          ); show result( "abundant ",  abundant count,  abundant example  );show result( "deficient", deficient count, deficient example );show result( "perfect  ",   perfect count,   perfect example   )
Output:
There are     4953 abundant  numbers up to 20000 e.g.: 20000
There are    15043 deficient numbers up to 20000 e.g.: 19999
There are        4 perfect   numbers up to 20000 e.g.: 8128


## AutoHotkey

Loop{    m := A_index    ; getting factors=====================    loop % floor(sqrt(m))    {        if ( mod(m, A_index) == "0" )        {            if ( A_index ** 2 == m )            {                list .= A_index . ":"                sum := sum + A_index                continue            }            if ( A_index != 1 )            {                list .= A_index . ":" . m//A_index . ":"                sum := sum + A_index + m//A_index            }            if ( A_index == "1" )            {                list .= A_index . ":"                sum := sum + A_index            }        }    }    ; Factors obtained above===============    if ( sum == m ) && ( sum != 1 )    {        result := "perfect"        perfect++    }    if ( sum > m )    {        result := "Abundant"        Abundant++    }    if ( sum < m ) or ( m == "1" )    {        result := "Deficient"        Deficient++    }    if ( m == 20000 )	    {        MsgBox % "number: " . m . "nFactors:n" . list . "nSum of Factors: " . Sum . "nResult: " . result . "n_______________________nTotals up to: " . m . "nPerfect: " . perfect . "nAbundant: " . Abundant . "nDeficient: " . Deficient         ExitApp    }    list := ""    sum := 0} esc::ExitApp 
Output:
number: 20000
Factors:
1:2:10000:4:5000:5:4000:8:2500:10:2000:16:1250:20:1000:25:800:32:625:40:500:50:400:80:250:100:200:125:160:
Sum of Factors: 29203
Result: Abundant
_______________________
Totals up to: 20000
Perfect: 4
Abundant: 4953
Deficient: 15043


## AWK

works with GNU Awk 3.1.5 and with BusyBox v1.21.1

 #!/bin/gawk -ffunction sumprop(num,   i,sum,root) {if (num == 1) return 0sum=1root=sqrt(num)for ( i=2; i < root; i++) {    if (num % i == 0 )    {     sum = sum + i + num/i    }    }if (num % root == 0)    {    sum = sum + root   }    return sum} BEGIN{limit = 20000abundant = 0defiecient =0 perfect = 0 for (j=1; j < limit+1; j++)    {    sump = sumprop(j)    if (sump < j) deficient = deficient + 1    if (sump == j) perfect = perfect + 1    if (sump > j) abundant = abundant + 1    }print "For 1 through " limitprint "Perfect: " perfectprint "Abundant: " abundantprint "Deficient: " deficient    } 
Output:
For 1 through 20000
Perfect: 4
Abundant: 4953
Deficient: 15043


## Batch File

As batch files aren't particularly well-suited to increasingly large arrays of data, this code will chew through processing power.

 @echo offsetlocal enabledelayedexpansion :_main for /l %%i in (1,1,20000) do (   echo Processing %%i   call:_P %%i  set Pn=!errorlevel!  if !Pn! lss %%i set /a deficient+=1  if !Pn!==%%i set /a perfect+=1  if !Pn! gtr %%i set /a abundant+=1  cls) echo Deficient - %deficient% ^| Perfect - %perfect% ^| Abundant - %abundant%pause>nul  :_Psetlocal enabledelayedexpansionset sumdivisers=0 set /a upperlimit=%1-1 for /l %%i in (1,1,%upperlimit%) do (  set /a isdiviser=%1 %% %%i  if !isdiviser!==0 set /a sumdivisers+=%%i) exit /b %sumdivisers% 

## Befunge

This is not a particularly efficient implementation, so unless you're using a compiler, you can expect it to take a good few minutes to complete. But you can always test with a shorter range of numbers by replacing the 20000 ("2":*8*) near the start of the first line.

p0"2":*8*>::2/\:2/\28*:*:**+>::28*:*:*/\28*:*:*%%#v_\:28*:*:*%v>00p:0\0\-1v++\1-:1#^_$:28*:*:*/\28*vv_^#<<<!%*:*:*82:-1\-1\<<<\+**:*:*82<+>*:*:**\2-!#+v"There are "0\g00+1%*:*:<>28*:*:*/\28*:*:*/:0\28*:*:**+-:!00g^^82!:g01\p01<>:#,_\." ,tneicifed">:#,_\." dna ,tcefrep">:#,_\.55+".srebmun tnadnuba">:#,[email protected] Output: There are 15043 deficient, 4 perfect, and 4953 abundant numbers. ## Bracmat Two solutions are given. The first solution first decomposes the current number into a multiset of prime factors and then constructs the proper divisors. The second solution finds proper divisors by checking all candidates from 1 up to the square root of the given number. The first solution is a few times faster, because establishing the prime factors of a small enough number (less than 2^32 or less than 2^64, depending on the bitness of Bracmat) is fast. ( clk$:?t0& ( multiples  =   prime multiplicity    .     !arg:(?prime.?multiplicity)        & !multiplicity:0        & 1      |   !prime^!multiplicity*(.!multiplicity)        + multiples$(!prime.-1+!multiplicity) )& ( P = primeFactors prime exp poly S . !arg^1/67:?primeFactors & ( !primeFactors:?^1/67&0 | 1:?poly & whl ' ( !primeFactors:%?prime^?exp*?primeFactors & !poly*multiples$(!prime.67*!exp):?poly              )          & -1+!poly+1:?poly          & 1:?S          & (   !poly              :   ?                + (#%@?s*?&!S+!s:?S&~)                + ?            | 1/2*!S            )        )  )& 0:?deficient:?perfect:?abundant& 0:?n&   whl  ' ( 1+!n:~>20000:?n    &   P$!n : ( <!n&1+!deficient:?deficient | !n&1+!perfect:?perfect | >!n&1+!abundant:?abundant ) )& out$(deficient !deficient perfect !perfect abundant !abundant)& clk$:?t1& out$(flt$(!t1+-1*!t0,2) sec)& clk$:?t2& ( P  =   f h S    .   0:?f      & 0:?S      &   whl        ' ( 1+!f:?f          & !f^2:~>!n          & (   !arg*!f^-1:~/:?g              & !S+!f:?S              & ( !g:~!f&!S+!g:?S                |                 )            |             )          )      & 1/2*!S  )& 0:?deficient:?perfect:?abundant& 0:?n&   whl  ' ( 1+!n:~>20000:?n    &   P$!n : ( <!n&1+!deficient:?deficient | !n&1+!perfect:?perfect | >!n&1+!abundant:?abundant ) )& out$(deficient !deficient perfect !perfect abundant !abundant)& clk$:?t3& out$(flt$(!t3+-1*!t2,2) sec)); Output: deficient 15043 perfect 4 abundant 4953 4,27*10E0 sec deficient 15043 perfect 4 abundant 4953 1,63*10E1 sec ## C  #include<stdio.h>#define de 0#define pe 1#define ab 2 int main(){ int sum = 0, i, j; int try_max = 0; //1 is deficient by default and can add it deficient list int count_list[3] = {1,0,0}; for(i=2; i <= 20000; i++){ //Set maximum to check for proper division try_max = i/2; //1 is in all proper division number sum = 1; for(j=2; j<try_max; j++){ //Check for proper division if (i % j) continue; //Pass if not proper division //Set new maximum for divisibility check try_max = i/j; //Add j to sum sum += j; if (j != try_max) sum += try_max; } //Categorize summation if (sum < i){ count_list[de]++; continue; } if (sum > i){ count_list[ab]++; continue; } count_list[pe]++; } printf("\nThere are %d deficient," ,count_list[de]); printf(" %d perfect," ,count_list[pe]); printf(" %d abundant numbers between 1 and 20000.\n" ,count_list[ab]);return 0;}  Output: There are 15043 deficient, 4 perfect, 4953 abundant numbers between 1 and 20000.  ## C# using System;using System.Linq; public class Program{ public static void Main() { int abundant, deficient, perfect; ClassifyNumbers.UsingSieve(20000, out abundant, out deficient, out perfect); Console.WriteLine($"Abundant: {abundant}, Deficient: {deficient}, Perfect: {perfect}");         ClassifyNumbers.UsingDivision(20000, out abundant, out deficient, out perfect);        Console.WriteLine($"Abundant: {abundant}, Deficient: {deficient}, Perfect: {perfect}"); }} public static class ClassifyNumbers{ //Fastest way public static void UsingSieve(int bound, out int abundant, out int deficient, out int perfect) { int a = 0, d = 0, p = 0; //For very large bounds, this array can get big. int[] sum = new int[bound + 1]; for (int divisor = 1; divisor <= bound / 2; divisor++) { for (int i = divisor + divisor; i <= bound; i += divisor) { sum[i] += divisor; } } for (int i = 1; i <= bound; i++) { if (sum[i] < i) d++; else if (sum[i] > i) a++; else p++; } abundant = a; deficient = d; perfect = p; } //Much slower, but doesn't use storage public static void UsingDivision(int bound, out int abundant, out int deficient, out int perfect) { int a = 0, d = 0, p = 0; for (int i = 1; i < 20001; i++) { int sum = Enumerable.Range(1, (i + 1) / 2) .Where(div => div != i && i % div == 0).Sum(); if (sum < i) d++; else if (sum > i) a++; else p++; } abundant = a; deficient = d; perfect = p; }} Output: Abundant: 4953, Deficient: 15043, Perfect: 4 Abundant: 4953, Deficient: 15043, Perfect: 4  ## C++ #include <iostream>#include <algorithm>#include <vector> std::vector<int> findProperDivisors ( int n ) { std::vector<int> divisors ; for ( int i = 1 ; i < n / 2 + 1 ; i++ ) { if ( n % i == 0 ) divisors.push_back( i ) ; } return divisors ;} int main( ) { std::vector<int> deficients , perfects , abundants , divisors ; for ( int n = 1 ; n < 20001 ; n++ ) { divisors = findProperDivisors( n ) ; int sum = std::accumulate( divisors.begin( ) , divisors.end( ) , 0 ) ; if ( sum < n ) { deficients.push_back( n ) ; } if ( sum == n ) { perfects.push_back( n ) ; } if ( sum > n ) { abundants.push_back( n ) ; } } std::cout << "Deficient : " << deficients.size( ) << std::endl ; std::cout << "Perfect : " << perfects.size( ) << std::endl ; std::cout << "Abundant : " << abundants.size( ) << std::endl ; return 0 ;} Output: Deficient : 15043 Perfect : 4 Abundant : 4953  ## Ceylon shared void run() { function divisors(Integer int) => if(int <= 1) then {} else (1..int / 2).filter((Integer element) => element.divides(int)); function classify(Integer int) => sum {0, *divisors(int)} <=> int; value counts = (1..20k).map(classify).frequencies(); print("deficient: counts[smaller] else "none""); print("perfect: counts[equal] else "none""); print("abundant: counts[larger] else "none"");} Output: deficient: 15043 perfect: 4 abundant: 4953 ## Clojure (defn pad-class [n] (let [divs (filter #(zero? (mod n %)) (range 1 n)) divs-sum (reduce + divs)] (cond (< divs-sum n) :deficient (= divs-sum n) :perfect (> divs-sum n) :abundant))) (def pad-classes (map pad-class (map inc (range)))) (defn count-classes [n] (let [classes (take n pad-classes)] {:perfect (count (filter #(= % :perfect) classes)) :abundant (count (filter #(= % :abundant) classes)) :deficient (count (filter #(= % :deficient) classes))})) Example: (count-classes 20000);=> {:perfect 4,; :abundant 4953,; :deficient 15043} ## Common Lisp (defun number-class (n) (let ((divisor-sum (sum-divisors n))) (cond ((< divisor-sum n) :deficient) ((= divisor-sum n) :perfect) ((> divisor-sum n) :abundant)))) (defun sum-divisors (n) (loop :for i :from 1 :to (/ n 2) :when (zerop (mod n i)) :sum i)) (defun classification () (loop :for n :from 1 :to 20000 :for class := (number-class n) :count (eq class :deficient) :into deficient :count (eq class :perfect) :into perfect :count (eq class :abundant) :into abundant :finally (return (values deficient perfect abundant)))) Output: CL-USER> (classification) 15043 4 4953 ## D void main() /*@safe*/ { import std.stdio, std.algorithm, std.range; static immutable properDivs = (in uint n) pure nothrow @safe /*@nogc*/ => iota(1, (n + 1) / 2 + 1).filter!(x => n % x == 0 && n != x); enum Class { deficient, perfect, abundant } static Class classify(in uint n) pure nothrow @safe /*@nogc*/ { immutable p = properDivs(n).sum; with (Class) return (p < n) ? deficient : ((p == n) ? perfect : abundant); } enum rangeMax = 20_000; //iota(1, 1 + rangeMax).map!classify.hashGroup.writeln; iota(1, 1 + rangeMax).map!classify.array.sort().group.writeln;} Output: [Tuple!(Class, uint)(deficient, 15043), Tuple!(Class, uint)(perfect, 4), Tuple!(Class, uint)(abundant, 4953)] ## EchoLisp  (lib 'math) ;; sum-divisors function (define-syntax-rule (++ a) (set! a (1+ a))) (define (abondance (N 20000)) (define-values (delta abondant deficient perfect) '(0 0 0 0)) (for ((n (in-range 1 (1+ N)))) (set! delta (- (sum-divisors n) n)) (cond ((< delta 0) (++ deficient)) ((> delta 0) (++ abondant)) (else (writeln 'perfect→ n) (++ perfect)))) (printf "In range 1.. %d" N) (for-each (lambda(x) (writeln x (eval x))) '(abondant deficient perfect))) (abondance) perfect→ 6 perfect→ 28 perfect→ 496 perfect→ 8128 In range 1.. 20000 abondant 4953 deficient 15043 perfect 4  ## Ela Translation of: Haskell open monad io number list divisors n = filter ((0 ==) << (n mod)) [1 .. (n div 2)] classOf n = compare (sum$ divisors n) n do  let classes = map classOf [1 .. 20000]  let printRes w c = putStrLn $w ++ (show << length$ filter (== c) classes)  printRes "deficient: " LT  printRes "perfect:   " EQ  printRes "abundant:  " GT
Output:
deficient: 15043
perfect:   4
abundant:  4953

## Elena

Translation of: C#

ELENA 4.x :

import extensions; classifyNumbers(int bound, ref int abundant, ref int deficient, ref int perfect){    int a := 0;    int d := 0;    int p := 0;    int[] sum := new int[](bound + 1);     for(int divisor := 1, divisor <= bound / 2, divisor += 1)    {        for(int i := divisor + divisor, i <= bound, i += divisor)        {            sum[i] := sum[i] + divisor        }    };     for(int i := 1, i <= bound, i += 1)    {        int t := sum[i];         if (sum[i]<i)        {            d += 1        }        else        {            if (sum[i]>i)            {                a += 1            }            else            {                p += 1            }        }    };     abundant := a;    deficient := d;    perfect := p} public program(){    int abundant := 0;    int deficient := 0;    int perfect := 0;    classifyNumbers(20000, ref abundant, ref deficient, ref perfect);    console.printLine("Abundant: ",abundant,", Deficient: ",deficient,", Perfect: ",perfect)}
Output:
Abundant: 4953, Deficient: 15043, Perfect: 4


## 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)]end {abundant, deficient, perfect} = Enum.reduce(1..20000, {0,0,0}, fn n,{a, d, p} ->  sum = Proper.divisors(n) |> Enum.sum  cond do    n < sum -> {a+1, d, p}    n > sum -> {a, d+1, p}    true    -> {a, d, p+1}  endend)IO.puts "Deficient: #{deficient}   Perfect: #{perfect}   Abundant: #{abundant}"
Output:
Deficient: 15043   Perfect: 4   Abundant: 4953


## Erlang

 -module(properdivs).-export([divs/1,sumdivs/1,class/1]). divs(0) -> [];divs(1) -> [];divs(N) -> lists:sort(divisors(1,N)). divisors(1,N) ->      [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,math:sqrt(N));divisors(K,N,_Q) when K * K  == N ->     [K] ++ divisors(K+1,N,math:sqrt(N));divisors(K,N,_Q) ->    [K, N div K] ++ divisors(K+1,N,math:sqrt(N)). sumdivs(N) -> lists:sum(divs(N)). class(Limit) -> class(0,0,0,sumdivs(2),2,Limit). class(D,P,A,_Sum,Acc,L) when Acc > L +1->     io:format("Deficient: ~w, Perfect: ~w, Abundant: ~w~n", [D,P,A]); class(D,P,A,Sum,Acc,L) when Acc < Sum ->                        class(D,P,A+1,sumdivs(Acc+1),Acc+1,L);      class(D,P,A,Sum,Acc,L) when Acc == Sum ->                       class(D,P+1,A,sumdivs(Acc+1),Acc+1,L);      class(D,P,A,Sum,Acc,L) when Acc > Sum  ->                       class(D+1,P,A,sumdivs(Acc+1),Acc+1,L).       
Output:
24> c(properdivs).
{ok,properdivs}
25> properdivs:class(20000).
Deficient: 15043, Perfect: 4, Abundant: 4953
ok


The version above is not tail-call recursive, and so cannot classify large ranges. Here is a more optimal solution.

 -module(proper_divisors).-export([classify_range/2]). classify_range(Start, Stop) ->    lists:foldl(fun (X, A) ->                  Class = classify(X),                  A#{Class => maps:get(Class, A, 0)+1} end,                #{},                lists:seq(Start, Stop)). classify(N) ->    SumPD = lists:sum(proper_divisors(N)),    if        SumPD  <  N -> deficient;        SumPD =:= N -> perfect;        SumPD  >  N -> abundant    end. proper_divisors(1) -> [];proper_divisors(N) when N > 1, is_integer(N) ->    proper_divisors(2, math:sqrt(N), N, [1]). proper_divisors(I, L, _, A) when I > L -> lists:sort(A);proper_divisors(I, L, N, A) when N rem I =/= 0 ->    proper_divisors(I+1, L, N, A);proper_divisors(I, L, N, A) when I * I =:= N ->    proper_divisors(I+1, L, N, [I|A]);proper_divisors(I, L, N, A) ->    proper_divisors(I+1, L, N, [N div I, I|A]). 

## F#

 let mutable a=0 let mutable b=0let mutable c=0let mutable d=0let mutable e=0let mutable f=0for i=1 to 20000 do    b <- 0    f <- i/2        for j=1 to f do        if i%j=0 then           b <- b+i    if b<i then       c <- c+1    if b=i then       d <- d+1    if b>i then       e <- e+1printfn " deficient %i"cprintfn "perfect %i"dprintfn "abundant %i"e 

An immutable solution.

 let deficient, perfect, abundant = 0,1,2 let classify n = ([1..n/2] |> List.filter (fun x->n % x = 0) |> List.sum) |> function  | x when x<n -> deficient | x when x>n -> abundant | _ -> perfect let incClass xs n =  let cn = n |> classify  xs |> List.mapi (fun i x->if i=cn then x + 1 else x) [1..20000]|> List.fold incClass [0;0;0]|> List.zip [ "deficient"; "perfect"; "abundant" ]|> List.iter (fun (label, count) -> printfn "%s: %d" label count) 

## Factor

 USING: fry math.primes.factors math.ranges ;: psum     ( n -- m )   divisors but-last sum ;: pcompare ( n -- <=> ) dup psum swap <=> ;: classify ( -- seq )   20,000 [1,b] [ pcompare ] map ;: pcount   ( <=> -- n ) '[ _ = ] count ;classify [ +lt+ pcount "Deficient: " write . ]         [ +eq+ pcount "Perfect: "   write . ]         [ +gt+ pcount "Abundant: "  write . ] tri 
Output:
Deficient: 15043
Perfect: 4
Abundant: 4953


## Forth

Works with: Gforth version 0.7.3
CREATE A 0 ,: SLOT ( x y -- 0|1|2)  OVER OVER < -ROT > -  1+ ;: CLASSIFY ( n -- n')  \ 0 == deficient, 1 == perfect, 2 == abundant   DUP A !  \ we'll be accessing this often, so save somewhere convenient   2 / >R   \ upper bound   1        \ starting sum, 1 is always a divisor   2        \ current check   BEGIN DUP [email protected] < WHILE     A @ OVER /MOD SWAP ( s c d m)     IF DROP ELSE       R> DROP DUP >R  ( R: d n)       OVER TUCK OVER <> * -  ( s c c+?d)       ROT + SWAP ( s' c)     THEN 1+   REPEAT  DROP R> DROP A @  ( sum n)  SLOT ; CREATE COUNTS 0 , 0 , 0 ,: INIT   COUNTS 3 CELLS ERASE  1 COUNTS ! ;: CLASSIFY-NUMBERS ( n --)  INIT   BEGIN DUP WHILE      1 OVER CLASSIFY  CELLS COUNTS + +!  1-   REPEAT  DROP ;: .COUNTS   ." Deficient : " [ COUNTS ]L           @ . CR   ." Perfect   : " [ COUNTS 1 CELLS + ]L @ . CR   ." Abundant  : " [ COUNTS 2 CELLS + ]L @ . CR ;20000 CLASSIFY-NUMBERS .COUNTS BYE
Output:
Deficient : 15043
Perfect   : 5
Abundant  : 4953

## Fortran

Although Fortran offers an intrinsic function SIGN(a,b) which returns the absolute value of a with the sign of b, it does not recognise zero as a special case, instead distinguishing only the two conditions b < 0 and b >= 0. Rather than a mess such as SIGN(a*b,b), a suitable SIGN3 function is needed. For it to be acceptable in whole-array expressions, it must have the PURE attribute asserted (signifying that it it may be treated as having a value dependent only on its explicit parameters) and further, that parameters must be declared with the (verbose) new protocol that enables the use of INTENT(IN) as further assurance to the compiler. Finally, such a function must be associated with INTERFACE arrangements, easily done here merely by placing it within a MODULE.

Alternatively, an explicit DO-loop could simply inspect the KnownSum array and maintain three counts, moreover, doing so in a single pass rather than the three passes needed for the three COUNT statements.

Output:

Inspecting sums of proper divisors for 1 to       20000
Deficient       15043
Perfect!            4
Abundant         4953

       MODULE FACTORSTUFF	!This protocol evades the need for multiple parameters, or COMMON, or one shapeless main line...Concocted by R.N.McLean, MMXV.       INTEGER LOTS		!The span..       PARAMETER (LOTS = 20000)!Nor is computer storage infinite.       INTEGER KNOWNSUM(LOTS)	!Calculate these once.       CONTAINS		!Assistants.        SUBROUTINE PREPARESUMF	!Initialise the KNOWNSUM array.Convert the Sieve of Eratoshenes to have each slot contain the sum of the proper divisors of its slot number.Changes to instead count the number of factors, or prime factors, etc. would be simple enough.         INTEGER F		!A factor for numbers such as 2F, 3F, 4F, 5F, ...          KNOWNSUM(1) = 0		!Proper divisors of N do not include N.          KNOWNSUM(2:LOTS) = 1		!So, although 1 divides all N without remainder, 1 is excluded for itself.          DO F = 2,LOTS/2		!Step through all the possible divisors of numbers not exceeding LOTS.            FORALL(I = F + F:LOTS:F) KNOWNSUM(I) = KNOWNSUM(I) + F	!And augment each corresponding slot.          END DO			!Different divisors can hit the same slot. For instance, 6 by 2 and also by 3.        END SUBROUTINE PREPARESUMF	!Could alternatively generate all products of prime numbers.           PURE INTEGER FUNCTION SIGN3(N)	!Returns -1, 0, +1 according to the sign of N.Confounded by the intrinsic function SIGN distinguishing only two states: < 0 from >= 0. NOT three-way.         INTEGER, INTENT(IN):: N	!The number.          IF (N) 1,2,3	!A three-way result calls for a three-way test.    1     SIGN3 = -1	!Negative.          RETURN    2     SIGN3 = 0	!Zero.          RETURN    3     SIGN3 = +1	!Positive.        END FUNCTION SIGN3	!Rather basic.      END MODULE FACTORSTUFF	!Enough assistants.        PROGRAM THREEWAYS	!Classify N against the sum of proper divisors of N, for N up to 20,000.       USE FACTORSTUFF		!This should help.       INTEGER I		!Stepper.       INTEGER TEST(LOTS)	!Assesses the three states in one pass.        WRITE (6,*) "Inspecting sums of proper divisors for 1 to",LOTS        CALL PREPARESUMF		!Values for every N up to the search limit will be called for at least once.        FORALL(I = 1:LOTS) TEST(I) = SIGN3(KNOWNSUM(I) - I)	!How does KnownSum(i) compare to i?        WRITE (6,*) "Deficient",COUNT(TEST .LT. 0)	!This means one pass through the array        WRITE (6,*) "Perfect! ",COUNT(TEST .EQ. 0)	!For each of three types.        WRITE (6,*) "Abundant ",COUNT(TEST .GT. 0)	!Alternatively, make one pass with three counts.      END			!Done. 

## FreeBASIC

 ' FreeBASIC v1.05.0 win64 Function SumProperDivisors(number As Integer) As Integer  If number < 2 Then Return 0  Dim sum As Integer = 0  For i As Integer = 1 To number \ 2    If number Mod i = 0 Then sum += i  Next  Return sumEnd Function Dim As Integer sum, deficient, perfect, abundant For n As Integer = 1 To 20000  sum = SumProperDivisors(n)  If sum < n Then    deficient += 1  ElseIf sum = n Then    perfect += 1  Else    abundant += 1  EndIfNext Print "The classification of the numbers from 1 to 20,000 is as follows : "PrintPrint "Deficient = "; deficientPrint "Perfect   = "; perfectPrint "Abundant  = "; abundantPrintPrint "Press any key to exit the program"SleepEnd 
Output:
The classification of the numbers from 1 to 20,000 is as follows :

Deficient =  15043
Perfect   =  4
Abundant  =  4953


## GFA Basic

 num_deficient%=0num_perfect%=0num_abundant%=0'FOR current%=1 TO 20000  sum_divisors%[email protected]_proper_divisors(current%)  IF sum_divisors%<current%    num_deficient%=num_deficient%+1  ELSE IF sum_divisors%=current%    num_perfect%=num_perfect%+1  ELSE ! sum_divisors%>current%    num_abundant%=num_abundant%+1  ENDIFNEXT current%'' Display results on a window'OPENW 1CLEARW 1PRINT "Number deficient ";num_deficient%PRINT "Number perfect   ";num_perfect%PRINT "Number abundant  ";num_abundant%~INP(2)CLOSEW 1'' Compute the sum of proper divisors of given number'FUNCTION sum_proper_divisors(n%)  LOCAL i%,sum%,root%  '  IF n%>1 ! n% must be 2 or higher    sum%=1 ! start with 1    root%=SQR(n%) ! note that root% is an integer    ' check possible factors, up to sqrt    FOR i%=2 TO root%      IF n% MOD i%=0        sum%=sum%+i% ! i% is a factor        IF i%*i%<>n% ! check i% is not actual square root of n%          sum%=sum%+n%/i% ! so n%/i% will also be a factor        ENDIF      ENDIF    NEXT i%  ENDIF  RETURN sum%ENDFUNC 

Output is:

Number deficient 15043
Number perfect   4
Number abundant  4953


## Go

package main import "fmt" func pfacSum(i int) int {    sum := 0    for p := 1; p <= i/2; p++ {        if i%p == 0 {            sum += p        }    }    return sum} func main() {    var d, a, p = 0, 0, 0    for i := 1; i <= 20000; i++ {        j := pfacSum(i)        if j < i {            d++        } else if j == i {            p++        } else {            a++        }    }    fmt.Printf("There are %d deficient numbers between 1 and 20000\n", d)    fmt.Printf("There are %d abundant numbers  between 1 and 20000\n", a)    fmt.Printf("There are %d perfect numbers between 1 and 20000\n", p)}
Output:
There are 15043 deficient numbers between 1 and 20000
There are 4953 abundant numbers  between 1 and 20000
There are 4 perfect numbers between 1 and 20000


## Groovy

##### Solution:

Uses the "factorize" closure from Factors of an integer

def dpaCalc = { factors ->    def n = factors.pop()    def fSum = factors.sum()    fSum < n        ? 'deficient'        : fSum > n            ? 'abundant'            : 'perfect'} (1..20000).inject([deficient:0, perfect:0, abundant:0]) { map, n ->    map[dpaCalc(factorize(n))]++    map}.each { e -> println e }
Output:
deficient=15043
perfect=4
abundant=4953

divisors :: (Integral a) => a -> [a]divisors n = filter ((0 ==) . (n mod)) [1 .. (n div 2)] classOf :: (Integral a) => a -> OrderingclassOf n = compare (sum $divisors n) n main :: IO ()main = do let classes = map classOf [1 .. 20000 :: Int] printRes w c = putStrLn$ w ++ (show . length $filter (== c) classes) printRes "deficient: " LT printRes "perfect: " EQ printRes "abundant: " GT Output: deficient: 15043 perfect: 4 abundant: 4953 ## J factors=: [: /:[email protected], */&>@{@((^ [email protected]>:)&.>/)@q:~&__properDivisors=: factors -. ] We can subtract the sum of a number's proper divisors from itself to classify the number:  (- +/@properDivisors&>) 1+i.101 1 2 1 4 0 6 1 5 2 Except, we are only concerned with the sign of this difference:  *(- +/@properDivisors&>) 1+i.301 1 1 1 1 0 1 1 1 1 1 _1 1 1 1 1 1 _1 1 _1 1 1 1 _1 1 1 1 0 1 _1 Also, we do not care about the individual classification but only about how many numbers fall in each category:  #/.~ *(- +/@properDivisors&>) 1+i.2000015043 4 4953 So: 15043 deficient, 4 perfect and 4953 abundant numbers in this range. How do we know which is which? We look at the unique values (which are arranged by their first appearance, scanning the list left to right):  ~. *(- +/@properDivisors&>) 1+i.200001 0 _1 The sign of the difference is negative for the abundant case - where the sum is greater than the number. And we rely on order being preserved in sequences (this happens to be a fundamental property of computer memory, also). ## Java Works with: Java version 8 import java.util.stream.LongStream; public class NumberClassifications { public static void main(String[] args) { int deficient = 0; int perfect = 0; int abundant = 0; for (long i = 1; i <= 20_000; i++) { long sum = properDivsSum(i); if (sum < i) deficient++; else if (sum == i) perfect++; else abundant++; } System.out.println("Deficient: " + deficient); System.out.println("Perfect: " + perfect); System.out.println("Abundant: " + abundant); } public static long properDivsSum(long n) { return LongStream.rangeClosed(1, (n + 1) / 2).filter(i -> n != i && n % i == 0).sum(); }} Deficient: 15043 Perfect: 4 Abundant: 4953 ## JavaScript ### ES5 for (var dpa=[1,0,0], n=2; n<=20000; n+=1) { for (var ds=0, d=1, e=n/2+1; d<e; d+=1) if (n%d==0) ds+=d dpa[ds<n ? 0 : ds==n ? 1 : 2]+=1}document.write('Deficient:',dpa[0], ', Perfect:',dpa[1], ', Abundant:',dpa[2], '<br>' ) Or: for (var dpa=[1,0,0], n=2; n<=20000; n+=1) { for (var ds=1, d=2, e=Math.sqrt(n); d<e; d+=1) if (n%d==0) ds+=d+n/d if (n%e==0) ds+=e dpa[ds<n ? 0 : ds==n ? 1 : 2]+=1}document.write('Deficient:',dpa[0], ', Perfect:',dpa[1], ', Abundant:',dpa[2], '<br>' ) Or: function primes(t) { var ps = {2:true, 3:true} next: for (var n=5, i=2; n<=t; n+=i, i=6-i) { var s = Math.sqrt( n ) for ( var p in ps ) { if ( p > s ) break if ( n % p ) continue continue next } ps[n] = true } return ps} function factorize(f, t) { var cs = {}, ps = primes(t) for (var n=f; n<=t; n++) if (!ps[n]) cs[n] = factors(n) return cs function factors(n) { for ( var p in ps ) if ( n % p == 0 ) break var ts = {} ts[p] = 1 if ( ps[n /= p] ) { if ( !ts[n]++ ) ts[n]=1 } else { var fs = cs[n] if ( !fs ) fs = cs[n] = factors(n) for ( var e in fs ) ts[e] = fs[e] + (e==p) } return ts }} function pContrib(p, e) { for (var pc=1, n=1, i=1; i<=e; i+=1) pc+=n*=p; return pc} for (var dpa=[1,0,0], t=20000, cs=factorize(2,t), n=2; n<=t; n+=1) { var ds=1, fs=cs[n] if (fs) { for (var p in fs) ds *= pContrib(p, fs[p]) ds -= n } dpa[ds<n ? 0 : ds==n ? 1 : 2]+=1}document.write('Deficient:',dpa[0], ', Perfect:',dpa[1], ', Abundant:',dpa[2], '<br>' ) Output: Deficient:15043, Perfect:4, Abundant:4953 ### ES6 Translation of: Haskell (() => { 'use strict'; const // divisors :: (Integral a) => a -> [a] divisors = n => range(1, Math.floor(n / 2)) .filter(x => n % x === 0), // classOf :: (Integral a) => a -> Ordering classOf = n => compare(divisors(n) .reduce((a, b) => a + b, 0), n), classTypes = { deficient: -1, perfect: 0, abundant: 1 }; // GENERIC FUNCTIONS const // compare :: Ord a => a -> a -> Ordering compare = (a, b) => a < b ? -1 : (a > b ? 1 : 0), // range :: Int -> Int -> [Int] range = (m, n) => Array.from({ length: Math.floor(n - m) + 1 }, (_, i) => m + i); // TEST // classes :: [Ordering] const classes = range(1, 20000) .map(classOf); return Object.keys(classTypes) .map(k => k + ": " + classes .filter(x => x === classTypes[k]) .length.toString()) .join('\n');})(); Output: deficient: 15043 perfect: 4 abundant: 4953 ## Jsish From Javascript ES5 entry. /* Classify Deficient, Perfect and Abdundant integers */function classifyDPA(stop:number, start:number=0, step:number=1):array { var dpa = [1, 0, 0]; for (var n=start; n<=stop; n+=step) { for (var ds=0, d=1, e=n/2+1; d<e; d+=1) if (n%d == 0) ds += d; dpa[ds < n ? 0 : ds==n ? 1 : 2] += 1; } return dpa;} var dpa = classifyDPA(20000, 2);printf('Deficient: %d, Perfect: %d, Abundant: %d\n', dpa[0], dpa[1], dpa[2]); Output: prompt$ jsish classifyDPA.jsi
Deficient: 15043, Perfect: 4, Abundant: 4953

## Julia

This post was created with Julia version 0.3.6. The code uses no exotic features and should work for a wide range of Julia versions.

The Math

A natural number can be written as a product of powers of its prime factors, ${\displaystyle \prod _{i}p_{i}^{a_{i}}}$. Handily Julia has the factor function, which provides these parameters. The sum of n's divisors (n inclusive) is ${\displaystyle \prod _{i}{\frac {p_{i}^{a_{i}+1}-1}{p_{i}-1}}=\prod _{i}p_{i}^{a_{i}}+p_{i}^{a_{i}-1}+\cdots +p_{i}+1}$.

Functions

divisorsum calculates the sum of aliquot divisors. It uses pcontrib to calculate the contribution of each prime factor.

 function pcontrib(p::Int64, a::Int64)    n = one(p)    pcon = one(p)    for i in 1:a        n *= p        pcon += n    end    return pconend function divisorsum(n::Int64)    dsum = one(n)    for (p, a) in factor(n)        dsum *= pcontrib(p, a)    end    dsum -= nend 

Perhaps pcontrib could be made more efficient by caching results to avoid repeated calculations.

Main

Use a three element array, iclass, rather than three separate variables to tally the classifications. Take advantage of the fact that the sign of divisorsum(n) - n depends upon its class to increment iclass. 1 is a difficult case, it is deficient by convention, so I manually add its contribution and start the accumulation with 2. All primes are deficient, so I test for those and tally accordingly, bypassing divisorsum.

 const L = 2*10^4iclasslabel = ["Deficient", "Perfect", "Abundant"]iclass = zeros(Int64, 3)iclass[1] = one(Int64) #by convention 1 is deficient for n in 2:L    if isprime(n)        iclass[1] += 1    else        iclass[sign(divisorsum(n)-n)+2] += 1    endend println("Classification of integers from 1 to ", L)for i in 1:3    println("   ", iclasslabel[i], ", ", iclass[i])end 
Output:

 

 Classification of integers from 1 to 20000 Deficient, 15043 Perfect, 4 Abundant, 4953 

## jq

Works with: jq version 1.4

The definition of proper_divisors is taken from Proper_divisors#jq:

# unordereddef proper_divisors:  . as $n | if$n > 1 then 1,      ( range(2; 1 + (sqrt|floor)) as $i | if ($n % $i) == 0 then$i,            (($n /$i) | if . == $i then empty else . end) else empty end) else empty end; The task: def sum(stream): reduce stream as$i (0; . + $i); def classify: . as$n  | sum(proper_divisors)  | if . < $n then "deficient" elif . ==$n then "perfect" else "abundant" end; reduce (range(1; 20001) | classify) as $c ({}; .[$c] += 1 )
Output:
$jq -n -c -f AbundantDeficientPerfect.jq{"deficient":15043,"perfect":4,"abundant":4953} ## Kotlin Translation of: FreeBASIC // version 1.1 fun sumProperDivisors(n: Int) = if (n < 2) 0 else (1..n / 2).filter { (n % it) == 0 }.sum() fun main(args: Array<String>) { var sum: Int var deficient = 0 var perfect = 0 var abundant = 0 for (n in 1..20000) { sum = sumProperDivisors(n) when { sum < n -> deficient++ sum == n -> perfect++ sum > n -> abundant++ } } println("The classification of the numbers from 1 to 20,000 is as follows:\n") println("Deficient =$deficient")    println("Perfect   = $perfect") println("Abundant =$abundant")}
Output:
The classification of the numbers from 1 to 20,000 is as follows:

Deficient = 15043
Perfect   = 4
Abundant  = 4953


 /Classification of numbers into abundant, perfect and deficient/ numclass.k /return 0,1 or -1 if perfect or abundant or deficient respectivelynumclass: {s:(+/&~x!'!1+x)-x; :[s>x;:1;:[s<x;:-1;:0]]}/classify numbers from 1 to 20000 into respective groupsc: =numclass' 1+!20000/print statistics0: ,"Deficient = ", $(#c[0])0: ,"Perfect = ",$(#c[1])0: ,"Abundant  = ", $(#c[2])  Output: Deficient = 15043 Perfect = 4 Abundant = 4953  ## Liberty BASIC  print "ROSETTA CODE - Abundant, deficient and perfect number classifications"printfor x=1 to 20000 x$=NumberClassification$(x) select case x$        case "deficient": de=de+1        case "perfect": pe=pe+1: print x; " is a perfect number"        case "abundant": ab=ab+1    end select    select case x        case 2000: print "Checking the number classifications of 20,000 integers..."        case 4000: print "Please be patient."        case 7000: print "7,000"        case 10000: print "10,000"        case 12000: print "12,000"        case 14000: print "14,000"        case 16000: print "16,000"        case 18000: print "18,000"        case 19000: print "Almost done..."    end selectnext xprint "Deficient numbers = "; deprint "Perfect numbers = "; peprint "Abundant numbers = "; abprint "TOTAL = "; pe+de+ab[Quit]print "Program complete."end function NumberClassification$(n) x=ProperDivisorCount(n) for y=1 to x PDtotal=PDtotal+ProperDivisor(y) next y if PDtotal=n then NumberClassification$="perfect": exit function    if PDtotal<n then NumberClassification$="deficient": exit function if PDtotal>n then NumberClassification$="abundant": exit functionend function function ProperDivisorCount(n)    n=abs(int(n)): if n=0 or n>20000 then exit function    dim ProperDivisor(100)    for y=2 to n        if (n mod y)=0 then            ProperDivisorCount=ProperDivisorCount+1            ProperDivisor(ProperDivisorCount)=n/y        end if    next yend function
Output:
ROSETTA CODE - Abundant, deficient and perfect number classifications

6 is a perfect number
28 is a perfect number
496 is a perfect number
Checking the number classifications of 20,000 integers...
7,000
8128 is a perfect number
10,000
12,000
14,000
16,000
18,000
Almost done...
Deficient numbers = 15043
Perfect numbers = 4
Abundant numbers = 4953
TOTAL = 20000
Program complete.


## Lua

function sumDivs (n)    if n < 2 then return 0 end    local sum, sr = 1, math.sqrt(n)    for d = 2, sr do        if n % d == 0 then            sum = sum + d            if d ~= sr then sum = sum + n / d end        end    end    return sumend local a, d, p, Pn = 0, 0, 0for n = 1, 20000 do    Pn = sumDivs(n)    if Pn > n then a = a + 1 end    if Pn < n then d = d + 1 end    if Pn == n then p = p + 1 endendprint("Abundant:", a)print("Deficient:", d)print("Perfect:", p)
Output:
Abundant:       4953
Deficient:      15043
Perfect:        4

## Mathematica / Wolfram Language

classify[n_Integer] := Sign[Total[[email protected]@n] - n] StringJoin[ Flatten[Tally[     Table[classify[n], {n, 20000}]] /. {-1 -> "deficient: ",      0 -> "  perfect: ", 1 -> "  abundant: "}] /.   n_Integer :> ToString[n]]
Output:
deficient: 15043  perfect: 4  abundant: 4953

## ML

### mLite

fun proper		(number, count, limit, remainder, results) where (count > limit) = rev results	|	(number, count, limit, remainder, results) = 			proper (number, count + 1, limit, number rem (count+1), if remainder = 0 then 				count :: results			else 				results)	|	number = (proper (number, 1, number div 2, 0, [])); fun is_abundant  number = number < (fold (op +, 0)  proper number);fun is_deficient number = number > (fold (op +, 0)  proper number);fun is_perfect   number = number = (fold (op +, 0)  proper number); val one_to_20000 = iota 20000; print "Abundant numbers between 1 and 20000: ";println  fold (op +, 0)  map ((fn n = if n then 1 else 0) o is_abundant) one_to_20000; print "Deficient numbers between 1 and 20000: ";println  fold (op +, 0)  map ((fn n = if n then 1 else 0) o is_deficient) one_to_20000; print "Perfect numbers between 1 and 20000: ";println  fold (op +, 0)  map ((fn n = if n then 1 else 0) o is_perfect) one_to_20000; 

Output

Abundant numbers between 1 and 20000: 4953
Deficient numbers between 1 and 20000: 15043
Perfect numbers between 1 and 20000: 4


## Modula-2

MODULE ADP;FROM FormatString IMPORT FormatString;FROM Terminal IMPORT WriteString,WriteLn,ReadChar; PROCEDURE ProperDivisorSum(n : INTEGER) : INTEGER;VAR i,sum : INTEGER;BEGIN    sum := 0;    IF n<2 THEN        RETURN 0    END;    FOR i:=1 TO (n DIV 2) DO        IF n MOD i = 0 THEN            INC(sum,i)        END    END;    RETURN sumEND ProperDivisorSum; VAR    buf : ARRAY[0..63] OF CHAR;    n : INTEGER;    d,p,a : INTEGER = 0;    sum : INTEGER;BEGIN    FOR n:=1 TO 20000 DO        sum := ProperDivisorSum(n);        IF sum<n THEN            INC(d)        ELSIF sum=n THEN            INC(p)        ELSIF sum>n THEN            INC(a)        END    END;     WriteString("The classification of the numbers from 1 to 20,000 is as follows:");    WriteLn;     FormatString("Deficient = %i\n", buf, d);    WriteString(buf);    FormatString("Perfect = %i\n", buf, p);    WriteString(buf);    FormatString("Abundant = %i\n", buf, a);    WriteString(buf);    ReadCharEND ADP.

## Nim

 proc sumProperDivisors(number: int) : int =  if number < 2 : return 0  for i in 1 .. number div 2 :    if number mod i == 0 : result += i var   sum : int  deficient = 0  perfect = 0  abundant = 0 for n in 1 .. 20000 :  sum = sumProperDivisors(n)  if sum < n :    inc(deficient)  elif sum == n :    inc(perfect)  else :     inc(abundant) echo "The classification of the numbers between 1 and 20,000 is as follows :\n"echo "  Deficient = " , deficientecho "  Perfect   = " , perfectecho "  Abundant  = " , abundant  
Output:
The classification of the numbers between 1 and 20,000 is as follows :

Deficient = 15043
Perfect   = 4
Abundant  = 4953


## Oforth

import: mapping Integer method: properDivs -- []    self 2 / seq  filter( #[ self swap mod 0 == ] ) ; : numberClasses| i deficient perfect s |   0 0 ->deficient ->perfect    0 20000 loop: i [      0 #+ i properDivs apply ->s      s i <  ifTrue: [ deficient 1+ ->deficient continue ]      s i == ifTrue: [ perfect 1+ ->perfect continue ]      1+      ]   "Deficients :" . deficient .cr   "Perfects   :" . perfect   .cr   "Abundant   :" . .cr ; 
Output:
numberClasses
Deficients : 15043
Perfects   : 4
Abundant   : 4953


## PARI/GP

classify(k)={  my(v=[0,0,0],t);  for(n=1,k,    t=sigma(n,-1);    if(t<2,v[1]++,t>2,v[3]++,v[2]++)  );  v;}classify(20000)
Output:
%1 = [15043, 4, 4953]

## Pascal

using the slightly modified http://rosettacode.org/wiki/Amicable_pairs#Alternative

program AmicablePairs;{find amicable pairs in a limited region 2..MAXbeware that >both< numbers must be smaller than MAXthere are 455 amicable pairs up to 524*1000*1000correct up to#437 460122410}//optimized for freepascal 2.6.4 32-Bit{$IFDEF FPC} {$MODE DELPHI}   {$OPTIMIZATION ON,peephole,cse,asmcse,regvar} {$CODEALIGN loop=1,proc=8}{$ELSE} {$APPTYPE CONSOLE}{$ENDIF} uses sysutils;const MAX = 20000;//{$IFDEF UNIX} MAX = 524*1000*1000;{$ELSE}MAX = 499*1000*1000;{$ENDIF}type  tValue = LongWord;  tpValue = ^tValue;  tPower = array[0..31] of tValue;  tIndex = record             idxI,             idxS : tValue;           end;  tdpa   = array[0..2] of LongWord;var  power        : tPower;  PowerFac     : tPower;  DivSumField  : array[0..MAX] of tValue;  Indices      : array[0..511] of tIndex;  DpaCnt       : tdpa; procedure Init;var  i : LongInt;begin  DivSumField[0]:= 0;  For i := 1 to MAX do    DivSumField[i]:= 1;end; procedure ProperDivs(n: tValue);//Only for output, normally a factorication would dovar  su,so : string;  i,q : tValue;begin  su:= '1';  so:= '';  i := 2;  while i*i <= n do  begin    q := n div i;    IF q*i -n = 0 then    begin      su:= su+','+IntToStr(i);      IF q <> i then        so:= ','+IntToStr(q)+so;    end;    inc(i);  end;  writeln('  [',su+so,']');end; procedure AmPairOutput(cnt:tValue);var  i : tValue;  r : double;begin  r := 1.0;  For i := 0 to cnt-1 do  with Indices[i] do  begin    writeln(i+1:4,IdxI:12,IDxS:12,' ratio ',IdxS/IDxI:10:7);    if r < IdxS/IDxI then      r := IdxS/IDxI;      IF cnt < 20 then      begin        ProperDivs(IdxI);        ProperDivs(IdxS);      end;  end;  writeln(' max ratio ',r:10:4);end; function Check:tValue;var  i,s,n : tValue;begin  fillchar(DpaCnt,SizeOf(dpaCnt),#0);  n := 0;  For i := 1 to MAX do  begin    //s = sum of proper divs (I)  == sum of divs (I) - I    s := DivSumField[i]-i;    IF (s <=MAX) AND (s>i) then    begin      IF DivSumField[s]-s = i then      begin        With indices[n] do        begin          idxI := i;          idxS := s;        end;        inc(n);      end;    end;    inc(DpaCnt[Ord(s>=i)-Ord(s<=i)+1]);  end;  result := n;end; Procedure CalcPotfactor(prim:tValue);//PowerFac[k] = (prim^(k+1)-1)/(prim-1) == Sum (i=1..k) prim^ivar  k: tValue;  Pot,       //== prim^k  PFac : Int64;begin  Pot := prim;  PFac := 1;  For k := 0 to High(PowerFac) do  begin    PFac := PFac+Pot;    IF (POT > MAX) then      BREAK;    PowerFac[k] := PFac;    Pot := Pot*prim;  end;end; procedure InitPW(prim:tValue);begin  fillchar(power,SizeOf(power),#0);  CalcPotfactor(prim);end; function NextPotCnt(p: tValue):tValue;inline;//return the first power <> 0//power == n to base primvar  i : tValue;begin  result := 0;  repeat    i := power[result];    Inc(i);    IF i < p then      BREAK    else    begin      i := 0;      power[result]  := 0;      inc(result);    end;  until false;  power[result] := i;end; function Sieve(prim: tValue):tValue;//simple versionvar  actNumber : tValue;begin  while prim <= MAX do  begin    InitPW(prim);    //actNumber = actual number = n*prim    //power == n to base prim    actNumber := prim;    while actNumber < MAX do    begin      DivSumField[actNumber] := DivSumField[actNumber] *PowerFac[NextPotCnt(prim)];      inc(actNumber,prim);    end;    //next prime    repeat      inc(prim);    until (DivSumField[prim] = 1);  end;  result := prim;end; var  T2,T1,T0: TDatetime;  APcnt: tValue; begin  T0:= time;  Init;  Sieve(2);  T1:= time;  APCnt := Check;  T2:= time;   //AmPairOutput(APCnt);  writeln(Max:10,' upper limit');  writeln(DpaCnt[0]:10,' deficient');  writeln(DpaCnt[1]:10,' perfect');  writeln(DpaCnt[2]:10,' abundant');  writeln(DpaCnt[2]/Max:14:10,' ratio abundant/upper Limit ');  writeln(DpaCnt[0]/Max:14:10,' ratio abundant/upper Limit ');  writeln(DpaCnt[2]/DpaCnt[0]:14:10,' ratio abundant/deficient   ');    writeln('Time to calc sum of divs    ',FormatDateTime('HH:NN:SS.ZZZ' ,T1-T0));  writeln('Time to find amicable pairs ',FormatDateTime('HH:NN:SS.ZZZ' ,T2-T1));  {$IFNDEF UNIX} readln; {$ENDIF}end. 

output

     20000 upper limit
15043 deficient
4 perfect
4953 abundant
0.2476500000 ratio abundant/upper Limit
0.7521500000 ratio abundant/upper Limit
0.3292561324 ratio abundant/deficient
Time to calc sum of divs    00:00:00.000
Time to find amicable pairs 00:00:00.000

...
524000000 upper limit
394250308 deficient
5 perfect
129749687 abundant
0.2476139065 ratio abundant/upper Limit
0.7523860840 ratio abundant/upper Limit
0.3291048463 ratio abundant/deficient
Time to calc sum of divs    00:00:12.597
Time to find amicable pairs 00:00:04.064


## Perl

### Using a module

Library: ntheory

Use the <=> operator to return a comparison of -1, 0, or 1, which classifies the results. 1 is classified as a deficient number, 6 is a perfect number, 12 is an abundant number. As per task spec, also showing the totals for the first 20,000 numbers.

use ntheory qw/divisor_sum/;my @type = <Perfect Abundant Deficient>;say join "\n", map { sprintf "%2d %s", $_,$type[divisor_sum($_)-$_ <=> $_] } 1..12;my %h;$h{divisor_sum($_)-$_ <=> $_}++ for 1..20000;say "Perfect:$h{0}    Deficient: $h{-1} Abundant:$h{1}";
Output:
 1 Deficient
2 Deficient
3 Deficient
4 Deficient
5 Deficient
6 Perfect
7 Deficient
8 Deficient
9 Deficient
10 Deficient
11 Deficient
12 Abundant

Perfect: 4    Deficient: 15043    Abundant: 4953

### Not using a module

Everything as above, but done more slowly with div_sum providing sum of proper divisors.

sub div_sum {    my($n) = @_; my$sum = 0;    map { $sum +=$_ unless $n %$_ } 1 .. $n-1;$sum;} my @type = <Perfect Abundant Deficient>;say join "\n", map { sprintf "%2d %s", $_,$type[div_sum($_) <=>$_] } 1..12;my %h;$h{div_sum($_) <=> $_}++ for 1..20000;say "Perfect:$h{0}    Deficient: $h{-1} Abundant:$h{1}";

## Perl 6

Works with: rakudo version 2018.12
sub propdivsum (\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 }    }    sum @l} say bag (1..20000).map: { propdivsum($_) <=>$_ }
Output:
Bag(Less(15043), More(4953), Same(4))

## Phix

I cheated a little and added a new factors() builtin, but it's there for good now.

integer deficient=0, perfect=0, abundant=0, Nfor i=1 to 20000 do    N = sum(factors(i))+(i!=1)    if N=i then        perfect += 1    elsif N<i then        deficient += 1    else        abundant += 1    end ifend forprintf(1,"deficient:%d, perfect:%d, abundant:%d\n",{deficient, perfect, abundant})
Output:
deficient:15043, perfect:4, abundant:4953


## PicoLisp

(de accud (Var Key)   (if (assoc Key (val Var))      (con @ (inc (cdr @)))      (push Var (cons Key 1)) )   Key )(de factor-sum (N)   (if (=1 N)      0      (let         (R NIL            D 2            L (1 2 2 . (4 2 4 2 4 6 2 6 .))            M (sqrt N)            N1 N            S 1 )         (while (>= M D)            (if (=0 (% N1 D))               (setq M                  (sqrt (setq N1 (/ N1 (accud 'R D)))) )               (inc 'D (pop 'L)) ) )         (accud 'R N1)         (for I R            (one D)            (one M)            (for J (cdr I)               (setq M (* M (car I)))               (inc 'D M) )               (setq S (* S D)) )         (- S N) ) ) )(bench   (let      (A 0         D 0         P 0 )      (for I 20000         (setq @@ (factor-sum I))         (cond            ((< @@ I) (inc 'D))            ((= @@ I) (inc 'P))            ((> @@ I) (inc 'A)) ) )      (println D P A) ) )(bye)
Output:
15043 4 4953
0.593 sec


## PL/I

*process source xref; apd: 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); Do x=1 To 20000;   Call proper_divisors(x,pd,npd);   sumpd=sum(pd,npd);   Select;     When(x<sumpd) cnt(1)+=1; /* abundant  */     When(x=sumpd) cnt(2)+=1; /* perfect   */     Otherwise     cnt(3)+=1; /* deficient */     End;   End;  Put Edit('In the range 1 - 20000')(Skip,a); Put Edit(cnt(1),' numbers are abundant ')(Skip,f(5),a); Put Edit(cnt(2),' numbers are perfect  ')(Skip,f(5),a); Put Edit(cnt(3),' numbers are deficient')(Skip,f(5),a); 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;  sum: Proc(pd,npd) Returns(Bin Fixed(31)); Dcl (pd(300),npd) Bin Fixed(31); Dcl sum Bin Fixed(31) Init(0); Dcl i   Bin Fixed(31); Do i=1 To npd;   sum+=pd(i);   End; Return(sum); End;  End;
Output:
In the range 1 - 20000
4953 numbers are abundant
4 numbers are perfect
15043 numbers are deficient
0.560 seconds elapsed


## PowerShell

Works with: PowerShell version 2
 function Get-ProperDivisorSum ( [int]$N ) { If ($N -lt 2 ) { return 0 }     $Sum = 1 If ($N -gt 3 )        {        $SqrtN = [math]::Sqrt($N )        ForEach ( $Divisor in 2..$SqrtN )            {            If ( $N %$Divisor -eq 0 ) { $Sum +=$Divisor + $N /$Divisor }            }        If ( $N %$SqrtN -eq 0 ) { $Sum -=$SqrtN }        }    return $Sum }$Deficient = $Perfect =$Abundant = 0 ForEach ( $N in 1..20000 ) { Switch ( [math]::Sign( ( Get-ProperDivisorSum$N ) - $N ) ) { -1 {$Deficient++ }         0 { $Perfect++ } 1 {$Abundant++  }        }    } "Deficient: $Deficient""Perfect :$Perfect""Abundant : $Abundant"  Output: Deficient: 15043 Perfect : 4 Abundant : 4953  ### As a single function Using the Get-ProperDivisorSum as a helper function in an advanced function:  function Get-NumberClassification{ [CmdletBinding()] [OutputType([PSCustomObject])] Param ( [Parameter(Mandatory=$true,                   ValueFromPipeline=$true, ValueFromPipelineByPropertyName=$true,                   Position=0)]        [int]        $Number ) Begin { function Get-ProperDivisorSum ([int]$Number)        {            if ($Number -lt 2) {return 0}$sum = 1             if ($Number -gt 3) {$sqrtNumber = [Math]::Sqrt($Number) foreach ($divisor in 2..$sqrtNumber) { if ($Number % $divisor -eq 0) {$sum += $divisor +$Number / $divisor} } if ($Number % $sqrtNumber -eq 0) {$sum -= $sqrtNumber} }$sum        }         [System.Collections.ArrayList]$numbers = @() } Process { switch ([Math]::Sign((Get-ProperDivisorSum$Number) - $Number)) { -1 { [void]$numbers.Add([PSCustomObject]@{Class="Deficient"; Number=$Number}) } 0 { [void]$numbers.Add([PSCustomObject]@{Class="Perfect"  ; Number=$Number}) } 1 { [void]$numbers.Add([PSCustomObject]@{Class="Abundant" ; Number=$Number}) } } } End {$numbers | Group-Object  -Property Class |                   Select-Object -Property Count,                                           @{Name='Class' ; Expression={$_.Name}}, @{Name='Number'; Expression={$_.Group.Number}}    }} 
 1..20000 | Get-NumberClassification 
Output:
Count Class     Number
----- -----     ------
15043 Deficient {1, 2, 3, 4...}
4 Perfect   {6, 28, 496, 8128}
4953 Abundant  {12, 18, 20, 24...}


## Prolog

 proper_divisors(1, []) :- !.proper_divisors(N, [1|L]) :-	FSQRTN is floor(sqrt(N)),	proper_divisors(2, FSQRTN, N, L). proper_divisors(M, FSQRTN, _, []) :-	M > FSQRTN,	!.proper_divisors(M, FSQRTN, N, L) :-	N mod M =:= 0, !,	MO is N//M, % must be integer	L = [M,MO|L1], % both proper divisors	M1 is M+1,	proper_divisors(M1, FSQRTN, N, L1).proper_divisors(M, FSQRTN, N, L) :-	M1 is M+1,	proper_divisors(M1, FSQRTN, N, L). dpa(1, [1], [], []) :-	!.dpa(N, D, P, A) :-	N > 1,	proper_divisors(N, PN),	sum_list(PN, SPN),	compare(VGL, SPN, N),	dpa(VGL, N, D, P, A). dpa(<, N, [N|D], P, A) :- N1 is N-1, dpa(N1, D, P, A).dpa(=, N, D, [N|P], A) :- N1 is N-1, dpa(N1, D, P, A).dpa(>, N, D, P, [N|A]) :- N1 is N-1, dpa(N1, D, P, A).  dpa(N) :-	T0 is cputime,	dpa(N, D, P, A),	Dur is cputime-T0,	length(D, LD),	length(P, LP),	length(A, LA),	format("deficient: ~d~n abundant: ~d~n  perfect: ~d~n",		   [LD, LA, LP]),	format("took ~f seconds~n", [Dur]). 
Output:
?- dpa(20000).
deficient: 15036
abundant: 4960
perfect: 4
took 0.802559 seconds


## PureBasic

 EnableExplicit Procedure.i SumProperDivisors(Number)  If Number < 2 : ProcedureReturn 0 : EndIf  Protected i, sum = 0  For i = 1 To Number / 2    If Number % i = 0      sum + i    EndIf  Next  ProcedureReturn sumEndProcedure Define n, sum, deficient, perfect, abundant If OpenConsole()  For n = 1 To 20000    sum = SumProperDivisors(n)    If sum < n      deficient + 1    ElseIf sum = n      perfect + 1    Else      abundant + 1    EndIf  Next  PrintN("The breakdown for the numbers 1 to 20,000 is as follows : ")  PrintN("")  PrintN("Deficient = " + deficient)  PrintN("Pefect    = " + perfect)  PrintN("Abundant  = " + abundant)  PrintN("")  PrintN("Press any key to close the console")  Repeat: Delay(10) : Until Inkey() <> ""  CloseConsole()EndIf 
Output:
The breakdown for the numbers 1 to 20,000 is as follows :

Deficient = 15043
Pefect    = 4
Abundant  = 4953


## Python

Importing Proper divisors from prime factors:

>>> from proper_divisors import proper_divs>>> from collections import Counter>>> >>> rangemax = 20000>>> >>> def pdsum(n):...     return sum(proper_divs(n))... >>> def classify(n, p):...     return 'perfect' if n == p else 'abundant' if p > n else 'deficient'... >>> classes = Counter(classify(n, pdsum(n)) for n in range(1, 1 + rangemax))>>> classes.most_common()[('deficient', 15043), ('abundant', 4953), ('perfect', 4)]>>> 
Output:
Between 1 and 20000:
4953 abundant numbers
15043 deficient numbers
4 perfect numbers


## R

Works with: R version 3.3.2 and above
 # Abundant, deficient and perfect number classifications. 12/10/16 aevrequire(numbers);propdivcls <- function(n) {  V <- sapply(1:n, Sigma, proper = TRUE);  c1 <- c2 <- c3 <- 0;  for(i in 1:n){    if(V[i]<i){c1 = c1 +1} else if(V[i]==i){c2 = c2 +1} else{c3 = c3 +1}  }   cat(" *** Between 1 and ", n, ":\n");  cat("   * ", c1, "deficient numbers\n");  cat("   * ", c2, "perfect numbers\n");  cat("   * ", c3, "abundant numbers\n");}propdivcls(20000); 
Output:
> require(numbers)
> propdivcls(20000);
*** Between 1 and  20000 :
*  15043 deficient numbers
*  4 perfect numbers
*  4953 abundant numbers
>


## Racket

#lang racket(require math)(define (proper-divisors n) (drop-right (divisors n) 1))(define classes '(deficient perfect abundant))(define (classify n)  (list-ref classes (add1 (sgn (- (apply + (proper-divisors n)) n))))) (let ([N 20000])  (define t (make-hasheq))  (for ([i (in-range 1 (add1 N))])    (define c (classify i))    (hash-set! t c (add1 (hash-ref t c 0))))  (printf "The range between 1 and ~a has:\n" N)  (for ([c classes]) (printf "  ~a ~a numbers\n" (hash-ref t c 0) c)))
Output:
The range between 1 and 20000 has:
15043 deficient numbers
4 perfect numbers
4953 abundant numbers


## REXX

/*REXX program counts the number of  abundant/deficient/perfect  numbers within a range.*/parse arg low high .                             /*obtain optional arguments from the CL*/high=word(high low 20000,1);  low= word(low 1,1) /*obtain the   LOW  and  HIGH   values.*/say center('integers from '   low   " to "   high,  45,  "═")        /*display a header.*/!.= 0                                            /*define all types of  sums  to zero.  */      do j=low  to high;           $= sigma(j) /*get sigma for an integer in a range. */ if$<j  then               !.d= !.d + 1    /*Less?      It's a  deficient  number.*/              else if $>j then !.a= !.a + 1 /*Greater? " " abundant " */ else !.p= !.p + 1 /*Equal? " " perfect " */ end /*j*/ /* [↑] IFs are coded as per likelihood*/ say ' the number of perfect numbers: ' right(!.p, length(high) )say ' the number of abundant numbers: ' right(!.a, length(high) )say ' the number of deficient numbers: ' right(!.d, length(high) )exit /*stick a fork in it, we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/sigma: procedure; parse arg x; if x<2 then return 0; odd=x // 2 /* // ◄──remainder.*/ s= 1 /* [↓] only use EVEN or ODD integers.*/ do k=2+odd by 1+odd while k*k<x /*divide by all integers up to √x. */ if x//k==0 then s= s + k + x % k /*add the two divisors to (sigma) sum. */ end /*k*/ /* [↑] % is the REXX integer division*/ if k*k==x then return s + k /*Was X a square? If so, add √ x */ return s /*return (sigma) sum of the divisors. */ output when using the default input: ═════════integers from 1 to 20000═════════ the number of perfect numbers: 4 the number of abundant numbers: 4953 the number of deficient numbers: 15043  ### version 1.5 This version is pretty much identical to the 1st version but uses an integer square root calculation to find the limit of the do loop in the sigma function.  For 20k integers, it's approximately 6% faster. " 100k " " " 20% " " 1m " " " 30% "  /*REXX program counts the number of abundant/deficient/perfect numbers within a range.*/parse arg low high . /*obtain optional arguments from the CL*/high=word(high low 20000,1); low=word(low 1, 1) /*obtain the LOW and HIGH values.*/say center('integers from ' low " to " high, 45, "═") /*display a header.*/!.= 0 /*define all types of sums to zero. */ do j=low to high;$= sigma(j)   /*get sigma for an integer in a range. */      if $<j then !.d= !.d + 1 /*Less? It's a deficient number.*/ else if$>j  then  !.a= !.a + 1    /*Greater?     "  "  abundant      "   */                           else  !.p= !.p + 1    /*Equal?       "  "  perfect       "   */      end  /*j*/                                 /* [↑]  IFs are coded as per likelihood*/ say '   the number of perfect   numbers: '       right(!.p, length(high) )say '   the number of abundant  numbers: '       right(!.a, length(high) )say '   the number of deficient numbers: '       right(!.d, length(high) )exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/sigma: procedure; parse arg x 1 z;  if x<5  then return max(0, x-1)  /*sets X&Z to arg1.*/       q=1;  do  while q<=z;  q= q * 4;     end  /* ◄──↓  compute integer sqrt of Z (=R)*/       r=0;  do  while q>1; q=q%4; _=z-r-q; r=r%2; if _>=0  then do; z=_; r=r+q; end;  end       odd= x//2                                 /* [↓]  only use EVEN | ODD ints.   ___*/       s= 1;     do k=2+odd  by 1+odd  to r      /*divide by  all  integers up to   √ x */                 if x//k==0  then  s=s + k + x%k /*add the two divisors to (sigma) sum. */                 end   /*k*/                     /* [↑]  %  is the REXX integer division*/       if r*r==x  then  return s - k             /*Was X a square?  If so, subtract √ x */                        return s                 /*return (sigma) sum of the divisors.  */
output   is identical to the 1st REXX version.

### version 2

/* REXX */Call time 'R'cnt.=0Do x=1 To 20000  pd=proper_divisors(x)  sumpd=sum(pd)  Select    When x<sumpd Then cnt.abundant =cnt.abundant +1    When x=sumpd Then cnt.perfect  =cnt.perfect  +1    Otherwise         cnt.deficient=cnt.deficient+1    End  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 'In the range 1 - 20000'Say format(cnt.abundant ,5) 'numbers are abundant  'Say format(cnt.perfect  ,5) 'numbers are perfect   'Say format(cnt.deficient,5) 'numbers are deficient 'Say time('E') 'seconds elapsed'Exit proper_divisors: ProcedureParse Arg nPd=''If n=1 Then Return ''If n//2=1 Then  /* odd number  */  delta=2Else            /* even number */  delta=1Do d=1 To n%2 By delta  If n//d=0 Then    pd=pd d  EndReturn space(pd) sum: ProcedureParse Arg listsum=0Do i=1 To words(list)  sum=sum+word(list,i)  EndReturn sum
Output:
In the range 1 - 20000
4953 numbers are abundant
4 numbers are perfect
15043 numbers are deficient
28.392000 seconds elapsed

## Ring

 n = 30perfect(n) func perfect nfor i = 1 to n    sum = 0    for j = 1 to i - 1        if i % j = 0 sum = sum + j ok    next    see i    if sum = i see " is a perfect number" + nl    but sum < i see " is a deficient number" + nl    else see " is a abundant number" + nl ok   next 

## Rust

With proper_divisors#Rust in place:

fn main() {    // deficient starts at 1 because 1 is deficient but proper_divisors returns    // and empty Vec    let (mut abundant, mut deficient, mut perfect) = (0u32, 1u32, 0u32);    for i in 1..20_001 {        if let Some(divisors) = i.proper_divisors() {            let sum: u64 = divisors.iter().sum();            if sum < i {                deficient += 1            } else if sum > i {                abundant += 1            } else {                perfect += 1            }        }    }    println!("deficient:\t{:5}\nperfect:\t{:5}\nabundant:\t{:5}",             deficient, perfect, abundant);} 
Output:
deficient:      15043
perfect:            4
abundant:        4953


## Ruby

With proper_divisors#Ruby in place:

res = Hash.new(0)(1 .. 20_000).each{|n| res[n.proper_divisors.sum <=> n] += 1}puts "Deficient: #{res[-1]}   Perfect: #{res[0]}   Abundant: #{res[1]}" 
Output:

Deficient: 15043   Perfect: 4   Abundant: 4953



## Scala

def properDivisors(n: Int) = (1 to n/2).filter(i => n % i == 0)def classifier(i: Int) = properDivisors(i).sum compare ival groups = (1 to 20000).groupBy( classifier )println("Deficient: " + groups(-1).length)println("Abundant: " + groups(1).length)println("Perfect: " + groups(0).length + " (" + groups(0).mkString(",") + ")")
Output:
Deficient: 15043
Abundant: 4953
Perfect: 4 (6,28,496,8128)

## Scheme

 (define (classify n) (define (sum_of_factors x)  (cond ((= x 1) 1)        ((= (remainder n x) 0) (+ x (sum_of_factors (- x 1))))        (else (sum_of_factors (- x 1))))) (cond ((or (= n 1) (< (sum_of_factors (floor (/ n 2))) n)) -1)       ((= (sum_of_factors (floor (/ n 2))) n) 0)       (else 1)))(define n_perfect 0)(define n_abundant 0)(define n_deficient 0)(define (count n) (cond ((= n 1) (begin (display "perfect ")                       (display n_perfect)                       (newline)                       (display "abundant")                       (display n_abundant)                       (newline)                       (display "deficinet")                       (display n_perfect)                       (newline)))       ((equal? (classify n) 0) (begin (set! n_perfect (+ 1 n_perfect)) (display n) (newline) (count (- n 1))))       ((equal? (classify n) 1) (begin (set! n_abundant (+ 1 n_abundant)) (count (- n 1))))       ((equal? (classify n) -1) (begin (set! n_deficient (+ 1 n_deficient)) (count (- n 1)))))) 

$include "seed7_05.s7i"; const func integer: sumProperDivisors (in integer: number) is func result var integer: sum is 0; local var integer: num is 0; begin if number >= 2 then for num range 1 to number div 2 do if number rem num = 0 then sum +:= num; end if; end for; end if; end func; const proc: main is func local var integer: sum is 0; var integer: deficient is 0; var integer: perfect is 0; var integer: abundant is 0; var integer: number is 0; begin for number range 1 to 20000 do sum := sumProperDivisors(number); if sum < number then incr(deficient); elsif sum = number then incr(perfect); else incr(abundant); end if; end for; writeln("Deficient: " <& deficient); writeln("Perfect: " <& perfect); writeln("Abundant: " <& abundant); end func; Output: Deficient: 15043 Perfect: 4 Abundant: 4953  ## Sidef func propdivsum(n) { n.sigma - n } var h = Hash(){|i| ++(h{propdivsum(i) <=> i} := 0) } << 1..20000say "Perfect: #{h{0}} Deficient: #{h{-1}} Abundant: #{h{1}}" Output: Perfect: 4 Deficient: 15043 Abundant: 4953  ## Swift Translation of: C var deficients = 0 // sumPd < nvar perfects = 0 // sumPd = nvar abundants = 0 // sumPd > n // 1 is deficient (no proper divisor)deficients++ for i in 2...20000 { var sumPd = 1 // 1 is a proper divisor of all integer above 1 var maxPdToTest = i/2 // the max divisor to test for var j = 2; j < maxPdToTest; j++ { if (i%j) == 0 { // j is a proper divisor sumPd += j // New maximum for divisibility check maxPdToTest = i / j // To add to sum of proper divisors unless already done if maxPdToTest != j { sumPd += maxPdToTest } } } // Select type according to sum of Proper divisors if sumPd < i { deficients++ } else if sumPd > i { abundants++ } else { perfects++ }} println("There are \(deficients) deficient, \(perfects) perfect and \(abundants) abundant integers from 1 to 20000.") Output: There are 15043 deficient, 4 perfect and 4953 abundant integers from 1 to 20000. ## Tcl proc ProperDivisors {n} { if {$n == 1} {return 0}    set divs 1    set sum 1    for {set i 2} {$i*$i <= $n} {incr i} { if {! ($n % $i)} { lappend divs$i            incr sum $i if {$i*$i<$n} {                lappend divs [set d [expr {$n /$i}]]                incr sum $d } } } list$sum $divs} proc cmp {i j} { ;# analogous to [string compare], but for numbers if {$i == $j} {return 0} if {$i > $j} {return 1} return -1} proc classify {k} { lassign [ProperDivisors$k] p    ;# we only care about the first part of the result    dict get {        1   abundant        0   perfect       -1   deficient    } [cmp $k$p]} puts "Classifying the integers in $1, 20_000$:"set classes {}    ;# this will be a dict for {set i 1} {$i <= 20000} {incr i} { set class [classify$i]    dict incr classes $class} # using [lsort] to order the dictionary by value:foreach {kind count} [lsort -stride 2 -index 1 -integer$classes] {    puts "$kind:$count"}
Output:
Classifying the integers in [1, 20_000]:
perfect: 4
deficient: 4953
abundant: 15043

## TypeScript

function integer_classification(){
var sum:number=0, i:number,j:number;
var try:number=0;
var number_list:number[]={1,0,0};
for(i=2;i<=20000;i++){
try=i/2;
sum=1;
for(j=2;j<try;j++){
if (i%j)
continue;
try=i/j;
sum+=j;
if (j!=try)
sum+=try;
}
if (sum<i){
number_list[d]++;
continue;
}
else if (sum>i){
number_list[a]++;
continue;
}
number_list[p]++;
}
console.log('There are '+number_list[d]+ ' deficient , ' + 'number_list[p] + ' perfect and '+ number_list[a]+ ' abundant numbers
between 1 and 20000');
}


## uBasic/4tH

This is about the limit of what is feasible with uBasic/4tH performance wise, since a full run takes over 5 minutes.

P = 0 : D = 0 : A = 0 For n= 1 to 20000  s = FUNC(_SumDivisors(n))-n  If s = n Then P = P + 1  If s < n Then D = D + 1  If s > n Then A = A + 1Next Print "Perfect: ";P;" Deficient: ";D;" Abundant: ";AEnd ' Return the least power of [email protected] that does not divide [email protected] _LeastPower Param(2)  Local(1)   [email protected] = [email protected]  Do While ([email protected] % [email protected]) = 0    [email protected] = [email protected] * [email protected]  Loop Return ([email protected])  ' Return the sum of the proper divisors of [email protected] _SumDivisors Param(1)  Local(4)   [email protected] = [email protected]  [email protected] = 1   ' Handle two specially   [email protected] = FUNC(_LeastPower (2,[email protected]))  [email protected] = [email protected] * ([email protected] - 1)  [email protected] = [email protected] / ([email protected] / 2)   ' Handle odd factors   For [email protected] = 3 Step 2 While ([email protected]*[email protected]) < ([email protected]+1)    [email protected] = FUNC(_LeastPower ([email protected],[email protected]))    [email protected] = [email protected] * (([email protected] - 1) / ([email protected] - 1))    [email protected] = [email protected] / ([email protected] / [email protected])  Loop   ' At this point, t must be one or prime   If ([email protected] > 1) [email protected] = [email protected] * ([email protected]+1)Return ([email protected])
Output:
Perfect: 4 Deficient: 15043 Abundant: 4953

0 OK, 0:210

## VBA

 Option Explicit Public Sub Nb_Classifications()Dim A As New Collection, D As New Collection, P As New CollectionDim n As Long, l As Long, s As String, t As Single     t = Timer    'Start    For n = 1 To 20000        l = SumPropers(n): s = CStr(n)        Select Case n            Case Is > l: D.Add s, s            Case Is < l: A.Add s, s            Case l: P.Add s, s        End Select    Next     'End. Return :    Debug.Print "Execution Time : " & Timer - t & " seconds."    Debug.Print "-------------------------------------------"    Debug.Print "Deficient := " & D.Count    Debug.Print "Perfect := " & P.Count    Debug.Print "Abundant := " & A.CountEnd Sub Private Function SumPropers(n As Long) As Long'returns the sum of the proper divisors of nDim j As Long    For j = 1 To n \ 2        If n Mod j = 0 Then SumPropers = j + SumPropers    NextEnd Function
Output:
Execution Time : 2,6875 seconds.
-------------------------------------------
Deficient := 15043
Perfect := 4
Abundant := 4953

## VBScript

Deficient = 0Perfect = 0Abundant = 0For i = 1 To 20000	sum = 0	For n = 1 To 20000		If n < i Then			If i Mod n = 0 Then				sum = sum + n			End If		End If	Next	If sum < i Then		Deficient = Deficient + 1	ElseIf sum = i Then		Perfect = Perfect + 1	ElseIf sum > i Then		Abundant = Abundant + 1	End IfNextWScript.Echo "Deficient = " & Deficient & vbCrLf &_			 "Perfect = " & Perfect & vbCrLf &_			 "Abundant = " & Abundant
Output:
Deficient = 15043
Perfect = 4
Abundant = 4953

## Visual Basic .NET

Translation of: FreeBASIC
Module Module1     Function SumProperDivisors(number As Integer) As Integer        If number < 2 Then Return 0        Dim sum As Integer = 0        For i As Integer = 1 To number \ 2            If number Mod i = 0 Then sum += i        Next        Return sum    End Function     Sub Main()        Dim sum, deficient, perfect, abundant As Integer         For n As Integer = 1 To 20000            sum = SumProperDivisors(n)            If sum < n Then                deficient += 1            ElseIf sum = n Then                perfect += 1            Else                abundant += 1            End If        Next         Console.WriteLine("The classification of the numbers from 1 to 20,000 is as follows : ")        Console.WriteLine()        Console.WriteLine("Deficient = {0}", deficient)        Console.WriteLine("Perfect   = {0}", perfect)        Console.WriteLine("Abundant  = {0}", abundant)    End Sub End Module
Output:
The classification of the numbers from 1 to 20,000 is as follows :

Deficient = 15043
Perfect   = 4
Abundant  = 4953

## Yabasic

Translation of: AWK
clear screen Deficient = 0Perfect = 0Abundant = 0For j=1 to 20000	sump = sumprop(j)	If sump < j Then		Deficient = Deficient + 1	ElseIf sump = j Then		Perfect = Perfect + 1	ElseIf sump > j Then		Abundant = Abundant + 1	End IfNext j PRINT "Number deficient: ",DeficientPRINT "Number perfect:   ",PerfectPRINT "Number abundant:  ",Abundant sub sumprop(num)	local i, sum, root 	if num>1 then		sum=1		root=sqrt(num)		for i=2 to root			if mod(num,i) = 0 then				sum=sum+i				if (i*i)<>num sum=sum+num/i			end if		next i	end if	return sumend sub

## zkl

Translation of: D
fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) } fcn classify(n){   p:=properDivs(n).sum();   return(if(p<n) -1 else if(p==n) 0 else 1);} const rangeMax=20_000;classified:=[1..rangeMax].apply(classify);perfect   :=classified.filter('==(0)).len();abundant  :=classified.filter('==(1)).len();println("Deficient=%d, perfect=%d, abundant=%d".fmt(   classified.len()-perfect-abundant, perfect, abundant));
Output:
Deficient=15043, perfect=4, abundant=4953

## ZX Spectrum Basic

Solution 1:

  10 LET nd=1: LET np=0: LET na=0  20 FOR i=2 TO 20000  30 LET sum=1  40 LET max=i/2  50 LET n=2: LET l=max-1  60 IF n>l THEN GO TO 90  70 IF i/n=INT (i/n) THEN LET sum=sum+n: LET max=i/n: IF max<>n THEN LET sum=sum+max: LET l=max-1  80 LET n=n+1: GO TO 60  90 IF sum<i THEN LET nd=nd+1: GO TO 120 100 IF sum=i THEN LET np=np+1: GO TO 120 110 LET na=na+1 120 NEXT i 130 PRINT "Number deficient: ";nd 140 PRINT "Number perfect:   ";np 150 PRINT "Number abundant:  ";na

Solution 2 (more efficient):

  10 LET abundant=0: LET deficient=0: LET perfect=0  20 FOR j=1 TO 20000  30 GO SUB 120  40 IF sump<j THEN LET deficient=deficient+1: GO TO 70  50 IF sump=j THEN LET perfect=perfect+1: GO TO 70  60 LET abundant=abundant+1  70 NEXT j  80 PRINT "Perfect: ";perfect  90 PRINT "Abundant: ";abundant 100 PRINT "Deficient: ";deficient 110 STOP 120 IF j=1 THEN LET sump=0: RETURN 130 LET sum=1 140 LET root=SQR j 150 FOR i=2 TO root 160 IF j/i=INT (j/i) THEN LET sum=sum+i: IF (i*i)<>j THEN LET sum=sum+j/i 170 NEXT i 180 LET sump=sum 190 RETURN`