Amicable pairs: Difference between revisions

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=={{header|Java}}==
=={{header|Java}}==
{{works with|Java|8}}
{{works with|Java|8}}
<lang java>import static java.util.stream.IntStream.range;
<lang java>import static java.util.stream.LongStream.range;


public class AmicablePairs {
public class AmicablePairs {

Revision as of 13:05, 20 February 2016

Task
Amicable pairs
You are encouraged to solve this task according to the task description, using any language you may know.

Two integers and are said to be amicable pairs if and the sum of the proper divisors of () as well as .

For example 1184 and 1210 are an amicable pair (with proper divisors 1, 2, 4, 8, 16, 32, 37, 74, 148, 296, 592 and 1, 2, 5, 10, 11, 22, 55, 110, 121, 242, 605 respectively).

Task

Calculate and show here the Amicable pairs below 20,000; (there are eight).

Cf.

AutoHotkey

<lang d>SetBatchLines -1 Loop, 20000 { m := A_index

; Getting factors loop % floor(sqrt(m)) { if ( mod(m, A_index) = 0 ) { if ( A_index ** 2 == m ) { sum += A_index continue } else if ( A_index != 1 ) { sum += A_index + m//A_index } else if ( A_index = 1 ) { sum += A_index } } } ; Factors obtained

; Checking factors of sum if ( sum > 1 ) { loop % floor(sqrt(sum)) { if ( mod(sum, A_index) = 0 ) { if ( A_index ** 2 == sum ) { sum2 += A_index continue } else if ( A_index != 1 ) { sum2 += A_index + sum//A_index } else if ( A_index = 1 ) { sum2 += A_index } } } if ( m = sum2 ) && ( m != sum ) && ( m < sum ) final .= m . ":" . sum . "`n" } ; Checked

sum := 0 sum2 := 0 } MsgBox % final ExitApp</lang>

Output:
220:284
1184:1210
2620:2924
5020:5564
6232:6368
10744:10856
12285:14595
17296:18416

AWK

<lang awk>

  1. !/bin/awk -f

function sumprop(num, i,sum,root) { if (num < 2) return 0 sum=1 root=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=20000 print "Amicable pairs < ",limit for (n=1; n < limit+1; n++) 

   {
   m=sumprop(n)
   if (n == sumprop(m) && n < m) print n,m
   }

} }</lang>

Output:
# ./amicable 
Amicable pairs < 20000
220 284
1184 1210
2620 2924
5020 5564
6232 6368
10744 10856
12285 14595
17296 18416

C

Remark: Look at Pascal Alternative [[1]].You are using the same principle, so here too both numbers of the pair must be < top.

The program will overflow and error in all sorts of ways when given a commandline argument >= UINT_MAX/2 (generally 2^31) <lang c>#include <stdio.h>

  1. include <stdlib.h>

typedef unsigned int uint;

int main(int argc, char **argv) { 

 uint top = atoi(argv[1]);
 uint *divsum = malloc((top + 1) * sizeof(*divsum));
 uint pows[32] = {1, 0};

 for (uint i = 0; i <= top; i++) divsum[i] = 1;

 // sieve
 // only sieve within lower half , the modification starts at 2*p
 for (uint p = 2; p+p <= top; p++) {
   if (divsum[p] > 1) {
     divsum[p] -= p;// subtract number itself from divisor sum ('proper')
     continue;}     // p not prime

   uint x; // highest power of p we need
   //checking x <= top/y instead of x*y <= top to avoid overflow
   for (x = 1; pows[x - 1] <= top/p; x++)
     pows[x] = p*pows[x - 1];

   //counter where n is not a*p with a = ?*p, useful for most p.
   //think of p>31 seldom divisions or p>sqrt(top) than no division is needed
   //n = 2*p, so the prime itself is left unchanged => k=p-1
   uint k= p-1;
   for (uint n = p+p; n <= top; n += p) {
     uint s=1+pows[1];
     k--;
     // search the right power only if needed
     if ( k==0) {
       for (uint i = 2; i < x && !(n%pows[i]); s += pows[i++]);
       k = p; }
     divsum[n] *= s;
   }
 }

 //now correct the upper half
 for (uint p = (top >> 1)+1; p <= top; p++) {
   if (divsum[p] > 1){
     divsum[p] -= p;}
 }

 uint cnt = 0;
 for (uint a = 1; a <= top; a++) {
   uint b = divsum[a];
   if (b > a && b <= top && divsum[b] == a){
     printf("%u %u\n", a, b);
     cnt++;}
 }
 printf("\nTop %u count : %u\n",top,cnt);
 return 0;

}</lang>

Output:
% ./a.out 20000
220 284
1184 1210
2620 2924
5020 5564
6232 6368
10744 10856
12285 14595
17296 18416

Top 20000 count : 8


% ./a.out 524000000
..
475838415 514823985
491373104 511419856
509379344 523679536

Top 524000000 count : 442

real  0m16.285s
user  0m16.156s

D

Translation of: Python

<lang d>void main() @safe /*@nogc*/ { 

   import std.stdio, std.algorithm, std.range, std.typecons, std.array;

   immutable properDivs = (in uint n) pure nothrow @safe /*@nogc*/ =>
       iota(1, (n + 1) / 2 + 1).filter!(x => n % x == 0);

   enum rangeMax = 20_000;
   auto n2d = iota(1, rangeMax + 1).map!(n => properDivs(n).sum);

   foreach (immutable n, immutable divSum; n2d.enumerate(1))
       if (n < divSum && divSum <= rangeMax && n2d[divSum - 1] == n)
           writefln("Amicable pair: %d and %d with proper divisors:\n    %s\n    %s",
                    n, divSum, properDivs(n), properDivs(divSum));

}</lang>

Output:
Amicable pair: 220 and 284 with proper divisors:
    [1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110]
    [1, 2, 4, 71, 142]
Amicable pair: 1184 and 1210 with proper divisors:
    [1, 2, 4, 8, 16, 32, 37, 74, 148, 296, 592]
    [1, 2, 5, 10, 11, 22, 55, 110, 121, 242, 605]
Amicable pair: 2620 and 2924 with proper divisors:
    [1, 2, 4, 5, 10, 20, 131, 262, 524, 655, 1310]
    [1, 2, 4, 17, 34, 43, 68, 86, 172, 731, 1462]
Amicable pair: 5020 and 5564 with proper divisors:
    [1, 2, 4, 5, 10, 20, 251, 502, 1004, 1255, 2510]
    [1, 2, 4, 13, 26, 52, 107, 214, 428, 1391, 2782]
Amicable pair: 6232 and 6368 with proper divisors:
    [1, 2, 4, 8, 19, 38, 41, 76, 82, 152, 164, 328, 779, 1558, 3116]
    [1, 2, 4, 8, 16, 32, 199, 398, 796, 1592, 3184]
Amicable pair: 10744 and 10856 with proper divisors:
    [1, 2, 4, 8, 17, 34, 68, 79, 136, 158, 316, 632, 1343, 2686, 5372]
    [1, 2, 4, 8, 23, 46, 59, 92, 118, 184, 236, 472, 1357, 2714, 5428]
Amicable pair: 12285 and 14595 with proper divisors:
    [1, 3, 5, 7, 9, 13, 15, 21, 27, 35, 39, 45, 63, 65, 91, 105, 117, 135, 189, 195, 273, 315, 351, 455, 585, 819, 945, 1365, 1755, 2457, 4095]
    [1, 3, 5, 7, 15, 21, 35, 105, 139, 417, 695, 973, 2085, 2919, 4865]
Amicable pair: 17296 and 18416 with proper divisors:
    [1, 2, 4, 8, 16, 23, 46, 47, 92, 94, 184, 188, 368, 376, 752, 1081, 2162, 4324, 8648]
    [1, 2, 4, 8, 16, 1151, 2302, 4604, 9208]

EchoLisp

<lang scheme>

using (sum-divisors) from math.lib

(lib 'math) (define (amicable N) (define n 0) (for/list ((m (in-range 2 N))) (set! n (sum-divisors m)) #:continue (>= n (* 1.5 m))  ;; assume n/m < 1.5 #:continue (<= n m) ;; prevent perfect numbers #:continue (!= (sum-divisors n) m) (cons m n)))

(amicable 20000) 

   → ((220 . 284) (1184 . 1210) (2620 . 2924) (5020 . 5564) (6232 . 6368) (10744 . 10856) (12285 . 14595) (17296 . 18416))

(amicable 1_000_000) ;; 42 pairs 

  → (... (802725 . 863835) (879712 . 901424) (898216 . 980984) (947835 . 1125765) (998104 . 1043096))

</lang>

Elixir

With proper_divisors#Elixir in place: <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)]

end

map = Enum.into(1..20000, %{}, fn n -> {n, Proper.divisors(n) |> Enum.sum} end) Enum.filter(map, fn {n,sum} -> map[sum] == n and n < sum end) |> Enum.sort |> Enum.each(fn {i,j} -> IO.puts "#{i} and #{j}" end)</lang>

Output:
220 and 284
1184 and 1210
2620 and 2924
5020 and 5564
6232 and 6368
10744 and 10856
12285 and 14595
17296 and 18416

Erlang

Erlang, slow

Very slow solution. Same functions by and large as in proper divisors and co.

<lang erlang> -module(properdivs). -export([amicable/1,divs/1,sumdivs/1]).

amicable(Limit) -> amicable(Limit,[],3,2).

amicable(Limit,List,_Current,Acc) when Acc >= Limit -> List; amicable(Limit,List,Current,Acc) when Current =< Acc/2 -> 

   amicable(Limit,List,Acc,Acc+1);

amicable(Limit,List,Current,Acc) -> 

   CS = sumdivs(Current),                                                         
   AS = sumdivs(Acc),                                                             
   if                                                                             
       CS == Acc andalso AS == Current andalso Acc =/= Current ->                 
         io:format("A: ~w, B: ~w, ~nL: ~w~w~n",  [Current,Acc,divs(Current),divs(Acc)]),                                                                                
         NL = List ++ [{Current,Acc}],                                            
         amicable(Limit,NL,Acc+1,Acc+1);                                          
       true -> 
         amicable(Limit,List,Current-1,Acc) end.

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)). </lang>

Output:
3> properdivs:amicable(20000).
A: 220, B: 284, 
L: [1,2,4,5,10,11,20,22,44,55,110][1,2,4,71,142]
A: 1184, B: 1210, 
L: [1,2,4,8,16,32,37,74,148,296,592][1,2,5,10,11,22,55,110,121,242,605]
A: 2620, B: 2924, 
L: [1,2,4,5,10,20,131,262,524,655,1310][1,2,4,17,34,43,68,86,172,731,1462]
A: 5020, B: 5564, 
L: [1,2,4,5,10,20,251,502,1004,1255,2510][1,2,4,13,26,52,107,214,428,1391,2782]
A: 6232, B: 6368, 
L: [1,2,4,8,19,38,41,76,82,152,164,328,779,1558,3116][1,2,4,8,16,32,199,398,796,1592,3184]
A: 10744, B: 10856, 
L: [1,2,4,8,17,34,68,79,136,158,316,632,1343,2686,5372][1,2,4,8,23,46,59,92,118,184,236,472,1357,2714,5428]
A: 12285, B: 14595, 
L: [1,3,5,7,9,13,15,21,27,35,39,45,63,65,91,105,117,135,189,195,273,315,351,455,585,819,945,1365,1755,2457,4095][1,3,5,7,15,21,35,105,139,417,695,973,2085,2919,4865]
A: 17296, B: 18416, 
L: [1,2,4,8,16,23,46,47,92,94,184,188,368,376,752,1081,2162,4324,8648][1,2,4,8,16,1151,2302,4604,9208]
[{220,284},
 {1184,1210},
 {2620,2924},
 {5020,5564},
 {6232,6368},
 {10744,10856},
 {12285,14595},
 {17296,18416}]

Erlang, faster

This is lazy AND depends on the fun fact that we're not really identifying pairs. They just happen to order. Probably, this answer is false in some sense. But a good deal faster :) As above with the additional function.

[See the talk section   amicable pairs, out of order   for this Rosetta Code task.]

<lang erlang> friendly(Limit) -> 

   List = [{X,properdivs:sumdivs(X)} || X <- lists:seq(3,Limit)],
   Final = [ X || 
       X <- lists:seq(3,Limit), 
       X == properdivs:sumdivs(proplists:get_value(X,List)) 
       andalso X =/= proplists:get_value(X,List)],
   io:format("L: ~w~n", [Final]).

</lang>

Output:
45> properdivs:friendly(20000).
L: [220,284,1184,1210,2620,2924,5020,5564,6232,6368,10744,10856,12285,14595,17296,18416]
ok

We might answer a challenge by saying: <lang erlang> friendly(Limit) -> 

   List = [{X,properdivs:sumdivs(X)} || X <- lists:seq(3,Limit)],                 
   Final = [ X || X <- lists:seq(3,Limit), X == properdivs:sumdivs(proplists:get_value(X,List)) 
           andalso X =/= proplists:get_value(X,List)],
   findfriendlies(Final,[]).

findfriendlies(List,Acc) when length(List) =< 0 -> Acc; findfriendlies(List,Acc) -> 

   A = lists:nth(1,List),                                                         
   AS = sumdivs(A),                                                               
   B = lists:nth(2,List),                                                         
   BS = sumdivs(B),                                                               
   if                                                                             
       AS == B andalso BS == A ->                                                 
         {_,BL} = lists:split(2,List),                                            
         findfriendlies(BL,Acc++[{A,B}]);                                         
       true -> false                                                              
   end.

</lang>

Output:
94>  properdivs:friendly(20000).
[{220,284},
 {1184,1210},
 {2620,2924},
 {5020,5564},
 {6232,6368},
 {10744,10856},
 {12285,14595},
 {17296,18416}]

In either case, it's a lot faster than the recursion in my first example.

ERRE

<lang ERRE>PROGRAM AMICABLE

CONST LIMIT=20000

PROCEDURE SUMPROP(NUM->M) 

 IF NUM<2 THEN M=0 EXIT PROCEDURE
 SUM=1
 ROOT=SQR(NUM)
 FOR I=2 TO ROOT-1 DO
    IF (NUM=I*INT(NUM/I)) THEN
        SUM=SUM+I+NUM/I
    END IF
    IF (NUM=ROOT*INT(NUM/ROOT)) THEN
        SUM=SUM+ROOT
    END IF
 END FOR
 M=SUM

END PROCEDURE

BEGIN 

 PRINT(CHR$(12);) ! CLS
 PRINT("Amicable pairs < ";LIMIT)
 FOR N=1 TO LIMIT DO
   SUMPROP(N->M1)
   SUMPROP(M1->M2)
   IF (N=M2 AND N<M1) THEN PRINT(N,M1)
 END FOR

END PROGRAM</lang>

Output:
Amicable pairs <  20000
 220           284
 1184          1210
 2620          2924
 5020          5564
 6232          6368
 10744         10856
 12285         14595
 17296         18416

Fortran

This version uses some latter-day facilities such as array assignment that could be replaced by an ordinary DO-loop, as could the FOR ALL statement that for two adds two to every second element, for three adds three to every third, etc. Each FORALL statement applies its DO-given increment to all the selected array elements potentially in any order or even simultaneously. Likewise, the "MODULE" protocol could be abandoned, which would mean that the KNOWNSUM array would have to be declared COMMON for access across routines - or the whole re-written as a single mainline. And if the PARAMETER statements were replaced appropriately, this source could be compiled using Fortran 77.

Output: 

Perfect!!           6
Perfect!!          28
Amicable!         220         284
Perfect!!         496
Amicable!        1184        1210
Amicable!        2620        2924
Amicable!        5020        5564
Amicable!        6232        6368
Perfect!!        8128
Amicable!       10744       10856
Amicable!       12285       14595
Amicable!       17296       18416

<lang FORTRAN> 

     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,ILIMIT		!Some bounds.
      PARAMETER (ILIMIT = 2147483647)	!Computer arithmetic is not with real numbers.
      PARAMETER (LOTS = 22000)	!Nor is computer storage infinite.
      INTEGER KNOWNSUM(LOTS)		!Calculate these once as multiple references are expected.
      CONTAINS			!Assistants.
       INTEGER FUNCTION SUMF(N)	!Sum of the proper divisors of N.
        INTEGER N			!The number in question.
        INTEGER S,F,F2,INC,BOOST	!Assistants.
         IF (N.LE.LOTS) THEN		!If we're within reach,
           SUMF = KNOWNSUM(N)			!The result is to hand.
          ELSE			!Otherwise, some on-the-spot effort ensues.

Could use SUMF in place of S, but some compilers have been confused by such usage. 

           S = 1			!1 is always a factor of N, but N is deemed not.
           F = 1			!Prepare a crude search for factors.
           INC = 1			!One by plodding one.
           IF (MOD(N,2) .EQ. 1) INC = 2!Ah, but an odd number cannot have an even number as a divisor.
   1       F = F + INC			!So half the time we can doubleplod.
           F2 = F*F				!Up to F2 < N rather than F < SQRT(N) and worries over inexact arithmetic.
           IF (F2 .LT. N) THEN			!F2 = N handled below.
             IF (MOD(N,F) .EQ. 0) THEN		!Does F divide N?
               BOOST = F + N/F			!Yes. The divisor and its counterpart.
               IF (S .GT. ILIMIT - BOOST) GO TO 666	!Would their augmentation cause an overflow?
               S = S + BOOST			!No, so count in the two divisors just discovered.
             END IF				!So much for a divisor discovered.
             GO TO 1				!Try for another.
           END IF			!So much for the horde.
           IF (F2 .EQ. N) THEN	!Special case: N may be a perfect square, not necessarily of a prime number.
             IF (S .GT. ILIMIT - F) GO TO 666	!It is. And it too might cause overflow.
             S = S + F			!But if not, count F once only.
           END IF			!All done.
           SUMF = S			!This is the result.
         END IF			!Whichever way obtained,
        RETURN			!Done.

Cannot calculate the sum, because it exceeds the integer limit. 

 666     SUMF = -666		!An expression of dismay that the caller will notice.
       END FUNCTION SUMF	!Alternatively, find the prime factors, and combine them...  
        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 is a proper divisor of all N, 1 is excluded for itself.
         DO F = 2,LOTS/2		!Step through all the possible divisors of numbers not exceeding LOTS.
           FOR ALL(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.
     END MODULE FACTORSTUFF	!Enough assistants. 
      PROGRAM AMICABLE		!Seek N such that SumF(SumF(N)) = N, for N up to 20,000.
      USE FACTORSTUFF		!This should help.
      INTEGER I,N		!Steppers.
      INTEGER S1,S2		!Sums of factors.
       CALL PREPARESUMF		!Values for every N up to the search limit will be called for at least once.

c WRITE (6,66) (I,KNOWNSUM(I), I = 1,48) c 66 FORMAT (10(I3,":",I5,"|")) 

       DO N = 2,20000		!Step through the specified search space.
         S1 = SUMF(N)			!Only even numbers appear in the results, but check every one anyway.
         IF (S1 .EQ. N) THEN		!Catch a tight loop.
           WRITE (6,*) "Perfect!!",N	!Self amicable! Would otherwise appear as Amicable! n,n.
         ELSE IF (S1 .GT. N) THEN	!Look for a pair going upwards only.
           S2 = SUMF(S1)		!Since otherwise each would appear twice.
           IF (S2.EQ.N) WRITE (6,*) "Amicable!",N,S1	!Aha!
         END IF			!So much for that candidate.
       END DO			!On to the next.
     END			!Done.

</lang>

Haskell

<lang Haskell>divisors :: (Integral a) => a -> [a] divisors n = filter ((0 ==) . (n `mod`)) [1 .. (n `div` 2)]

main :: IO () main = do 

 let range = [1 .. 20000 :: Int]
     divs = zip range $ map (sum . divisors) range
     pairs = [(n, m) | (n, nd) <- divs, (m, md) <- divs,
              n < m, nd == m, md == n]
 print pairs</lang>
Output:
[(220,284),(1184,1210),(2620,2924),(5020,5564),(6232,6368),(10744,10856),(12285,14595),(17296,18416)]

J

Proper Divisor implementation:

<lang J>factors=: [: /:~@, */&>@{@((^ i.@>:)&.>/)@q:~&__ properDivisors=: factors -. -.&1</lang>

Amicable pairs:

<lang J> 1 + 0 20000 #: I. ,(</~@i.@# * (* |:))(=/ +/@properDivisors@>) 1 + i.20000 

 220   284
1184  1210
2620  2924
5020  5564
6232  6368

10744 10856 12285 14595 17296 18416</lang>

Java

Works with: Java version 8

<lang java>import static java.util.stream.LongStream.range;

public class AmicablePairs { 

   public static void main(String[] args) {
       final int limit = 20_000;

       for (int n = 1; n < limit; n++) {
           long m = properDivsSum(n);
           if (n == m || m > limit || m < n || properDivsSum((int) m) != n)
               continue;
           System.out.printf("%s %s %n", n, m);
       }
   }

   public static long properDivsSum(int n) {
       return range(1, (n + 1) / 2 + 1).filter(i -> n % i == 0).sum();
   }

}</lang> 

220 284 
1184 1210 
2620 2924 
5020 5564 
6232 6368 
10744 10856 
12285 14595 
17296 18416

JavaScript

<lang JavaScript>(function (max) { 

   // 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;
   }

   // Filter an array of proper divisor sums,
   // reading the array index as a function of N (N-1)
   // and the sum of proper divisors as a potential M

   var pairs = range(1, max).map(function (x) {
       return properDivisors(x).reduce(function (a, d) {
           return a + d;
       }, 0)
   }).reduce(function (a, m, i, lst) {
       var n = i + 1;

       return (m > n) && lst[m - 1] === n ? a.concat(n, m) : a;
   }, []);

   // 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(
       'N', 'M'.concat(pairs),
       true,
       'text-align:center'
   ) + '\n\n' + JSON.stringify(pairs);

})(20000);</lang>

Output:
N M
220 284
1184 1210
2620 2924
5020 5564
6232 6368
10744 10856
12285 14595
17296 18416

<lang JavaScript>[[220,284],[1184,1210],[2620,2924],[5020,5564], 

[6232,6368],[10744,10856],[12285,14595],[17296,18416]]</lang>

jq

<lang jq># unordered def proper_divisors: 

 . as $n
 | if $n > 1 then 1,
     (sqrt|floor as $s
     | range(2; $s+1) as $i
     | if ($n % $i) == 0 then $i,
          (if $i * $i == $n then empty else ($n / $i) end)

else empty end) 

   else empty
   end;

def addup(stream): reduce stream as $i (0; . + $i);

def task(n): 

 (reduce range(0; n+1) as $n
   ( [];  . + [$n | addup(proper_divisors)] )) as $listing
 | range(1;n+1) as $j
 | range(1;$j) as $k
 | if $listing[$j] == $k and $listing[$k] == $j
   then "\($k) and \($j) are amicable"
   else empty
   end ;

task(20000)</lang>

Output:

<lang sh>$ jq -c -n -f amicable_pairs.jq 220 and 284 are amicable 1184 and 1210 are amicable 2620 and 2924 are amicable 5020 and 5564 are amicable 6232 and 6368 are amicable 10744 and 10856 are amicable 12285 and 14595 are amicable 17296 and 18416 are amicable</lang>

Julia

Given factor, it is not necessary to calculate the individual divisors to compute their sum. See Abundant, deficient and perfect number classifications for the details.

Functions <lang Julia> function pcontrib(p::Int64, a::Int64) 

   n = one(p)
   pcon = one(p)
   for i in 1:a
       n *= p
       pcon += n
   end
   return pcon

end

function divisorsum(n::Int64) 

   dsum = one(n)
   for (p, a) in factor(n)
       dsum *= pcontrib(p, a)
   end
   dsum -= n

end </lang> pcontrib is a good candidate for Memoization should the performance of divisorsum become an issue.

Main

It is safe to exclude primes from consideration; their proper divisor sum is always 1. Also, this code uses a minor trick to ensure that none of the numbers identified are above the limit. All numbers in the range are checked for an amicable partner, but the pair is cataloged only when the greater member is reached. <lang Julia> const L = 2*10^4 acnt = 0

println("Amicable pairs not greater than ", L)

for i in 2:L 

   !isprime(i) || continue
   j = divisorsum(i)
   j < i && divisorsum(j) == i || continue
   acnt += 1
   println(@sprintf("%4d", acnt), " => ", j, ", ", i)

end </lang>

Output:
Amicable pairs not greater than 20000
   1 => 220, 284
   2 => 1184, 1210
   3 => 2620, 2924
   4 => 5020, 5564
   5 => 6232, 6368
   6 => 10744, 10856
   7 => 12285, 14595
   8 => 17296, 18416

Lua

0.02 of a second in 16 lines of code. The vital trick is to just set m to the sum of n's proper divisors each time. That way you only have to test the reverse, dividing your run time by half the loop limit (ie. 10,000)! <lang lua>function sumDivs (n) 

   local sum = 1
   for d = 2, math.sqrt(n) do
       if n % d == 0 then
           sum = sum + d
           sum = sum + n / d
       end
   end
   return sum

end

for n = 2, 20000 do 

   m = sumDivs(n)
   if m > n then
       if sumDivs(m) == n then print(n, m) end
   end

end</lang>

Output:


220	284
1184	1210
2620	2924
5020	5564
6232	6368
10744	10856
12285	14595
17296	18416

Mathematica / Wolfram Language

<lang Mathematica>amicableQ[n_] := 

Module[{sum = Total[Most@Divisors@n]},
 sum != n && n == Total[Most@Divisors@sum]]

Grid@Partition[Cases[Range[4, 20000], _?(amicableQ@# &)], 2]</lang>

Output:


220	284
1184	1210
2620	2924
5020	5564
6232	6368
10744	10856
12285	14595
17296	18416

Nim

Being a novice, I submitted my code to the Nim community for review and received much feedback and advice. They were instrumental in fine-tuning this code for style and readability, I can't thank them enough. <lang Nim> from math import sqrt

const N = 524_000_000.int32

proc sumProperDivisors(someNum: int32, chk4less: bool): int32 = 

   result = 1
   let maxPD = sqrt(someNum.float).int32
   let offset = someNum mod 2
   for divNum in countup(2 + offset, maxPD, 1 + offset):
       if someNum mod divNum == 0:
           result += divNum + someNum div divNum
           if chk4less and result >= someNum:
               return 0

for n in countdown(N, 2): 

   let m = sumProperDivisors(n, true)
   if m != 0 and n == sumProperDivisors(m, false):
       echo $n, " ", $m

</lang>

Output:
523679536 509379344
511419856 491373104
514823985 475838415
...... ......
..... .....
18416 17296
14595 12285
10856 10744
6368 6232
5564 5020
2924 2620
1210 1184
284 220

Total number of pairs is 442, on my machine the code takes ~389 minutes to run.
Here's a second version that uses a large amount of memory but runs in 2m32seconds. Again, thanks to the Nim community <lang Nim> from math import sqrt

const N = 524_000_000.int32 var x = newSeq[int32](N+1)

for i in 2..sqrt(N.float).int32: 

 var p = i*i
 x[p] += i
 var j = i + i
 while (p += i; p <= N):
   j.inc
   x[p] += j

for m in 4..N: 

 let n = x[m] + 1
 if n < m and n != 0 and m == x[n] + 1:
     echo n, " ", m

</lang>

Output:
220 284
1184 1210
2620 2924
5020 5564
6232 6368
10744 10856
12285 14595
17296 18416
..... .....
...... ......
426191535 514780497
475838415 514823985
509379344 523679536

Oberon-2

<lang Oberon2> MODULE AmicablePairs; IMPORT 

 Out;

CONST 

 max = 20000;

VAR 

 i,j: INTEGER;
 pd: ARRAY max + 1 OF LONGINT;

PROCEDURE ProperDivisorsSum(n: LONGINT): LONGINT; VAR 

  i,sum: LONGINT;

BEGIN 

 sum := 0;
 IF n > 1 THEN 
   INC(sum,1);i := 2;
   WHILE (i < n) DO
     IF (n MOD i) = 0 THEN INC(sum,i) END;
     INC(i)
   END
 END;
 RETURN sum

END ProperDivisorsSum;

BEGIN 

 FOR i := 0 TO max DO
   pd[i] := ProperDivisorsSum(i)
 END;

 FOR i := 2 TO max DO
   FOR j := i + 1 TO max DO
     IF (pd[i] = j) & (pd[j] = i) THEN
        Out.Char('[');Out.Int(i,0);Out.Char(',');Out.Int(j,0);Out.Char("]");Out.Ln
     END
   END 
 END

END AmicablePairs. </lang>

Output:
[220,284]
[1184,1210]
[2620,2924]
[5020,5564]
[6232,6368]
[10744,10856]
[12285,14595]
[17296,18416]

Oforth

Using properDivs implementation tasks without optimization (calculating proper divisors sum without returning a list for instance) :

<lang Oforth>Integer method: properDivs self 2 / seq filter(#[ self swap mod 0 == ]) ;

amicables

| i j | 

  ListBuffer new
  20000 loop: i [
     i properDivs sum dup ->j i <= ifTrue: [ continue ]
     j properDivs sum i <> ifTrue: [ continue ]
     [ i, j ] over add
     ] ;</lang>
Output:
amicables .
[[220, 284], [1184, 1210], [2620, 2924], [5020, 5564], [6232, 6368], [10744, 10856], [12285, 14595], [17296, 18416]]

PARI/GP

<lang parigp>for(x=1,20000,my(y=sigma(x)-x); if(y>x && x == sigma(y)-y,print(x" "y)))</lang>

Output:
220 284
1184 1210
2620 2924
5020 5564
6232 6368
10744 10856
12285 14595
17296 18416

Pascal

Direct approach

Works with: Turbo Pascal
Works with: Free Pascal

This version mutates the Sieve of Eratoshenes from striking out factors into summing factors. The Pascal source compiles with Turbo Pascal (7, patched to avoid the zero divide problem for cpu speeds better than ~150MHz) except that the array limit is too large: 15,000 works but does not reach 20,000. The Free Pascal compiler however can handle an array of 20,000 elements. Because the sum of factors of N can exceed N an ad-hoc SumF procedure is provided, thus the search could continue past the table limit, but at a cost in calculation time.

Output is 

Chasing Chains of Sums of Factors of Numbers.
Perfect!! 6,
Perfect!! 28,
Amicable! 220,284,
Perfect!! 496,
Amicable! 1184,1210,
Amicable! 2620,2924,
Amicable! 5020,5564,
Amicable! 6232,6368,
Perfect!! 8128,
Amicable! 10744,10856,
Amicable! 12285,14595,
Sociable: 12496,14288,15472,14536,14264,
Sociable: 14316,19116,31704,47616,83328,177792,295488,629072,589786,294896,358336,418904,366556,274924,275444,243760,376736,381028,285778,152990,122410,97946,48976,45946,22976,22744,19916,17716,
Amicable! 17296,18416,

Source file:<lang pascal> 

Program SumOfFactors; uses crt; {Perpetrated by R.N.McLean, December MCMXCV}

//{$DEFINE ShowOverflow} {$IFDEF FPC} 

 {$MODE DELPHI}//tested with lots = 524*1000*1000 takes 75 secs generating KnownSum

{$ENDIF} 

 var outf: text;
 const Limit = 2147483647;
 const lots = 20000;       {This should be much bigger, but problems apply.}
 var KnownSum: array[1..lots] of longint;
 Function SumF(N: Longint): Longint;
  var f,f2,s,ulp: longint;
  Begin
   if n <= lots then SumF:=KnownSum[N] {Hurrah!}
    else
     begin      {This is really crude...}
      s:=1;     {1 is always a factor, but N is not.}
      f:=2;
      f2:=f*f;
      while f2 < N do
       begin
        if N mod f = 0 then
         begin  {We have a divisor, and its friend.}
          ulp:=f + (N div f);
          if s > Limit - ulp then begin SumF:=-666; exit; end;
          s:=s + ulp;
         end;
        f:=f + 1;
        f2:=f*f;
       end;
       if f2 = N then {A perfect square gets its factor in once only.}
        if s <= Limit - f then s:=s + f
         else begin SumF:=-667; exit; end;
      SumF:=s;
     end;
  End;
 var i,j,l,sf,fs: LongInt;
 const enuff = 666; {Only so much sociability.}
 var trail: array[0..enuff] of longint;
 BEGIN
  ClrScr;
  WriteLn('Chasing Chains of Sums of Factors of Numbers.');
  for i:=1 to lots do KnownSum[i]:=1; {Sigh. KnownSum:=1;}

{start summing every divisor } 

  for i:=2 to lots do
   begin
    j:=i + i;
    While j <= lots do    {Sigh. For j:=i + i:Lots:i do KnownSum[j]:=KnownSum[j] + i;}
    begin
      KnownSum[j]:=KnownSum[j] + i;
      j:=j + i;
     end;
   end; 

{Enough preparation.}
  Assign(outf,'Factors.txt'); ReWrite(Outf);
  WriteLn(Outf,'Chasing Chains of Sums of Factors of Numbers.');

  for i:=2 to lots do    {Search.}
   begin
    l:=0;
    sf:=SumF(i);
    while (sf > i) and (l < enuff) do
     begin
      l:=l + 1;
      trail[l]:=sf;
      sf:=SumF(sf);
     end;
    if l >= enuff then writeln('Rope ran out! ',i);

{$IFDEF ShowOverflow} 

    if sf < 0 then writeln('Overflow with ',i);

{$ENDIF} 

    if i = sf then      {A loop?}
     begin              {Yes. Reveal its members.}
      trail[0]:=i;      {The first.}
      if l = 0 then write('Perfect!! ')
       else if l = 1 then write('Amicable! ')
        else write('Sociable: ');
      for j:=0 to l do Write(Trail[j],',');
      WriteLn;
      if l = 0 then write(outf,'Perfect!! ')
       else if l = 1 then write(outf,'Amicable! ')
        else write(outf,'Sociable: ');
      for j:=0 to l do write(outf,Trail[j],',');
      WriteLn(outf);
     end;
   end;
  Close (outf);
 END.</lang>

More expansive.

a "normal" Version. Nearly fast as perl using nTheory. <lang pascal>program AmicablePairs; {$IFDEF FPC} 

  {$MODE DELPHI}
  {$H+}

{$ELSE} 

 {$APPTYPE CONSOLE}

{$ENDIF} uses 

 sysutils;

const 

 MAX = 20000;

//MAX = 20*1000*1000; type 

 tValue = LongWord;
 tpValue = ^tValue;
 tPower = array[0..31] of tValue;
 tIndex = record
            idxI,
            idxS : Uint64;
          end;

var 

 Indices      : array[0..511] of tIndex;
 //primes up to 65536 enough until 2^32
 primes       : array[0..6542] of tValue;

procedure InitPrimes; // sieve of erathosthenes without multiples of 2 type 

 tSieve = array[0..(65536-1) div 2] of char;

var 

 ESieve : ^tSieve;
 idx,i,j,p : LongINt;

Begin 

 new(ESieve);
 fillchar(ESieve^[0],SizeOF(tSieve),#1);
 primes[0] := 2;
 idx := 1;

 //sieving
 j := 1;
 p := 2*j+1;
 repeat
   if Esieve^[j] = #1 then
   begin
     i := (2*j+2)*j;// i := (sqr(p) -1) div 2;
     if i > High(tSieve) then
       BREAK;
     repeat
       ESIeve^[i] := #0;
       inc(i,p);
     until i > High(tSieve);
   end;
   inc(j);
   inc(p,2);
 until j >High(tSieve);

 //collecting
 For i := 1 to High(tSieve) do
   IF Esieve^[i] = #1 then
   Begin
     primes[idx] := 2*i+1;
     inc(idx);
     IF idx>High(primes) then
       BREAK;
   end;
 dispose(Esieve);

end;

procedure Su_append(n,factor:tValue;var su:string); var 

 q,p : tValue;

begin 

 p := 0;
 repeat
   q := n div factor;
   IF q*factor<>n then
     Break;
   inc(p);
   n := q;
 until false;
 IF p > 0 then
   IF p= 1 then
     su:= su+IntToStr(factor)+'*'
   else
     su:= su+IntToStr(factor)+'^'+IntToStr(p)+'*';

end;

procedure ProperDivs(n: Uint64); //output of prime factorization var 

 su : string;
 primNo : tValue;
 p:tValue;

begin 

 str(n:8,su);
 su:= su +' [';
 primNo := 0;
 p := primes[0];
 repeat
   Su_Append(n,p,su);
   inc(primNo);
   p := primes[primNo];
 until (p=0) OR (p*p >= n);
 p := n;
 Su_Append(n,p,su);
 su[length(su)] := ']';
 writeln(su);

end;

procedure AmPairOutput(cnt:tValue); var 

 i : tValue;
 r_max,r_min,r : double;

begin 

 r_max := 1.0;
 r_min := 16.0;
 For i := 0 to cnt-1 do
   with Indices[i] do
   begin
     r := IdxS/IDxI;
     writeln(i+1:4,IdxI:16,IDxS:16,' ratio ',r:10:7);
     IF r < 1 then
     begin
       writeln(i);
       readln;
       halt;
     end;
     if r_max < r then
       r_max := r
     else
       if r_min > r then
         r_min := r;
   IF cnt < 20 then
     begin
       ProperDivs(IdxI);
       ProperDivs(IdxS);
     end;
   end;
 writeln(' min ratio ',r_min:12:10);  writeln(' max ratio ',r_max:12:10);

end;

procedure SumOFProperDiv(n: tValue;var SumOfProperDivs:tValue); // calculated by prime factorization var 

 i,q, primNo, Prime,pot : tValue;
 SumOfDivs: tValue;

begin 

 i := N;
 SumOfDivs := 1;
 primNo := 0;
 Prime := Primes[0];
 q := i DIV Prime;
 repeat
   if q*Prime = i then
   Begin
     pot := 1;
     repeat
       i := q;
       q := i div Prime;
       Pot := Pot * Prime+1;
     until q*Prime <> i;
     SumOfDivs := SumOfDivs * pot;
   end;
   Inc(primNo);
   Prime := Primes[primNo];
   q := i DIV Prime;

   {check if i already prime}
   if Prime > q then
   begin
     prime := i;
     q := 1;
   end;
 until i = 1;
 SumOfProperDivs := SumOfDivs - N;

end;

function Check:tValue; const 

 //going backwards
 DIV23 : array[0..5] of byte =
          //== 5,4,3,2,1,0
              (1,0,0,0,1,0);

var 

 i,s,k,n : tValue;
 idx : nativeInt;

begin 

 n := 0;
 idx := 3;
 For i := 2 to MAX do
 begin
   //must be divisble by 2 or 3 ( n < High(tValue) < 1e14 )
   IF DIV23[idx] = 0 then
   begin
     SumOFProperDiv(i,s);
     //only 24.7...%
     IF s>i then
     Begin
       SumOFProperDiv(s,k);
       IF k = i then
       begin
         With indices[n] do
         begin
           idxI := i;
           idxS := s;
         end;
         inc(n);
       end;
     end;
   end;
   dec(idx);
   IF idx < 0 then
     idx := high(DIV23);
 end;
 result := n;

end;

var 

 T2,T1: TDatetime;
 APcnt: tValue;

begin 

 InitPrimes;
 T1:= time;
 APCnt:= Check;
 T2:= time;
 AmPairOutput(APCnt);
 writeln('Time to find amicable pairs ',FormatDateTime('HH:NN:SS.ZZZ' ,T2-T1));
 {$IFNDEF UNIX} readln;{$ENDIF}

end.</lang> Output 

   1             220             284 ratio  1.2909091
     220 [2^2*5*11*220]
     284 [2^2*284]
   2            1184            1210 ratio  1.0219595
    1184 [2^5*1184]
    1210 [2*5*11^2*1210]
   3            2620            2924 ratio  1.1160305
    2620 [2^2*5*2620]
    2924 [2^2*17*43*2924]
   4            5020            5564 ratio  1.1083665
    5020 [2^2*5*5020]
    5564 [2^2*13*5564]
   5            6232            6368 ratio  1.0218228
    6232 [2^3*19*41*6232]
    6368 [2^5*6368]
   6           10744           10856 ratio  1.0104244
   10744 [2^3*17*79*10744]
   10856 [2^3*23*59*10856]
   7           12285           14595 ratio  1.1880342
   12285 [3^3*5*7*13*12285]
   14595 [3*5*7*14595]
   8           17296           18416 ratio  1.0647549
   17296 [2^4*23*47*17296]
   18416 [2^4*18416]

Alternative

about 25-times faster. This will not output the amicable number unless both! numbers are under the given limit.

So there will be differences to "Table of n, a(n) for n=1..39374" https://oeis.org/A002025/b002025.txt Up to 524'000'000 the pairs found are only correct by number up to no. 437 460122410 and only 442 out of 455 are found, because some pairs exceed the limit. The limits of the ratio between the numbers of the amicable pair up to 1E14 are, based on b002025.txt: 

No.    lower            upper         
31447  52326552030976  52326637800704 ratio  1.0000016 
52326552030976 [2^8*563*6079*59723]
52326637800704 [2^8*797*1439*178223]

38336  92371445691525 154378742017851 ratio  1.6712821
 92371445691525 [3^2*5^2*7^2*11*13^2*23*29^2*233]
154378742017851 [3^2*13^2*53*337*5682671]


The distance check is being corrected, the lower number is now not limited. The used method is not useful for very high limits.

n = p[1]^a[1]*p[2]^a[2]*...p[l]^a[l]

sum of divisors(n) = s(n) = (p[1]^(a[1]+1) -1) / (p[1] -1) * ... * (p[l]^(a[l]+1) -1) / (p[l] -1) with

p[k]^(a[k]+1) -1) / (p[k] -1) = sum (i= [1..a[k]])(p[k]^i)

Using "Sieve of Erathosthenes"-style

<lang pascal>program AmicPair; {find amicable pairs in a limited region 2..MAX beware that >both< numbers must be smaller than MAX there are 455 amicable pairs up to 524*1000*1000 correct up to

  1. 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;

type 

 tValue = LongWord;
 tpValue = ^tValue;
 tDivSum = array[0..0] of tValue;// evil, but dynamic arrays are slower
 tpDivSum = ^tDivSum;
 tPower = array[0..31] of tValue;
 tIndex = record
            idxI,
            idxS : tValue;
          end;

var 

 power,
 PowerFac     : tPower;
 ds           : array of tValue;
 Indices      : array[0..511] of tIndex;
 DivSumField  : tpDivSum;
 MAX : tValue;

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 do var 

 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 

 n := 0;
 For i := 1 to MAX do
 begin
   //s = sum of proper divs (I)  == sum of divs (I) - I
   s := DivSumField^[i];
   IF (s <=MAX) AND (s>i) AND (DivSumField^[s]= i)then
   begin
     With indices[n] do
     begin
       idxI := i;
       idxS := s;
     end;
     inc(n);
   end;
 end;
 result := n;

end;

Procedure CalcPotfactor(prim:tValue); //PowerFac[k] = (prim^(k+1)-1)/(prim-1) == Sum (i=0..k) prim^i var 

 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; //return the first power <> 0 //power == n to base prim var 

 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;

procedure Sieve(prim: tValue); var 

 actNumber,idx : tValue;

begin 

 //sieve with "small" primes
 while prim*prim <= MAX do
 begin
   InitPW(prim);
   Begin
     //actNumber = actual number = n*prim
     actNumber := prim;
     idx := prim;
     while actNumber <= MAX do
     begin
       dec(idx);
       IF idx > 0 then
         DivSumField^[actNumber] *= PowerFac[0]
       else
       Begin
         DivSumField^[actNumber] *= PowerFac[NextPotCnt(prim)+1];
         idx := Prim;
       end;
       inc(actNumber,prim);
     end;
   end;
   //next prime
   repeat
     inc(prim);
   until DivSumField^[prim]= 1;//(DivSumField[prim] = 1);
 end;

 //sieve with "big" primes, only one factor is possible
 while 2*prim <= MAX do
 begin
   InitPW(prim);
   Begin
     actNumber := prim;
     idx := PowerFac[0];
     while actNumber <= MAX do
     begin
       DivSumField^[actNumber] *= idx;
       inc(actNumber,prim);
     end;
   end;
   repeat
     inc(prim);
   until DivSumField^[prim]= 1;
 end;

 For idx := 2 to MAX do
   dec(DivSumField^[idx],idx);

end;

var 

 T2,T1,T0: TDatetime;
 APcnt: tValue;
 i: NativeInt;

begin 

 MAX := 20000;
 IF  ParamCount > 0 then
   MAX := StrToInt(ParamStr(1));
 setlength(ds,MAX);
 DivSumField := @ds[0];
 T0:= time;
 For i := 1 to 1 do
 Begin
   Init;
   Sieve(2);
 end;
 T1:= time;

 APCnt := Check;
 T2:= time;
 AmPairOutput(APCnt);
 writeln(APCnt,' amicable pairs til ',MAX);
 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));
 setlength(ds,0);
 {$IFNDEF UNIX}
   readln;
 {$ENDIF}

end.</lang> output 

       220       284
  [1,2,4,5,10,11,20,22,44,55,110]
  [1,2,4,71,142]

      1184      1210
  [1,2,4,8,16,32,37,74,148,296,592]
  [1,2,5,10,11,22,55,110,121,242,605]

      2620      2924
  [1,2,4,5,10,20,131,262,524,655,1310]
  [1,2,4,17,34,43,68,86,172,731,1462]

      5020      5564
  [1,2,4,5,10,20,251,502,1004,1255,2510]
  [1,2,4,13,26,52,107,214,428,1391,2782]

      6232      6368
  [1,2,4,8,19,38,41,76,82,152,164,328,779,1558,3116]
  [1,2,4,8,16,32,199,398,796,1592,3184]

     10744     10856
  [1,2,4,8,17,34,68,79,136,158,316,632,1343,2686,5372]
  [1,2,4,8,23,46,59,92,118,184,236,472,1357,2714,5428]

     12285     14595
  [1,3,5,7,9,13,15,21,27,35,39,45,63,65,91,105,117,135,189,195,273,315,351,455,585,819,945,1365,1755,2457,4095]
  [1,3,5,7,15,21,35,105,139,417,695,973,2085,2919,4865]

     17296     18416
  [1,2,4,8,16,23,46,47,92,94,184,188,368,376,752,1081,2162,4324,8648]
  [1,2,4,8,16,1151,2302,4604,9208]

8 amicable numbers up to 20000
00:00:00.000

.... Test with 524*1000*1000 Linux32, FPC 3.0.1, i4330 3.5 Ghz //Win32 swaps first to allocate 2 GB )
 440   475838415   514823985 ratio  1.0819303
 441   491373104   511419856 ratio  1.0407974
 442   509379344   523679536 ratio  1.0280738
 max ratio     1.3537
442 amicable pairs til 524000000
Time to calc sum of divs    00:00:12.601
Time to find amicable pairs 00:00:02.557

Perl

Not particularly clever, but instant for this example, and does up to 20 million in 11 seconds.

Library: ntheory

<lang perl>use ntheory qw/divisor_sum/; for my $x (1..20000) { 

 my $y = divisor_sum($x)-$x;
 say "$x $y" if $y > $x && $x == divisor_sum($y)-$y;

}</lang>

Output:
220 284
1184 1210
2620 2924
5020 5564
6232 6368
10744 10856
12285 14595
17296 18416

Perl 6

Works with: rakudo version 2015-10-31

<lang perl6>sub propdivsum (\x) { 

   my @l = x > 1, gather for 2 .. x.sqrt.floor -> \d {
       my \y = x div d;
       if y * d == x { take d; take y unless y == d }
   }
   [+] gather @l.deepmap(*.take);

}

for 1..20000 -> $i { 

   my $j = propdivsum($i);
   say "$i $j" if $j > $i and $i == propdivsum($j);

}</lang>

Output:
220 284
1184 1210
2620 2924
5020 5564
6232 6368
10744 10856
12285 14595
17296 18416

Phix

<lang Phix>integer n for m=1 to 20000 do 

   n = sum(factors(m,-1))
   if m<n and m=sum(factors(n,-1)) then ?{m,n} end if

end for</lang>

Output:
{220,284}
{1184,1210}
{2620,2924}
{5020,5564}
{6232,6368}
{10744,10856}
{12285,14595}
{17296,18416}

PicoLisp

<lang 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 

  (for I 20000
     (let X (factor-sum I)
        (and
           (< I X)
           (= I (factor-sum X))
           (println I X) ) ) ) )</lang>
Output:
220 284
1184 1210
2620 2924
5020 5564
6232 6368
10744 10856
12285 14595
17296 18416
0.734 sec

PL/I

Translation of: REXX

<lang pli>*process source xref; 

ami: Proc Options(main);
p9a=time();
Dcl (p9a,p9b,p9c) Pic'(9)9';
Dcl sumpd(20000) Bin Fixed(31);
Dcl pd(300) Bin Fixed(31);
Dcl npd     Bin Fixed(31);
Dcl (x,y)   Bin Fixed(31);

Do x=1 To 20000;
  Call proper_divisors(x,pd,npd);
  sumpd(x)=sum(pd,npd);
  End;
p9b=time();
Put Edit('sum(pd) computed in',(p9b-p9a)/1000,' seconds elapsed')
        (Skip,col(7),a,f(6,3),a);

Do x=1 To 20000;
  Do y=x+1 To 20000;
    If y=sumpd(x) &
       x=sumpd(y) Then
      Put Edit(x,y,' found after ',elapsed(),' seconds')
              (Skip,2(f(6)),a,f(6,3),a);
    End;
  End;
Put Edit(elapsed(),' seconds total search time')(Skip,f(6,3),a);

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;

elapsed: Proc Returns(Dec Fixed(6,3));
p9c=time();
Return((p9c-p9b)/1000);
End;
End;</lang>
Output:
      sum(pd) computed in 0.510 seconds elapsed
   220   284 found after  0.010 seconds
  1184  1210 found after  0.060 seconds
  2620  2924 found after  0.110 seconds
  5020  5564 found after  0.210 seconds
  6232  6368 found after  0.260 seconds
 10744 10856 found after  2.110 seconds
 12285 14595 found after  2.150 seconds
 17296 18416 found after  2.240 seconds
 2.250 seconds total search time

Prolog

Works with: SWI-Prolog 7

With some guidance from other solutions here:

<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.

assoc_num_divsSum_in_range(Low, High, Assoc) :- 

   findall( Num-DivSum,
            ( between(Low, High, Num),
              aggregate_all( sum(D),
                             proper_divisor(Num, D),
                             DivSum )),
            Pairs ),
   list_to_assoc(Pairs, Assoc).

get_amicable_pair(Assoc, M-N) :- 

   gen_assoc(M, Assoc, N),
   M < N,
   get_assoc(N, Assoc, M).

amicable_pairs_under_20000(Pairs) :- 

   assoc_num_divsSum_in_range(1,20000, Assoc),
   findall(P, get_amicable_pair(Assoc, P), Pairs).</lang>

Output:

<lang prolog>?- amicable_pairs_under_20000(R). R = [220-284, 1184-1210, 2620-2924, 5020-5564, 6232-6368, 10744-10856, 12285-14595, 17296-18416].</lang>

Python

Importing Proper divisors from prime factors: <lang python>from proper_divisors import proper_divs

def amicable(rangemax=20000): 

   n2divsum = {n: sum(proper_divs(n)) for n in range(1, rangemax + 1)}
   for num, divsum in n2divsum.items():
       if num < divsum and divsum <= rangemax and n2divsum[divsum] == num:
           yield num, divsum

if __name__ == '__main__': 

   for num, divsum in amicable():
       print('Amicable pair: %i and %i With proper divisors:\n    %r\n    %r'
             % (num, divsum, sorted(proper_divs(num)), sorted(proper_divs(divsum))))</lang>
Output:
Amicable pair: 220 and 284 With proper divisors:
    [1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110]
    [1, 2, 4, 71, 142]
Amicable pair: 1184 and 1210 With proper divisors:
    [1, 2, 4, 8, 16, 32, 37, 74, 148, 296, 592]
    [1, 2, 5, 10, 11, 22, 55, 110, 121, 242, 605]
Amicable pair: 2620 and 2924 With proper divisors:
    [1, 2, 4, 5, 10, 20, 131, 262, 524, 655, 1310]
    [1, 2, 4, 17, 34, 43, 68, 86, 172, 731, 1462]
Amicable pair: 5020 and 5564 With proper divisors:
    [1, 2, 4, 5, 10, 20, 251, 502, 1004, 1255, 2510]
    [1, 2, 4, 13, 26, 52, 107, 214, 428, 1391, 2782]
Amicable pair: 6232 and 6368 With proper divisors:
    [1, 2, 4, 8, 19, 38, 41, 76, 82, 152, 164, 328, 779, 1558, 3116]
    [1, 2, 4, 8, 16, 32, 199, 398, 796, 1592, 3184]
Amicable pair: 10744 and 10856 With proper divisors:
    [1, 2, 4, 8, 17, 34, 68, 79, 136, 158, 316, 632, 1343, 2686, 5372]
    [1, 2, 4, 8, 23, 46, 59, 92, 118, 184, 236, 472, 1357, 2714, 5428]
Amicable pair: 12285 and 14595 With proper divisors:
    [1, 3, 5, 7, 9, 13, 15, 21, 27, 35, 39, 45, 63, 65, 91, 105, 117, 135, 189, 195, 273, 315, 351, 455, 585, 819, 945, 1365, 1755, 2457, 4095]
    [1, 3, 5, 7, 15, 21, 35, 105, 139, 417, 695, 973, 2085, 2919, 4865]
Amicable pair: 17296 and 18416 With proper divisors:
    [1, 2, 4, 8, 16, 23, 46, 47, 92, 94, 184, 188, 368, 376, 752, 1081, 2162, 4324, 8648]
    [1, 2, 4, 8, 16, 1151, 2302, 4604, 9208]

Racket

With Proper_divisors#Racket in place: <lang racket>#lang racket (require "proper-divisors.rkt") (define SCOPE 20000)

(define P 

 (let ((P-v (vector)))
   (λ (n)
     (set! P-v (fold-divisors P-v n 0 +))
     (vector-ref P-v n))))
returns #f if not an amicable number, amicable pairing otherwise

(define (amicable? n) 

 (define m (P n))
 (define m-sod (P m))
 (and (= m-sod n)
      (< m n) ; each pair exactly once, also eliminates perfect numbers
      m))

(void (amicable? SCOPE)) ; prime the memoisation

(for* ((n (in-range 1 (add1 SCOPE))) 

      (m (in-value (amicable? n)))
      #:when m)
 (printf #<<EOS

amicable pair: ~a, ~a 

 ~a: divisors: ~a
 ~a: divisors: ~a


EOS 

         n m n (proper-divisors n)  m (proper-divisors m)))

</lang>

Output:
amicable pair: 284, 220
  284: divisors: (1 2 4 71 142)
  220: divisors: (1 2 4 5 10 11 20 22 44 55 110)

amicable pair: 1210, 1184
  1210: divisors: (1 2 5 10 11 22 55 110 121 242 605)
  1184: divisors: (1 2 4 8 16 32 37 74 148 296 592)

amicable pair: 2924, 2620
  2924: divisors: (1 2 4 17 34 43 68 86 172 731 1462)
  2620: divisors: (1 2 4 5 10 20 131 262 524 655 1310)

amicable pair: 5564, 5020
  5564: divisors: (1 2 4 13 26 52 107 214 428 1391 2782)
  5020: divisors: (1 2 4 5 10 20 251 502 1004 1255 2510)

amicable pair: 6368, 6232
  6368: divisors: (1 2 4 8 16 32 199 398 796 1592 3184)
  6232: divisors: (1 2 4 8 19 38 41 76 82 152 164 328 779 1558 3116)

amicable pair: 10856, 10744
  10856: divisors: (1 2 4 8 23 46 59 92 118 184 236 472 1357 2714 5428)
  10744: divisors: (1 2 4 8 17 34 68 79 136 158 316 632 1343 2686 5372)

amicable pair: 14595, 12285
  14595: divisors: (1 3 5 7 15 21 35 105 139 417 695 973 2085 2919 4865)
  12285: divisors: (1 3 5 7 9 13 15 21 27 35 39 45 63 65 91 105 117 135 189 195 273 315 351 455 585 819 945 1365 1755 2457 4095)

amicable pair: 18416, 17296
  18416: divisors: (1 2 4 8 16 1151 2302 4604 9208)
  17296: divisors: (1 2 4 8 16 23 46 47 92 94 184 188 368 376 752 1081 2162 4324 8648)

REXX

version 1

<lang rexx>Call time 'R' Do x=1 To 20000 

 pd=proper_divisors(x)
 sumpd.x=sum(pd)
 End

Say 'sum(pd) computed in' time('E') 'seconds' Call time 'R' Do x=1 To 20000 

 /* If x//1000=0 Then Say x time() */
 Do y=x+1 To 20000
   If y=sumpd.x &,
      x=sumpd.y Then
   Say x y 'found after' time('E') 'seconds'
   End
 End

Say time('E') 'seconds total search time' Exit

proper_divisors: Procedure Parse Arg n Pd= If n=1 Then Return 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)

sum: Procedure Parse Arg list sum=0 Do i=1 To words(list) 

 sum=sum+word(list,i)
 End

Return sum</lang>

Output:
sum(pd) computed in 48.502000 seconds
220 284 found after 3.775000 seconds
1184 1210 found after 21.611000 seconds
2620 2924 found after 46.817000 seconds
5020 5564 found after 84.296000 seconds
6232 6368 found after 100.918000 seconds
10744 10856 found after 150.126000 seconds
12285 14595 found after 162.124000 seconds
17296 18416 found after 185.600000 seconds
188.836000 seconds total search time

version 2

This REXX version allows the specification of the upper limit (for the searching of amicable pairs).

Some optimization was incorporated by using a   sigma   function,   which was a re-coded   proper divisors   (Pdivs)   function,
which was taken from the REXX language entry for Rosetta code task   integer factors.

Other optimizations were incorporated which took advantage of several well-known generalizations about amicable pairs.

The generation/summation is about forty times faster;   searching is about two orders of magnitude faster. <lang rexx>/*REXX program finds and displays all amicable pairs up to a given number. */ parse arg H .; if H== then H=20000 /*get optional arguments (high limit).*/ w=length(H)  ; low=220 /*W: used for columnar output alignment*/ @.=. /* [↑] LOW is lowest amicable number. */ 

   do k=low  for H-low;  _=sigma(k)   /*generate sigma sums for a range of #s*/
   if _>=low  then @.k=_              /*only keep the pertinent sigma sums.  */
   end   /*k*/                        /* [↑]   process a range of integers.  */
  1. =0 /*number of amicable pairs found so far*/
   do   m=low  to  H;    n=@.m        /*start the search at the lowest number*/
   if n==.  then iterate              /*if not pertinent,  then ignore the #.*/
   if m==@.n  then do                 /*If equal, might be an amicable number*/
                   if m==n  then iterate        /*skip any perfect numbers.  */
                   #=#+1                        /*bump amicable pair counter.*/
                   say right(m,w)  ' and ' right(n,w)  " are an amicable pair."
                   m=n                          /*start M (DO index)  from N.*/
                   end
   end    /*m*/

say say # 'amicable pairs found up to' H /*display count of the amicable pairs. */ exit /*stick a fork in it, we're all done. */ /*────────────────────────────────────────────────────────────────────────────*/ sigma: procedure; parse arg x; od=x//2 /*use either EVEN or ODD integers. */ s=1 /*set initial sigma sum to one. ___*/ 

     do j=2+od  by 1+od  while  j*j<x /*divide by all integers up to the √ x */
     if x//j==0  then  s=s + j + x%j  /*add the two divisors to the sum.     */
     end   /*j*/                      /* [↑]  %  is REXX integer division.   */
                                      /* [↓]  adjust for square.          ___*/
           /*if j*j==x  then s=s+j*/  /*Was X a square?  If so, add the  √ x */
                                      /* [↑]  not needed for amicable #s.    */

return s /*return the sum of the divisors. */</lang> output   when using the default input: 

  220  and    284  are an amicable pair.
 1184  and   1210  are an amicable pair.
 2620  and   2924  are an amicable pair.
 5020  and   5564  are an amicable pair.
 6232  and   6368  are an amicable pair.
10744  and  10856  are an amicable pair.
12285  and  14595  are an amicable pair.
17296  and  18416  are an amicable pair.

8 amicable pairs found up to 20000

version 3

This REXX version is optimized to take advantage of the lowest ending-single-digit amicable number.

The optimization makes it about another 30% faster when searching for amicable numbers up to one million. <lang rexx>/*REXX program finds and displays all amicable pairs up to a given number. */ parse arg H .; if H== then H=20000 /*get optional arguments (high limit).*/ w=length(H)  ; low=220 /*W: used for columnar output alignment*/ x=220 34765731 6232 87633 284 12285 10856 36939357 6368 5684679 /*S minimums.*/ y=220 34765731 6232 69615 220 12285 10744 34765731 6232 5357625 /*D minimums.*/ 

  do i=0 for 10; $.i=word(x,i+1); L.i=word(y,i+1); end /*minimum amicable #s.*/
  1. =0 /*number of amicable pairs found so far*/

@.= /* [↑] LOW is lowest amicable number. */ 

    do k=low  for H-low               /*generate sigma sums for a range of #s*/
    parse var k    -1  D            /*obtain last decimal digit of   K.    */
    if k<$.D    then iterate          /*if no need to compute, then skip it. */
       _=sigma(k)                     /*generate sigma sum for the number  K.*/
    if _<L.D    then iterate          /*if sigma sum is too low, then skip it*/
    @.k=_                             /*only keep the pertinent sigma sums.  */
    if k==@._  then do                       /*is it a possible amicable # ? */
                    if _==k  then iterate    /*ignore any perfect numbers.   */
                    #=#+1                    /*bump the amicable pair counter*/
                    say right(_,w) ' and ' right(k,w)  " are an amicable pair."
                    end
    end   /*k*/                       /* [↑]   process a range of integers.  */

say say # 'amicable pairs found up to' H /*display the count of amicable pairs. */ exit /*stick a fork in it, we're all done. */ /*────────────────────────────────────────────────────────────────────────────*/ sigma: procedure; parse arg x; od=x//2 /*use either EVEN or ODD integers. */ s=1 /*set initial sigma sum to one. ___*/ 

     do j=2+od  by 1+od  while  j*j<x /*divide by all integers up to the √ x */
     if x//j==0  then  s=s + j + x%j  /*add the two divisors to the sum.     */
     end   /*j*/                      /* [↑]  %  is REXX integer division.   */
                                      /* [↓]  adjust for square.          ___*/
           /*if j*j==x  then s=s+j*/  /*Was X a square?  If so, add the  √ x */
                                      /* [↑]  not needed for amicable #s.    */

return s /*return the sum of the divisors. */</lang> output   is the same as the 2nd REXX version.

version 4

This REXX version is optimized to use the   integer square root of X   in the   sigma   function   (instead of
computing the square of   J   to see if that value exceeds   X).

The optimization makes it about another 20% faster when searching for amicable numbers up to one million. <lang rexx>/*REXX program finds and displays all amicable pairs up to a given number. */ parse arg H .; if H== then H=20000 /*get optional arguments (high limit).*/ w=length(H)  ; low=220 /*W: used for columnar output alignment*/ x=220 34765731 6232 87633 284 12285 10856 36939357 6368 5684679 /*S minimums.*/ y=220 34765731 6232 69615 220 12285 10744 34765731 6232 5357625 /*D minimums.*/ 

  do i=0 for 10; $.i=word(x,i+1); L.i=word(y,i+1); end /*minimum amicable #s.*/
  1. =0 /*number of amicable pairs found so far*/

@.= /* [↑] LOW is lowest amicable number. */ 

    do k=low  for H-low               /*generate sigma sums for a range of #s*/
    parse var k    -1  D            /*obtain last decimal digit of   K.    */
    if k<$.D    then iterate          /*if no need to compute, then skip it. */
       _=sigma(k)                     /*generate sigma sum for the number  K.*/
    if _<L.D    then iterate          /*if sigma sum is too low, then skip it*/
    @.k=_                             /*only keep the pertinent sigma sums.  */
    if k==@._  then do                       /*is it a possible amicable # ? */
                    if _==k  then iterate    /*ignore any perfect numbers.   */
                    #=#+1                    /*bump the amicable pair counter*/
                    say right(_,w) ' and ' right(k,w)  " are an amicable pair."
                    end
    end   /*k*/                       /* [↑]   process a range of integers.  */

say say # 'amicable pairs found up to' H /*display the count of amicable pairs. */ 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

/*────────────────────────────────────────────────────────────────────────────*/ sigma: procedure; parse arg x; od=x//2 /*use either EVEN or ODD integers. */ s=1 /*set initial sigma sum to unity. ___*/ 

     do j=2+od  by 1+od  to iSqrt(x)  /*divide by all integers up to the √ x */
     if x//j==0  then  s=s + j + x%j  /*add the two divisors to the sum.     */
     end   /*j*/                      /* [↑]  %  is REXX integer division.   */

return s /*return the sum of the divisors. */</lang> output   is the same as the 2nd REXX version.

version 5

This REXX version is optimized by bringing the functions in-line   (which minimizes the overhead of invoking two
internal functions),   and by taking advantage of the fact that both numbers in the amicable pair must have the
same parity.

This method of coding has the disadvantage in that the code is less idiomatic and less readable.

The optimization makes it about another 15% faster when searching for amicable numbers up to one million. <lang rexx>/*REXX program finds and displays all amicable pairs up to a given number. */ parse arg H .; if H== then H=20000 /*get optional arguments (high limit).*/ w=length(H)  ; low=220 /*W: used for columnar output alignment*/ x=220 34765731 6232 87633 284 12285 10856 36939357 6368 5684679 /*S minimums.*/ y=220 34765731 6232 69615 220 12285 10744 34765731 6232 5357625 /*D minimums.*/ 

  do i=0 for 10; $.i=word(x,i+1); L.i=word(y,i+1); end /*minimum amicable #s.*/
  1. =0 /*number of amicable pairs found so far*/

@.= /* [↑] LOW is lowest amicable number. */ 

    do k=low  for H-low               /*generate sigma sums for a range of #s*/
    parse var k    -1  D            /*obtain last decimal digit of   K.    */
    if k<$.D     then iterate         /*if no need to compute, then skip it. */
    od=k//2                           /*OD:   set to  unity  if   K   is odd.*/
    z=k;  r=0;  q=1;      do while q<=z;  q=q*4;  end  /*R will be iSqrt of Z*/
        do while q>1; q=q%4; _=z-r-q; r=r%2; if _>=0 then do;z=_;r=r+q; end;end
    s=1                               /*set initial sigma sum to unity.   ___*/
        do j=2+od  by 1+od   to r     /*divide by all integers up to the √ K */
        if k//j==0  then s=s+ j + k%j /*add the two divisors to the sum.     */
        end   /*j*/                   /* [↑]  %  is REXX integer division.   */
    if s<L.D     then iterate         /*if sigma sum is too low, then skip it*/
    if od\==s//2 then iterate         /*if both don't have same parity, skip.*/
    @.k=s                             /*only keep the pertinent sigma sums.  */
    if k==@.s  then do                       /*is it a possible amicable # ? */
                    if s==k  then iterate    /*ignore any perfect numbers.   */
                    #=#+1                    /*bump the amicable pair counter*/
                    say right(s,w) ' and ' right(k,w)  " are an amicable pair."
                    end
    end   /*k*/                       /* [↑]   process a range of integers.  */

say say # 'amicable pairs found up to' H /*display the count of amicable pairs. */ 

                                      /*stick a fork in it,  we're all done. */</lang>

output   is the same as the 2nd REXX version.

Ruby

With proper_divisors#Ruby in place: <lang ruby>h = {} (1..20_000).each{|n| h[n] = n.proper_divisors.inject(:+)} h.select{|k,v| h[v] == k && k < v}.each do |key,val| # k<v filters out doubles and perfects 

 puts "#{key} and #{val}"

end </lang>

Output:


220 and 284
1184 and 1210
2620 and 2924
5020 and 5564
6232 and 6368
10744 and 10856
12285 and 14595
17296 and 18416

Rust

<lang rust>fn sum_of_divisors(val: u32) -> u32 { 

   (1..val/2+1).filter(|n| val % n == 0)
               .fold(0, |sum, n| sum + n)

}

fn main() { 

   let iter = (1..20_000).map(|i| (i, sum_of_divisors(i)))
                         .filter(|&(i, div_sum)| i > div_sum);

   for (i, sum1) in iter {
       if sum_of_divisors(sum1) == i {
          println!("{} {}", i, sum1);
       }
   }

}</lang>

Output:
284 220
1210 1184
2924 2620
5564 5020
6368 6232
10856 10744
14595 12285
18416 17296

Scala

<lang Scala>def properDivisors(n: Int) = (1 to n/2).filter(i => n % i == 0) val divisorsSum = (1 to 20000).map(i => i -> properDivisors(i).sum).toMap val result = divisorsSum.filter(v => v._1 < v._2 && divisorsSum.get(v._2) == Some(v._1))

println( result mkString ", " )</lang>

Output:
5020 -> 5564, 220 -> 284, 6232 -> 6368, 17296 -> 18416, 2620 -> 2924, 10744 -> 10856, 12285 -> 14595, 1184 -> 1210

Sidef

Translation of: Perl 6

<lang ruby>func propdivsum(x) { 

   gather {
       2.to(x.isqrt).each { |d|
           if (x %% d) {
               take(d)
               take(x/d) if !x.is_sqr
           }
       }
   }.sum(x > 0 ? 1 : 0)

}

1.to(20000).each { |i| 

   var j = propdivsum(i)
   say "#{i} #{j}" if (j>i && i==propdivsum(j))

}</lang>

Output:
220 284
1184 1210
2620 2924
5020 5564
6232 6368
10744 10856
12285 14595
17296 18416

Swift

<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)

}


func sumDivs(n:Int) -> Int { 

   struct Cache { static var sum = [Int:Int]() }
   
   if let sum = Cache.sum[n] { return sum }
   
   let sum = properDivs(n).reduce(0) { $0 + $1 }
   
   Cache.sum[n] = sum
   
   return sum

}

func amicable(n:Int, m:Int) -> Bool { 

   if n == m { return false }
   
   if sumDivs(n) != m || sumDivs(m) != n { return false }
   
   return true

}

var pairs = [(Int, Int)]()

for n in 1 ..< 20_000 { 

   for m in n+1 ... 20_000 {
       if amicable(n, m) {
           pairs.append(n, m)
           println("\(n, m)")
       }
   }

}</lang>

Alternative

about 800 times faster.<lang Swift>import func Darwin.sqrt

func sqrt(x:Int) -> Int { return Int(sqrt(Double(x))) }

func sigma(n: Int) -> Int { 

   if n == 1 { return 0 }          // definition of aliquot sum
   
   var result = 1
   let root = sqrt(n)
   
   for var div = 2; div <= root; ++div {
       if n % div == 0 {
           result += div + n/div
       }
       
   }
   if root*root == n { result -= root }
   
   return (result)

}

func amicables (upTo: Int) -> () { 

   var aliquot = Array(count: upTo+1, repeatedValue: 0)

   for i in 1 ... upTo {           // fill lookup array
       aliquot[i] = sigma(i)
   }
   
for i in 1 ... upTo {
       let a = aliquot[i]
       if a > upTo {continue}      //second part of pair out-of-bounds

       if a == i {continue}        //skip perfect numbers
       
       if i == aliquot[a] {
           print("\(i, a)")
           aliquot[a] = upTo+1     //prevent second display of pair
       }
   }

}

amicables(20_000)</lang>

Output:
(220, 284)
(1184, 1210)
(2620, 2924)
(5020, 5564)
(6232, 6368)
(10744, 10856)
(12285, 14595)
(17296, 18416)

Tcl

<lang tcl>proc properDivisors {n} { 

   if {$n == 1} return
   set divs 1
   set sum 1
   for {set i 2} {$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 } } 

   }
   return [list $sum $divs]

}

proc amicablePairs {limit} { 

   set result {}
   set sums [set divs {{}}]
   for {set n 1} {$n < $limit} {incr n} {

lassign [properDivisors $n] sum d lappend sums $sum lappend divs [lsort -integer $d] 

   }
   for {set n 1} {$n < $limit} {incr n} {

set nsum [lindex $sums $n] for {set m 1} {$m < $n} {incr m} { if {$n==[lindex $sums $m] && $m==$nsum} { lappend result $m $n [lindex $divs $m] [lindex $divs $n] } } 

   }
   return $result

}

foreach {m n md nd} [amicablePairs 20000] { 

   puts "$m and $n are an amicable pair with these proper divisors"
   puts "\t$m : $md"
   puts "\t$n : $nd"

}</lang>

Output:
220 and 284 are an amicable pair with these proper divisors
	220 : 1 2 4 5 10 11 20 22 44 55 110
	284 : 1 2 4 71 142
1184 and 1210 are an amicable pair with these proper divisors
	1184 : 1 2 4 8 16 32 37 74 148 296 592
	1210 : 1 2 5 10 11 22 55 110 121 242 605
2620 and 2924 are an amicable pair with these proper divisors
	2620 : 1 2 4 5 10 20 131 262 524 655 1310
	2924 : 1 2 4 17 34 43 68 86 172 731 1462
5020 and 5564 are an amicable pair with these proper divisors
	5020 : 1 2 4 5 10 20 251 502 1004 1255 2510
	5564 : 1 2 4 13 26 52 107 214 428 1391 2782
6232 and 6368 are an amicable pair with these proper divisors
	6232 : 1 2 4 8 19 38 41 76 82 152 164 328 779 1558 3116
	6368 : 1 2 4 8 16 32 199 398 796 1592 3184
10744 and 10856 are an amicable pair with these proper divisors
	10744 : 1 2 4 8 17 34 68 79 136 158 316 632 1343 2686 5372
	10856 : 1 2 4 8 23 46 59 92 118 184 236 472 1357 2714 5428
12285 and 14595 are an amicable pair with these proper divisors
	12285 : 1 3 5 7 9 13 15 21 27 35 39 45 63 65 91 105 117 135 189 195 273 315 351 455 585 819 945 1365 1755 2457 4095
	14595 : 1 3 5 7 15 21 35 105 139 417 695 973 2085 2919 4865
17296 and 18416 are an amicable pair with these proper divisors
	17296 : 1 2 4 8 16 23 46 47 92 94 184 188 368 376 752 1081 2162 4324 8648
	18416 : 1 2 4 8 16 1151 2302 4604 9208

VBScript

Not at all optimal. :-( <lang VBScript>start = Now Set nlookup = CreateObject("Scripting.Dictionary") Set uniquepair = CreateObject("Scripting.Dictionary")

For 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 nlookup.Add i,sum Next

For j = 1 To 20000 sum = 0 For m = 1 To 20000 If m < j Then If j Mod m = 0 Then sum = sum + m End If End If Next If nlookup.Exists(sum) And nlookup.Item(sum) = j And j <> sum _ And uniquepair.Exists(sum) = False Then uniquepair.Add j,sum End If Next

For Each key In uniquepair.Keys WScript.Echo key & ":" & uniquepair.Item(key) Next

WScript.Echo "Execution Time: " & DateDiff("s",Start,Now) & " seconds"</lang>

Output:
220:284
1184:1210
2620:2924
5020:5564
6232:6368
10744:10856
12285:14595
17296:18416
Execution Time: 162 seconds

zkl

Slooooow <lang zkl>fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) } const N=20000; sums:=[1..N].pump(T(-1),fcn(n){ properDivs(n).sum(0) }); [0..].zip(sums).filter('wrap([(n,s)]){ (n<s<=N) and sums[s]==n }).println();</lang>

Output:
L(L(220,284),L(1184,1210),L(2620,2924),L(5020,5564),L(6232,6368),L(10744,10856),L(12285,14595),L(17296,18416))
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