Proper divisors: Difference between revisions

From Rosetta Code
Content added Content deleted
(→‎{{header|ALGOL 68}}: Use a less pessimised algorithm)
(added GFA Basic example)
Line 766: Line 766:
from 1 to 20000
from 1 to 20000
15120 has max proper divisors: 79
15120 has max proper divisors: 79
</pre>

=={{header|GFA Basic}}==

<lang>
OPENW 1
CLEARW 1
'
' Array f% is used to hold the divisors
DIM f%(SQR(20000)) ! cannot redim arrays, so set size to largest needed
'
' 1. Show proper divisors of 1 to 10, inclusive
'
FOR i%=1 TO 10
num%=@proper_divisors(i%)
PRINT "Divisors for ";i%;":";
FOR j%=1 TO num%
PRINT " ";f%(j%);
NEXT j%
PRINT
NEXT i%
'
' 2. Find (smallest) number <= 20000 with largest number of proper divisors
'
result%=1 ! largest so far
number%=0 ! its number of divisors
FOR i%=1 TO 20000
num%=@proper_divisors(i%)
IF num%>number%
result%=i%
number%=num%
ENDIF
NEXT i%
PRINT "Largest number of divisors is ";number%;" for ";result%
'
~INP(2)
CLOSEW 1
'
' find the proper divisors of n%, placing results in f%
' and return the number found
'
FUNCTION proper_divisors(n%)
LOCAL i%,root%,count%
'
ARRAYFILL f%(),0
count%=1 ! index of next slot in f% to fill
'
IF n%>1
f%(count%)=1
count%=count%+1
root%=SQR(n%)
FOR i%=2 TO root%
IF n% MOD i%=0
f%(count%)=i%
count%=count%+1
IF i%*i%<>n% ! root% is an integer, so check if i% is actual square root of n%
f%(count%)=n%/i%
count%=count%+1
ENDIF
ENDIF
NEXT i%
ENDIF
'
RETURN count%-1
ENDFUNC
</lang>

Output is:
<pre>
Divisors for 1:
Divisors for 2: 1
Divisors for 3: 1
Divisors for 4: 1 2
Divisors for 5: 1
Divisors for 6: 1 2 3
Divisors for 7: 1
Divisors for 8: 1 2 4
Divisors for 9: 1 3
Divisors for 10: 1 2 5
Largest number of divisors is 79 for 15120
</pre>
</pre>


Revision as of 10:41, 11 May 2016

Proper divisors is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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

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


Examples

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


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


Show all output here.


Cf.



ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

<lang algol68># MODE to hold an element of a list of proper divisors # MODE DIVISORLIST = STRUCT( INT divisor, REF DIVISORLIST next );

  1. end of divisor list value #

REF DIVISORLIST nil divisor list = REF DIVISORLIST(NIL);

  1. resturns a DIVISORLIST containing the proper divisors of n #
  2. if n = 1, 0 or -1, we return no divisors #

PROC proper divisors = ( INT n )REF DIVISORLIST: 

    BEGIN
        REF DIVISORLIST result   := nil divisor list;
        REF DIVISORLIST end list := result;
        INT abs n  = ABS n;
        IF abs n > 1 THEN
            # build the list of divisors backeards, so they are  #
            # returned in ascending order                        #
            INT root n = ENTIER sqrt( abs n );
            FOR d FROM root n BY -1 TO 2 DO
                IF abs n MOD d = 0 THEN
                    # found another divisor                      #
                    result := HEAP DIVISORLIST
                           := DIVISORLIST( d, result );
                    IF end list IS nil divisor list THEN
                        # first result                           #
                        end list := result
                    FI;
                    IF d * d /= n THEN
                        # add the other divisor to the end of    #
                        # the list                               #
                        next OF end list := HEAP DIVISORLIST
                                         := DIVISORLIST( abs n OVER d, nil divisor list );
                        end list         := next OF end list
                    FI
                FI
            OD;
            # 1 is always a proper divisor of numbers > 1        #
            result := HEAP DIVISORLIST
                   := DIVISORLIST( 1, result )
        FI;
        result
    END # proper divisors # ;
  1. returns the number of divisors in a DIVISORLIST #

PROC count divisors = ( REF DIVISORLIST list )INT: 

    BEGIN
       INT result := 0;
       REF DIVISORLIST divisors := list;
       WHILE divisors ISNT nil divisor list DO
           result +:= 1;
           divisors := next OF divisors
       OD;
       result
    END # count divisors # ;
  1. find the proper divisors of 1 : 10 #

FOR n TO 10 DO 

   REF DIVISORLIST divisors := proper divisors( n );
   print( ( "Proper divisors of: ", whole( n, -2 ), ": " ) );
   WHILE divisors ISNT nil divisor list DO
       print( ( " ", whole( divisor OF divisors, 0 ) ) );
       divisors := next OF divisors
   OD;
   print( ( newline ) )

OD;

  1. find the first/only number in 1 : 20 000 with the most divisors #

INT max number = 20 000; INT max divisors  := 0; INT has max divisors  := 0; INT with max divisors := 0; FOR d TO max number DO 

   INT divisor count = count divisors( proper divisors( d ) );
   IF divisor count > max divisors THEN
       # found a number with more divisors than the previous max  #
       max divisors       := divisor count;
       has max divisors   := d;
       with max divisors  := 1
   ELIF divisor count = max divisors THEN
       # found another number with that many divisors             #
       with max divisors +:= 1
   FI

OD; print( ( whole( has max divisors, 0 ) 

      , " is the "
      , IF with max divisors < 2 THEN "only" ELSE "first" FI
      , " number upto "
      , whole( max number, 0 )
      , " with "
      , whole( max divisors, 0 )
      , " divisors"
      , newline
      ) )</lang>
Output:
Proper divisors of:  1:
Proper divisors of:  2:  1
Proper divisors of:  3:  1
Proper divisors of:  4:  1 2
Proper divisors of:  5:  1
Proper divisors of:  6:  1 2 3
Proper divisors of:  7:  1
Proper divisors of:  8:  1 2 4
Proper divisors of:  9:  1 3
Proper divisors of: 10:  1 2 5
15120 is the first number upto 20000 with 79 divisors

AWK

<lang AWK>

  1. syntax: GAWK -f PROPER_DIVISORS.AWK

BEGIN { 

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

} function divisors(n, i) { 

   if (n == 1) {
     Dcnt = 0
     Dstr = ""
     return
   }
   Dcnt = Dstr = 1
   for (i=2; i<n; i++) {
     if (n % i == 0) {
       Dcnt++
       Dstr = sprintf("%s %s",Dstr,i)
     }
   }
   return

} </lang>

output:

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

C

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

  1. include <stdio.h>
  2. include <stdbool.h>

int proper_divisors(const int n, bool print_flag) { 

   int count = 0;

   for (int i = 1; i < n; ++i) {
       if (n % i == 0) {
           count++;
           if (print_flag)
               printf("%d ", i);
       }
   }

   if (print_flag)
       printf("\n");

   return count;

}

int main(void) { 

   for (int i = 1; i <= 10; ++i) {
       printf("%d: ", i);
       proper_divisors(i, true);
   }

   int max = 0;
   int max_i = 1;

   for (int i = 1; i <= 20000; ++i) {
       int v = proper_divisors(i, false);
       if (v >= max) {
           max = v;
           max_i = i;
       }
   }

   printf("%d with %d divisors\n", max_i, max);
   return 0;

} </lang>

Output:
1: 
2: 1 
3: 1 
4: 1 2 
5: 1 
6: 1 2 3 
7: 1 
8: 1 2 4 
9: 1 3 
10: 1 2 5 
18480 with 79 divisors

C#

<lang csharp>namespace RosettaCode.ProperDivisors { 

   using System;
   using System.Collections.Generic;
   using System.Linq;

   internal static class Program
   {
       private static IEnumerable<int> ProperDivisors(int number)
       {
           return
               Enumerable.Range(1, number / 2)
                   .Where(divisor => number % divisor == 0);
       }

       private static void Main()
       {
           foreach (var number in Enumerable.Range(1, 10))
           {
               Console.WriteLine("{0}: {{{1}}}", number,
                   string.Join(", ", ProperDivisors(number)));
           }

           var record = Enumerable.Range(1, 20000).Select(number => new
           {
               Number = number,
               Count = ProperDivisors(number).Count()
           }).OrderByDescending(currentRecord => currentRecord.Count).First();
           Console.WriteLine("{0}: {1}", record.Number, record.Count);
       }
   }

}</lang>

Output:
1: {}
2: {1}
3: {1}
4: {1, 2}
5: {1}
6: {1, 2, 3}
7: {1}
8: {1, 2, 4}
9: {1, 3}
10: {1, 2, 5}
15120: 79

C++

<lang cpp>#include <vector>

  1. include <iostream>
  2. include <algorithm>

std::vector<int> properDivisors ( int number ) { 

  std::vector<int> divisors ;
  for ( int i = 1 ; i < number / 2 + 1 ; i++ )
     if ( number % i == 0 )

divisors.push_back( i ) ; 

  return divisors ;

}

int main( ) { 

  std::vector<int> divisors ;
  unsigned int maxdivisors = 0 ;
  int corresponding_number = 0 ;
  for ( int i = 1 ; i < 11 ; i++ ) {
     divisors =  properDivisors ( i ) ;
     std::cout << "Proper divisors of " << i << ":\n" ;
     for ( int number : divisors ) {

std::cout << number << " " ; 

     }
     std::cout << std::endl ;
     divisors.clear( ) ;
  }
  for ( int i = 11 ; i < 20001 ; i++ ) {
     divisors =  properDivisors ( i ) ;
     if ( divisors.size( ) > maxdivisors ) {

maxdivisors = divisors.size( ) ; corresponding_number = i ; 

     }
     divisors.clear( ) ;
  }

  std::cout << "Most divisors has " << corresponding_number <<
     " , it has " << maxdivisors << " divisors!\n" ; 
  return 0 ;

} </lang>

Output:
Proper divisors of 1:

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

Ceylon

<lang ceylon>shared void run() {

function divisors(Integer int) => if(int <= 1) then {} else (1..int / 2).filter((Integer element) => element.divides(int));

for(i in 1..10) { print("``i`` => ``divisors(i)``"); }

value start = 1; value end = 20k;

value mostDivisors = map {for(i in start..end) i->divisors(i).size} .inverse() .max(byKey(byIncreasing(Integer.magnitude)));

print("the number(s) with the most divisors between ``start`` and ``end`` is/are: ``mostDivisors?.item else "nothing"`` with ``mostDivisors?.key else "no"`` divisors"); }</lang>

Output:
1 => []
2 => { 1 }
3 => { 1 }
4 => { 1, 2 }
5 => { 1 }
6 => { 1, 2, 3 }
7 => { 1 }
8 => { 1, 2, 4 }
9 => { 1, 3 }
10 => { 1, 2, 5 }
the number(s) with the most divisors between 1 and 20000 is/are:
[15120, 18480] with 79 divisors

Common Lisp

Ideally, the smallest-divisor function would only try prime numbers instead of odd numbers. <lang lisp>(defun proper-divisors-recursive (product &optional (results '(1))) 

  "(int,list)->list::Function to find all proper divisors of a +ve integer."
 
  (defun smallest-divisor (x)
     "int->int::Find the smallest divisor of an integer > 1."
     (if (evenp x) 2
         (do ((lim (truncate (sqrt x)))
              (sd 3 (+ sd 2)))
             ((or (integerp (/ x sd)) (> sd lim)) (if (> sd lim) x sd)))))
  
  (defun pd-rec (fac)
     "(int,int)->nil::Recursive function to find proper divisors of a +ve integer"
     (when (not (member fac results))
        (push fac results)
        (let ((hifac (/ fac (smallest-divisor fac))))
           (pd-rec hifac)
           (pd-rec (/ product hifac)))))
   
  (pd-rec product)
  (butlast (sort (copy-list results) #'<)))

(defun task (method &optional (n 1) (most-pds '(0))) 

  (dotimes (i 19999)
     (let ((npds (length (funcall method (incf n))))
           (hiest (car most-pds)))
        (when (>= npds hiest)
           (if (> npds hiest)
               (setf most-pds (list npds (list n)))
               (setf most-pds (list npds (cons n (second most-pds))))))))
  most-pds)

(defun main () 

  (format t "Task 1:Proper Divisors of [1,10]:~%")
  (dotimes (i 10) (format t "~A:~A~%" (1+ i) (proper-divisors-recursive (1+ i))))
  (format t "Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:~%~A~%"
          (task #'proper-divisors-recursive)))</lang>
Output:
CL-USER(10): (main)
Task 1:Proper Divisors of [1,10]:
1:NIL
2:(1)
3:(1)
4:(1 2)
5:(1)
6:(1 2 3)
7:(1)
8:(1 2 4)
9:(1 3)
10:(1 2 5)
Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:
(79 (18480 15120))
NIL

D

Translation of: Python

Currently the lambda of the filter allocates a closure on the GC-managed heap. <lang d>void main() /*@safe*/ { 

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

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

   iota(1, 11).map!properDivs.writeln;
   iota(1, 20_001).map!(n => tuple(properDivs(n).count, n)).reduce!max.writeln;

}</lang>

Output:
[[], [1], [1], [1, 2], [1], [1, 2, 3], [1], [1, 2, 4], [1, 3], [1, 2, 5]]
Tuple!(uint, int)(79, 18480)

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

EchoLisp

<lang scheme> (lib 'list) ;; list-delete

let n = product p_i^a_i , p_i prime
number of divisors = product (a_i + 1) - 1

(define (numdivs n) 

   (1- (apply * (map (lambda(g) (1+ (length g))) (group (prime-factors n))))))

(remember 'numdivs)

prime powers
input
a list g of grouped prime factors ( 3 3 3 ..)
returns (1 3 9 27 ...)

(define (ppows g (mult 1)) (for/fold (ppows '(1)) ((a g)) (set! mult (* mult a)) (cons mult ppows)))

proper divisors
decomp n into ((2 2 ..) ( 3 3 ..) ) prime factors groups
combines (1 2 4 8 ..) (1 3 9 ..) lists
remove n from the list

(define (divs n) 

  (if (<= n 1) null 
   (list-delete
       (for/fold (divs'(1)) ((g (map  ppows (group (prime-factors n)))))

(for*/list ((a divs) (b g)) (* a b))) 

   n )))
find number(s) with max # of proper divisors
returns list of (n . maxdivs) for n in range 2..N

(define (most-proper N) 

   (define maxdivs 1)
   (define ndivs 0)
   (for/fold (most-proper null) ((n (in-range 2 N)))
      (set! ndivs (numdivs n))
       #:continue (< ndivs maxdivs)
       (when (> ndivs maxdivs)
       (set!-values (most-proper maxdivs) (values null ndivs)))
       (cons (cons n maxdivs) most-proper)))


</lang>

Output:

<lang scheme> (for ((i (in-range 1 11))) (writeln i (divs i))) 1 null 2 (1) 3 (1) 4 (2 1) 5 (1) 6 (2 3 1) 7 (1) 8 (4 2 1) 9 (3 1) 10 (2 5 1)


(most-proper 20000) 

   → ((18480 . 79) (15120 . 79))

(most-proper 1_000_000) 

   → ((997920 . 239) (982800 . 239) (942480 . 239) (831600 . 239) (720720 . 239))

(lib 'bigint) (numdivs 95952222101012742144) → 666 ;; 🎩 </lang>

Eiffel

<lang Eiffel> class APPLICATION

create make

feature

make -- Test the feature proper_divisors. local list: LINKED_LIST [INTEGER] count, number: INTEGER do across 1 |..| 10 as c loop list := proper_divisors (c.item) io.put_string (c.item.out + ": ") across list as l loop io.put_string (l.item.out + " ") end io.new_line end across 1 |..| 20000 as c loop list := proper_divisors (c.item) if list.count > count then count := list.count number := c.item end end io.put_string (number.out + " has with " + count.out + " divisors the highest number of proper divisors.") end

proper_divisors (n: INTEGER): LINKED_LIST [INTEGER] -- Proper divisors of 'n'. do create Result.make across 1 |..| (n - 1) as c loop if n \\ c.item = 0 then Result.extend (c.item) end end end

end </lang>

Output:
1:
2: 1
3: 1
4: 1 2
5: 1 
6: 1 2 3 
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
15120 has with 79 divisors the highest number of proper divisors.

Elixir

Translation of: Erlang

<lang elixir>defmodule Proper do 

 def divisors(1), do: []
 def divisors(n), do: [1 | divisors(2,n,:math.sqrt(n))] |> Enum.sort
 
 defp divisors(k,_n,q) when k>q, do: []
 defp divisors(k,n,q) when rem(n,k)>0, do: divisors(k+1,n,q)
 defp divisors(k,n,q) when k * k == n, do: [k | divisors(k+1,n,q)]
 defp divisors(k,n,q)                , do: [k,div(n,k) | divisors(k+1,n,q)]
 
 def most_divisors(limit) do
   {length,nums} = Enum.group_by(1..limit, fn n -> length(divisors(n)) end)
                   |> Enum.max_by(fn {length,_nums} -> length end)
   IO.puts "With #{length}, Number #{inspect nums} has the most divisors"
 end

end

Enum.each(1..10, fn n -> 

 IO.puts "#{n}: #{inspect Proper.divisors(n)}"

end) Proper.most_divisors(20000)</lang>

Output:
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
With 79, Number [18480, 15120] has the most divisors

Erlang

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

divs(0) -> []; divs(1) -> []; divs(N) -> lists:sort([1] ++ divisors(2,N,math:sqrt(N))).

divisors(K,_N,Q) when K > Q -> []; divisors(K,N,Q) when N rem K =/= 0 -> 

   divisors(K+1,N,Q);

divisors(K,N,Q) when K * K == N -> 

   [K] ++ divisors(K+1,N,Q);

divisors(K,N,Q) -> 

   [K, N div K] ++ divisors(K+1,N,Q).

sumdivs(N) -> lists:sum(divs(N)).

longest(Limit) -> longest(Limit,0,0,1).

longest(L,Current,CurLeng,Acc) when Acc >= L -> 

   io:format("With ~w, Number ~w has the most divisors~n", [CurLeng,Current]);

longest(L,Current,CurLeng,Acc) -> 

   A = length(divs(Acc)),
   if A > CurLeng ->
       longest(L,Acc,A,Acc+1);
       true -> longest(L,Current,CurLeng,Acc+1)
   end.</lang>
Output:
1> [io:format("X: ~w, N: ~w~n", [N,properdivs:divs(N)]) ||  N <- lists:seq(1,10)].                                      
X: 1, N: []
X: 2, N: [1]
X: 3, N: [1]
X: 4, N: [1,2]
X: 5, N: [1]
X: 6, N: [1,2,3]
X: 7, N: [1]
X: 8, N: [1,2,4]
X: 9, N: [1,3]
X: 10, N: [1,2,5]
[ok,ok,ok,ok,ok,ok,ok,ok,ok,ok]

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

Fortran

Compiled using G95 compiler, run on x86 system under Puppy Linux <lang Fortran> 

     function icntprop(num  )
     icnt=0
     do i=1 , num-1
         if (mod(num , i)  .eq. 0)  then
         icnt = icnt + 1
         if (num .lt. 11) print *,'    ',i
         end if
         end do
     icntprop =  icnt
     end function

     limit = 20000
     maxcnt = 0
     print *,'N   divisors'
     do j=1,limit,1
     if (j .lt. 11) print *,j
     icnt = icntprop(j)

     if (icnt .gt. maxcnt) then
     maxcnt = icnt
     maxj = j
     end if

     end do

     print *,' '
     print *,' from 1 to ',limit
     print *,maxj,' has max proper divisors: ',maxcnt
     end

</lang>

Output:

 N   divisors
 1
 2
      1
 3
      1
 4
      1
      2
 5
      1
 6
      1
      2
      3
 7
      1
 8
      1
      2
      4
 9
      1
      3
 10
      1
      2
      5
 
  from 1 to  20000
 15120  has max proper divisors:  79

GFA Basic

<lang> OPENW 1 CLEARW 1 ' ' Array f% is used to hold the divisors DIM f%(SQR(20000)) ! cannot redim arrays, so set size to largest needed ' ' 1. Show proper divisors of 1 to 10, inclusive ' FOR i%=1 TO 10 

 num%=@proper_divisors(i%)
 PRINT "Divisors for ";i%;":";
 FOR j%=1 TO num%
   PRINT " ";f%(j%);
 NEXT j%
 PRINT

NEXT i% ' ' 2. Find (smallest) number <= 20000 with largest number of proper divisors ' result%=1 ! largest so far number%=0 ! its number of divisors FOR i%=1 TO 20000 

 num%=@proper_divisors(i%)
 IF num%>number%
   result%=i%
   number%=num%
 ENDIF

NEXT i% PRINT "Largest number of divisors is ";number%;" for ";result% ' ~INP(2) CLOSEW 1 ' ' find the proper divisors of n%, placing results in f% ' and return the number found ' FUNCTION proper_divisors(n%) 

 LOCAL i%,root%,count%
 '
 ARRAYFILL f%(),0
 count%=1 ! index of next slot in f% to fill
 '
 IF n%>1
   f%(count%)=1
   count%=count%+1
   root%=SQR(n%)
   FOR i%=2 TO root%
     IF n% MOD i%=0
       f%(count%)=i%
       count%=count%+1
       IF i%*i%<>n% ! root% is an integer, so check if i% is actual square root of n%
         f%(count%)=n%/i%
         count%=count%+1
       ENDIF
     ENDIF
   NEXT i%
 ENDIF
 '
 RETURN count%-1

ENDFUNC </lang>

Output is: 

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

Haskell

<lang Haskell>import Data.Ord import Data.List

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

main :: IO () main = do 

 putStrLn "divisors of 1 to 10:"
 mapM_ (print . divisors) [1 .. 10]
 putStrLn "a number with the most divisors within 1 to 20000 (number, count):"
 print $ maximumBy (comparing snd)
   [(n, length $ divisors n) | n <- [1 .. 20000]]</lang>
Output:
divisors of 1 to 10:
[]
[1]
[1]
[1,2]
[1]
[1,2,3]
[1]
[1,2,4]
[1,3]
[1,2,5]
a number with the most divisors within 1 to 20000 (number, count):
(18480,79)

J

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

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

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

Proper divisors of numbers 1 through 10:

<lang J> (,&": ' -- ' ,&": properDivisors)&>1+i.10 1 -- 2 -- 1 3 -- 1 4 -- 1 2 5 -- 1 6 -- 1 2 3 7 -- 1 8 -- 1 2 4 9 -- 1 3 10 -- 1 2 5</lang>

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

<lang J> (, #@properDivisors)&> 1+I.(= >./) #@properDivisors@> 1+i.20000 15120 79 18480 79</lang>

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

<lang J> (, #@properDivisors)&> 1+I.(= >./) #@factors@> 1+i.20000 15120 79 18480 79</lang>

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

Java

Works with: Java version 1.5+

<lang java5>import java.util.Collections; import java.util.LinkedList; import java.util.List;

public class Proper{ 

   public static List<Integer> properDivs(int n){
       List<Integer> divs = new LinkedList<Integer>();
       if(n == 1) return divs;
       divs.add(1);
       for(int x = 2; x < n; x++){
           if(n % x == 0) divs.add(x);
       }
       
       Collections.sort(divs);
       
       return divs;
   }
   
   public static void main(String[] args){
       for(int x = 1; x <= 10; x++){
           System.out.println(x + ": " + properDivs(x));
       }
       
       int x = 0, count = 0;
       for(int n = 1; n <= 20000; n++){
           if(properDivs(n).size() > count){
               x = n;
               count = properDivs(n).size();
           }
       }
       System.out.println(x + ": " + count);
   }

}</lang>

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

JavaScript

<lang JavaScript>(function () { 

   // Proper divisors
   function properDivisors(n) {
       if (n < 2) return [];
       else {
           var rRoot = Math.sqrt(n),
               intRoot = Math.floor(rRoot),

               lows = range(1, intRoot).filter(function (x) {
                   return (n % x) === 0;
               });

           return lows.concat(lows.slice(1).map(function (x) {
               return n / x;
           }).reverse().slice((rRoot === intRoot) | 0));
       }
   }

   // [m..n]
   function range(m, n) {
       var a = Array(n - m + 1),
           i = n + 1;
       while (i--) a[i - 1] = i;
       return a;
   }

   var tblOneToTen = [
           ['Number', 'Proper Divisors', 'Count']
       ].concat(range(1, 10).map(function (x) {
           var ds = properDivisors(x);

           return [x, ds.join(', '), ds.length];
       })),

       dctMostBelow20k = range(1, 20000).reduce(function (a, x) {
           var lng = properDivisors(x).length;

           return lng > a.divisorCount ? {
               n: x,
               divisorCount: lng
           } : a;
       }, {
           n: 0,
           divisorCount: 0
       });



   // a -> bool -> s -> s
   function wikiTable(lstRows, blnHeaderRow, strStyle) {
       return '{| class="wikitable" ' + (
           strStyle ? 'style="' + strStyle + '"' : 
       ) + lstRows.map(function (lstRow, iRow) {
           var strDelim = ((blnHeaderRow && !iRow) ? '!' : '|');

           return '\n|-\n' + strDelim + ' ' + lstRow.map(function (v) {
               return typeof v === 'undefined' ? ' ' : v;
           }).join(' ' + strDelim + strDelim + ' ');
       }).join() + '\n|}';
   }

   return wikiTable(
       tblOneToTen,
       true
   ) + '\n\nMost proper divisors below 20,000:\n\n  ' + JSON.stringify(
       dctMostBelow20k
   );

})();</lang>

Output:
Number Proper Divisors Count
1 0
2 1 1
3 1 1
4 1, 2 2
5 1 1
6 1, 2, 3 3
7 1 1
8 1, 2, 4 3
9 1, 3 2
10 1, 2, 5 3

Most proper divisors below 20,000: 

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

jq

Works with: jq version 1.4

In the following, proper_divisors returns a stream. In order to count the number of items in the stream economically, we first define "count(stream)": <lang jq>def count(stream): reduce stream as $i (0; . + 1);

  1. unordered

def proper_divisors: 

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

end) 

   else empty
   end;
  1. The first integer in 1 .. n inclusive
  2. with the maximal number of proper divisors in that range:

def most_proper_divisors(n): 

 reduce range(1; n+1) as $i
   ( [null, 0];
     count( $i | proper_divisors ) as $count
     | if $count > .[1] then [$i, $count] else . end);</lang>

The tasks: <lang jq>"The proper divisors of the numbers 1 to 10 inclusive are:", (range(1;11) as $i | "\($i): \( [ $i | proper_divisors] )"), "", "The pair consisting of the least number in the range 1 to 20,000 with", "the maximal number proper divisors together with the corresponding", "count of proper divisors is:", most_proper_divisors(20000) </lang>

Output:

<lang sh>$ jq -n -c -r -f /Users/peter/jq/proper_divisors.jq The proper divisors of the numbers 1 to 10 inclusive are: 1: [] 2: [1] 3: [1] 4: [1,2] 5: [1] 6: [1,2,3] 7: [1] 8: [1,2,4] 9: [1,3] 10: [1,2,5]

The pair consisting of the least number in the range 1 to 20,000 with the maximal number proper divisors together with the corresponding count of proper divisors is: [15120,79]</lang>

Julia

Use factor to obtain the prime factorization of the target number. I adopted the argument handling style of factor in my properdivisors function. <lang Julia> function properdivisors{T<:Integer}(n::T) 

   0 < n || throw(ArgumentError("number to be factored must be ≥ 0, got $n"))
   1 < n || return T[]
   !isprime(n) || return T[one(T), n]
   f = factor(n)
   d = T[one(T)]
   for (k, v) in f
       c = T[k^i for i in 0:v]
       d = d*c'
       d = reshape(d, length(d))
   end
   sort!(d)
   return d[1:end-1]

end

lo = 1 hi = 10 println("List the proper divisors for ", lo, " through ", hi, ".") for i in lo:hi 

   println(@sprintf("%4d", i), " ", properdivisors(i))

end

hi = 2*10^4 println("\nFind the numbers within [", lo, ",", hi, "] having the most divisors.")

maxdiv = 0 nlst = Int[]

for i in lo:hi 

   ndiv = length(properdivisors(i))
   if ndiv > maxdiv
       maxdiv = ndiv
       nlst = [i]
   elseif ndiv == maxdiv
       push!(nlst, i)
   end

end

println(nlst, " have the maximum proper divisor count of ", maxdiv, ".") </lang>

Output:
List the proper divisors for 1 through 10.
   1 []
   2 [1,2]
   3 [1,3]
   4 [1,2]
   5 [1,5]
   6 [1,2,3]
   7 [1,7]
   8 [1,2,4]
   9 [1,3]
  10 [1,2,5]

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

Mathematica / Wolfram Language

A Function that yields the proper divisors of an integer n: <lang Mathematica>ProperDivisors[n_Integer /; n > 0] := Most@Divisors@n;</lang>

Proper divisors of n from 1 to 10: <lang Mathematica>Grid@Table[{n, ProperDivisors[n]}, {n, 1, 10}]</lang>

Output:
1	{}
2	{1}
3	{1}
4	{1,2}
5	{1}
6	{1,2,3}
7	{1}
8	{1,2,4}
9	{1,3}
10	{1,2,5}

The number with the most divisors between 1 and 20,000: <lang Mathematica>Fold[ 

Last[SortBy[{#1, {#2, Length@ProperDivisors[#2]}}, Last]] &,
{0, 0},
Range[20000]]</lang>
Output:
{18480, 79}

An alternate way to find the number with the most divisors between 1 and 20,000: <lang Mathematica>Last@SortBy[ 

 Table[
   {n, Length@ProperDivisors[n]},
   {n, 1, 20000}],
 Last]</lang>
Output:
{15120, 79}

Oberon-2

<lang oberon2> MODULE ProperDivisors; IMPORT 

 Out;

CONST 

   initialSize = 128;

TYPE 

 Result* = POINTER TO ResultDesc;
 ResultDesc = RECORD 
   found-: LONGINT; (* number of slots in pd *)
   pd-: POINTER TO ARRAY OF LONGINT;
   cap: LONGINT;   (* Capacity *)
 END;

VAR 

 i,found,max,idxMx: LONGINT;
 mx: ARRAY 32 OF LONGINT;
 rs: Result;

 PROCEDURE (r: Result) Init(size: LONGINT);
 BEGIN 
   r.found := 0;
   r.cap := size;
   NEW(r.pd,r.cap);
 END Init;

 PROCEDURE (r: Result) Add(n: LONGINT);
 BEGIN
   (* Out.String("--->");Out.LongInt(n,0);Out.String(" At: ");Out.LongInt(r.found,0);Out.Ln; *)
   IF (r.found < LEN(r.pd^) - 1) THEN 
     r.pd[r.found] := n;
   ELSE
     (* expand pd for more room *)
   END;
   INC(r.found);
 END Add;

 PROCEDURE (r:Result) Show();
 VAR
   i: LONGINT;
 BEGIN
     Out.String("(Result:");Out.LongInt(r.found + 1,0);(* Out.String("/");Out.LongInt(r.cap,0);*)
     Out.String("-");
     IF r.found > 0 THEN
       FOR i:= 0 TO r.found - 1 DO
         Out.LongInt(r.pd[i],0);
         IF i = r.found - 1 THEN Out.Char(')') ELSE Out.Char(',') END
       END
     END;
     Out.Ln
 END Show;

 PROCEDURE (r:Result) Reset();
 BEGIN 
   r.found := 0;
 END Reset;

 PROCEDURE GetFor(n: LONGINT;VAR rs: Result);
 VAR
   i: LONGINT;
 BEGIN
   IF n > 1 THEN 
     rs.Add(1);i := 2;
     WHILE (i < n) DO
       IF (n MOD i) = 0 THEN rs.Add(i) END;
       INC(i)
     END
   END;
 END GetFor;

BEGIN 

 NEW(rs);rs.Init(initialSize);
 FOR i := 1 TO 10 DO 
   Out.LongInt(i,4);Out.Char(':');
   GetFor(i,rs);
   rs.Show();
   rs.Reset();
 END;
 Out.LongInt(100,4);Out.Char(':');GetFor(100,rs);rs.Show();rs.Reset();
 max := 0;idxMx := 0;found := 0;
 FOR i := 1 TO 20000 DO
   GetFor(i,rs);
   IF rs.found > max THEN
     idxMx:= 0;mx[idxMx] := i;max := rs.found
   ELSIF rs.found = max THEN
     INC(idxMx);mx[idxMx] := i
   END;
   rs.Reset()
 END;
 Out.String("Found: ");Out.LongInt(idxMx + 1,0);
 Out.String(" Numbers with most proper divisors ");
 Out.LongInt(max,0);Out.String(": ");Out.Ln;
 FOR i := 0 TO idxMx DO
   Out.LongInt(mx[i],0);Out.Ln
 END

END ProperDivisors. </lang>

Output:
   1:(Result:1-
   2:(Result:2-1)
   3:(Result:2-1)
   4:(Result:3-1,2)
   5:(Result:2-1)
   6:(Result:4-1,2,3)
   7:(Result:2-1)
   8:(Result:4-1,2,4)
   9:(Result:3-1,3)
  10:(Result:4-1,2,5)
 100:(Result:9-1,2,4,5,10,20,25,50)
Found: 2 Numbers with most proper divisors 79: 
15120
18480

Oforth

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

10 seq apply(#[ dup print " : " print properDivs println ]) 20000 seq map(#[ dup properDivs size Pair new ]) reduce(#maxKey) println</lang>

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

PARI/GP

<lang parigp>proper(n)=if(n==1, [], my(d=divisors(n)); d[2..#d]); apply(proper, [1..10]) r=at=0; for(n=1,20000, t=numdiv(n); if(t>r, r=t; at=n)); [at, numdiv(t)-1]</lang>

Output:
%1 = [[], [2], [3], [2, 4], [5], [2, 3, 6], [7], [2, 4, 8], [3, 9], [2, 5, 10]]
%2 = [15120, 7]

Perl

Using a module for divisors

Library: ntheory

<lang perl>use ntheory qw/divisors/; sub proper_divisors { 

 my $n = shift;
 # Like Pari/GP, divisors(0) = (0,1) and divisors(1) = ()
 return 1 if $n == 0;
 my @d = divisors($n);
 pop @d;  # divisors are in sorted order, so last entry is the input
 @d;

} say "$_: ", join " ", proper_divisors($_) for 1..10;

  1. 1. For the max, we can do a traditional loop.

my($max,$ind) = (0,0); for (1..20000) { 

 my $nd = scalar proper_divisors($_);
($max,$ind) = ($nd,$_) if $nd > $max;

} say "$max $ind";

  1. 2. Or we can use List::Util's max with decoration (this exploits its implementation)

{ 

 use List::Util qw/max/;
 no warnings 'numeric';
 say max(map { scalar(proper_divisors($_)) . " $_" } 1..20000);

}</lang>

Output:
1: 
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
79 15120
79 18480

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

Perl 6

Works with: rakudo version 2015-10-31

<lang perl6>sub propdiv (\x) { 

   my @l =(1 if 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);

}

say "$_ ", propdiv($_) for 1..10;

my $max = 0; my @candidates; for 1..20000 { 

   my @pd = propdiv($_);
   my $pd = @pd.elems;
   if $pd > $max {
       @candidates = ();
       $max = $pd;
   }
   push @candidates, $_ if $pd == $max;

} say "max = $max, candidates = @candidates[]";</lang>

Output:
1	Nil 
2	1 
3	1 
4	1 2
5	1 
6	1 2 3
7	1 
8	1 2 4
9	1 3
10	1 2 5
max = 79, candidates = 15120 18480

Phix

The factors routine is an auto-include. The actual implementation of it, from builtins\pfactors.e is <lang Phix>global function factors(atom n, integer include1=0) -- returns a list of all integer factors of n -- if include1 is 0 (the default), result does not contain either 1 or n -- if include1 is 1, and n>1, the result contains 1 and n -- if include1 is -1, and n>1, the result contains 1 but not n sequence lfactors = {}, hfactors = {} atom hfactor integer p = 2, 

       lim = floor(sqrt(n))

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

end function</lang> The compiler knows where to find that, so the main program is just: <lang Phix>for i=1 to 10 do 

   ?{i,factors(i,-1)}

end for

integer maxd = 0, k sequence candidates = {} 

   for i=1 to 20000 do
       k = length(factors(i,-1))
       if k>=maxd then
           if k=maxd then
               candidates &= i
           else
               candidates = {i}
               maxd = k
           end if
       end if
   end for

   printf(1,"%d divisors:", maxd)
   ?candidates
   {} = wait_key()</lang>
Output:
{1,{}}
{2,{1}}
{3,{1}}
{4,{1,2}}
{5,{1}}
{6,{1,2,3}}
{7,{1}}
{8,{1,2,4}}
{9,{1,3}}
{10,{1,2,5}}
79 divisors:{15120,18480}

PHP

<lang php><?php function ProperDivisors($n) { 

 yield 1;
 $large_divisors = [];
 for ($i = 2; $i <= sqrt($n); $i++) {
   if ($n % $i == 0) {
     yield $i;
     if ($i*$i != $n) {
       $large_divisors[] = $n / $i;
     }
   }
 }
 foreach (array_reverse($large_divisors) as $i) {
   yield $i;
 }

}

assert([1, 2, 4, 5, 10, 20, 25, 50] == 

       iterator_to_array(ProperDivisors(100)));

foreach (range(1, 10) as $n) { 

 echo "$n =>";
 foreach (ProperDivisors($n) as $divisor) {
   echo " $divisor";
 }
 echo "\n";

}

$divisorsCount = []; for ($i = 1; $i < 20000; $i++) { 

 $divisorsCount[sizeof(iterator_to_array(ProperDivisors($i)))][] = $i;

} ksort($divisorsCount);

echo "Numbers with most divisors: ", implode(", ", end($divisorsCount)), ".\n"; echo "They have ", key($divisorsCount), " divisors.\n";

</lang>

Outputs: 

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

PicoLisp

<lang PicoLisp># Generate all proper divisors. (de propdiv (N) 

  (head -1 (filter
     '((X) (=0 (% N X)))
     (range 1 N) )) )
  1. Obtaining the values from 1 to 10 inclusive.

(mapcar propdiv (range 1 10))

  1. Output:
  2. (NIL (1) (1) (1 2) (1) (1 2 3) (1) (1 2 4) (1 3) (1 2 5))</lang>

Brute-force

<lang PicoLisp>(de propdiv (N) 

  (cdr
     (rot
        (make
           (for I N
              (and (=0 (% N I)) (link I)) ) ) ) ) )

(de countdiv (N) 

  (let C -1 
     (for I N
        (and (=0 (% N I)) (inc 'C)) )
     C ) )

(let F (-5 -8) 

  (tab F "N" "LIST")
  (for I 10
     (tab F
        I
        (glue " + " (propdiv I)) ) ) )

(println 

  (maxi
     countdiv
     (range 1 20000) ) )</lang>

Factorization

<lang PicoLisp>(de accu1 (Var Key) 

  (if (assoc Key (val Var))
     (con @ (inc (cdr @)))
     (push Var (cons Key 2)) )
  Key )

(de factor (N) 

  (let
     (R NIL
        D 2
        L (1 2 2 . (4 2 4 2 4 6 2 6 .))
        M (sqrt N) )
     (while (>= M D)
        (if (=0 (% N D))
           (setq M
              (sqrt (setq N (/ N (accu1 'R D)))) )
           (inc 'D (pop 'L)) ) )
     (accu1 'R N)
     (dec (apply * (mapcar cdr R))) ) )

(bench 

  (println
     (maxi
        factor
        (range 1 20000) ) 
     @@ ) )</lang>

Output: 

15120 79
0.081 sec

PL/I

<lang pli>*process source xref; 

(subrg):
cpd: Proc Options(main);
p9a=time();
Dcl (p9a,p9b) Pic'(9)9';
Dcl cnt(3) Bin Fixed(31) Init((3)0);
Dcl x Bin Fixed(31);
Dcl pd(300) Bin Fixed(31);
Dcl sumpd   Bin Fixed(31);
Dcl npd     Bin Fixed(31);
Dcl hi      Bin Fixed(31) Init(0);
Dcl (xl(10),xi) Bin Fixed(31);
Dcl i       Bin Fixed(31);
Do x=1 To 10;
  Call proper_divisors(x,pd,npd);
  Put Edit(x,' -> ',(pd(i) Do i=1 To npd))(Skip,f(2),a,10(f(2)));
  End;
xi=0;
Do x=1 To 20000;
  Call proper_divisors(x,pd,npd);
  Select;
    When(npd>hi) Do;
      xi=1;
      xl(1)=x;
      hi=npd;
      End;
    When(npd=hi) Do;
      xi+=1;
      xl(xi)=x;
      End;
    Otherwise;
    End;
  End;
Put Edit(hi,' -> ',(xl(i) Do i=1 To xi))(Skip,f(3),a,10(f(6)));

x= 166320; Call proper_divisors(x,pd,npd);
Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4));
x=1441440; Call proper_divisors(x,pd,npd);
Put Edit(x,' -> ',npd)(Skip,f(8),a,f(4));



p9b=time();
Put Edit((p9b-p9a)/1000,' seconds elapsed')(Skip,f(6,3),a);
Return;

proper_divisors: Proc(n,pd,npd);
Dcl (n,pd(300),npd) Bin Fixed(31);
Dcl (d,delta)       Bin Fixed(31);
npd=0;
If n>1 Then Do;
  If mod(n,2)=1 Then  /* odd number  */
    delta=2;
  Else                /* even number */
    delta=1;
  Do d=1 To n/2 By delta;
    If mod(n,d)=0 Then Do;
      npd+=1;
      pd(npd)=d;
      End;
    End;
  End;
End;

End;</lang>
Output:
 1 ->
 2 ->  1
 3 ->  1
 4 ->  1 2
 5 ->  1
 6 ->  1 2 3
 7 ->  1
 8 ->  1 2 4
 9 ->  1 3
10 ->  1 2 5
 79 ->  15120 18480
  166320 ->  159
 1441440 ->  287
 0.530 seconds elapsed

PowerShell

version 1

<lang PowerShell> function proper-divisor ($n) { 

   if($n -ge 2) {
       $lim = [Math]::Floor([Math]::Sqrt($n))
       $less, $greater = @(1), @()
       for($i = 2; $i -lt $lim; $i++){
           if($n%$i -eq 0) {
               $less += @($i)
               $greater = @($n/$i) + $greater
           }
       }
       if(($lim -ne 1) -and ($n%$lim -eq 0)) {$less += @($lim)}
       $($less + $greater)
   } else {@()}

} "$(proper-divisor 100)" "$(proper-divisor 496)" "$(proper-divisor 2048)" </lang>

version 2

<lang PowerShell> function proper-divisor ($n) { 

   if($n -ge 2) {
       $lim = [Math]::Floor($n/2)+1
       $proper = @(1)
       for($i = 2; $i -lt $lim; $i++){
           if($n%$i -eq 0) {
               $proper += @($i)
           }
       }
       $proper
   } else {@()}

} "$(proper-divisor 100)" "$(proper-divisor 496)" "$(proper-divisor 2048)" </lang>

version 3

<lang PowerShell> function eratosthenes ($n) { 

   if($n -gt 1){
       $prime = @(0..$n| foreach{$true})
       $m = [Math]::Floor([Math]::Sqrt($n))
       function multiple($i) {
           for($j = $i*$i; $j -le $n; $j += $i) {
               $prime[$j] = $false
           }
       }
       multiple 2
       for($i = 3; $i -le $m; $i += 2) {
           if($prime[$i]) {multiple $i}
       }
       2
       for($i = 3; $i -le $n; $i += 2) {
           if($prime[$i]) {$i}
       }
       
   } else {
       Write-Error "$n is not greater than 1"
   }

} function prime-decomposition ($n) { 

   $array = eratosthenes $n
   $prime = @()
   foreach($p in $array) {
       while($n%$p -eq 0) {
           $n /= $p
           $prime += @($p)
       }
   }
   $prime

} function proper-divisor ($n) { 

   if($n -ge 2) {
       $array = prime-decomposition $n
       $lim = $array.Count
       function state($res, $i){  
           if($i -lt $lim) {
               state ($res) ($i + 1)
               state ($res*$array[$i]) ($i + 1)   
           } elseif($res -lt $n) {$res}
       }
       state 1 0 | sort -Unique
   } else {@()}

} "$(proper-divisor 100)" "$(proper-divisor 496)" "$(proper-divisor 2048)" </lang> Output: 

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

Prolog

Works with: SWI-Prolog 7

Taking a cue from an SO answer:

<lang prolog>divisor(N, Divisor) :- 

   UpperBound is round(sqrt(N)),
   between(1, UpperBound, D),
   0 is N mod D,
   (
       Divisor = D
    ;
       LargerDivisor is N/D,
       LargerDivisor =\= D,
       Divisor = LargerDivisor
   ).

proper_divisor(N, D) :- 

   divisor(N, D),
   D =\= N.


%% Task 1 %

proper_divisors(N, Ds) :- 

   setof(D, proper_divisor(N, D), Ds).


%% Task 2 %

show_proper_divisors_of_range(Low, High) :- 

   findall( N:Ds,
            ( between(Low, High, N),
              proper_divisors(N, Ds) ),
            Results ),
   maplist(writeln, Results).


%% Task 3 %

proper_divisor_count(N, Count) :- 

   proper_divisors(N, Ds),
   length(Ds, Count).

find_most_proper_divisors_in_range(Low, High, Result) :- 

   aggregate_all( max(Count, N),
                  ( between(Low, High, N),
                    proper_divisor_count(N, Count) ),
                  max(MaxCount, Num) ),
   Result = (num(Num)-divisor_count(MaxCount)).</lang>

Output:

<lang prolog>?- show_proper_divisors_of_range(1,10). 2:[1] 3:[1] 4:[1,2] 5:[1] 6:[1,2,3] 7:[1] 8:[1,2,4] 9:[1,3] 10:[1,2,5] true.

?- find_most_proper_divisors_in_range(1,20000,Result). Result = num(15120)-divisor_count(79). </lang>

Python

Python: Literal

A very literal interpretation <lang python>>>> def proper_divs2(n): ... return {x for x in range(1, (n + 1) // 2 + 1) if n % x == 0 and n != x} ... >>> [proper_divs2(n) for n in range(1, 11)] [set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}] >>> >>> n, length = max(((n, len(proper_divs2(n))) for n in range(1, 20001)), key=lambda pd: pd[1]) >>> n 15120 >>> length 79 >>> </lang>


Python: From prime factors

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

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

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

This version is over an order of magnitude faster for generating the proper divisors of the first 20,000 integers; at the expense of simplicity. <lang python>from math import sqrt from functools import lru_cache, reduce from collections import Counter from itertools import product


MUL = int.__mul__


def prime_factors(n): 

   'Map prime factors to their multiplicity for n'
   d = _divs(n)
   d = [] if d == [n] else (d[:-1] if d[-1] == d else d)
   pf = Counter(d)
   return dict(pf)

@lru_cache(maxsize=None) def _divs(n): 

   'Memoized recursive function returning prime factors of n as a list'
   for i in range(2, int(sqrt(n)+1)):
       d, m  = divmod(n, i)
       if not m:
           return [i] + _divs(d)
   return [n]


def proper_divs(n): 

   Return the set of proper divisors of n.
   pf = prime_factors(n)
   pfactors, occurrences = pf.keys(), pf.values()
   multiplicities = product(*(range(oc + 1) for oc in occurrences))
   divs = {reduce(MUL, (pf**m for pf, m in zip(pfactors, multis)), 1)
           for multis in multiplicities}
   try:
       divs.remove(n)
   except KeyError:
       pass
   return divs or ({1} if n != 1 else set())


if __name__ == '__main__': 

   rangemax = 20000
   
   print([proper_divs(n) for n in range(1, 11)])
   print(*max(((n, len(proper_divs(n))) for n in range(1, 20001)), key=lambda pd: pd[1]))</lang>
Output:
[set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}]
15120 79

Racket

Short version

<lang racket>#lang racket (require math) (define (proper-divisors n) (drop-right (divisors n) 1)) (for ([n (in-range 1 (add1 10))]) 

 (printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))

(define most-under-20000 

 (for/fold ([best '(1)]) ([n (in-range 2 (add1 20000))])
   (define divs (proper-divisors n))
   (if (< (length (cdr best)) (length divs)) (cons n divs) best)))

(printf "~a has ~a proper divisors\n" 

       (car most-under-20000) (length (cdr most-under-20000)))</lang>
Output:
proper divisors of: 1	()
proper divisors of: 2	(1)
proper divisors of: 3	(1)
proper divisors of: 4	(1 2)
proper divisors of: 5	(1)
proper divisors of: 6	(1 2 3)
proper divisors of: 7	(1)
proper divisors of: 8	(1 2 4)
proper divisors of: 9	(1 3)
proper divisors of: 10	(1 2 5)
15120 has 79 proper divisors


Long version

The main module will only be executed when this file is executed. When used as a library, it will not be used. <lang racket>#lang racket/base (provide fold-divisors ; name as per "Abundant..." 

        proper-divisors)

(define (fold-divisors v n k0 kons) 

 (define n1 (add1 n))
 (cond
   [(>= n1 (vector-length v))
    (define rv (make-vector n1 k0))
    (for* ([n (in-range 1 n1)] [m (in-range (* 2 n) n1 n)])
      (vector-set! rv m (kons n (vector-ref rv m))))
    rv]
   [else v]))

(define proper-divisors 

 (let ([p.d-v (vector)])
   (λ (n)
     (set! p.d-v (reverse (fold-divisors p.d-v n null cons)))
     (vector-ref p.d-v n))))

(module+ main 

 (for ([n (in-range 1 (add1 10))])
   (printf "proper divisors of: ~a\t~a\n" n (proper-divisors n)))

 (define count-proper-divisors
   (let ([p.d-v (vector)])
     (λ(n) (set! p.d-v (fold-divisors p.d-v n 0 (λ (d n) (add1 n))))
           (vector-ref p.d-v n))))

 (void (count-proper-divisors 20000))

 (define-values [C I]
   (for*/fold ([C 0] [I 1])
              ([i (in-range 1 (add1 20000))]
               [c (in-value (count-proper-divisors i))]
               #:when [> c C])
     (values c i)))
 (printf "~a has ~a proper divisors\n" I C))</lang>

The output is the same as the short version above.

REXX

version 1

<lang rexx>Call time 'R' Do x=1 To 10 

 Say x '->' proper_divisors(x)
 End

hi=1 Do x=1 To 20000 

 /* If x//1000=0 Then Say x */
 npd=count_proper_divisors(x)
 Select
   When npd>hi Then Do
     list.npd=x
     hi=npd
     End
   When npd=hi Then
     list.hi=list.hi x
   Otherwise
     Nop
   End
 End

Say hi '->' list.hi

Say ' 166320 ->' count_proper_divisors(166320) Say '1441440 ->' count_proper_divisors(1441440)

Say time('E') 'seconds elapsed' Exit

proper_divisors: Procedure Parse Arg n If n=1 Then Return pd= /* Optimization reduces 37 seconds to 28 seconds */ If n//2=1 Then /* odd number */ 

 delta=2

Else /* even number */ 

 delta=1

Do d=1 To n%2 By delta 

 If n//d=0 Then
   pd=pd d
 End

Return space(pd)

count_proper_divisors: Procedure Parse Arg n Return words(proper_divisors(n))</lang>

Output:
1 ->
2 -> 1
3 -> 1
4 -> 1 2
5 -> 1
6 -> 1 2 3
7 -> 1
8 -> 1 2 4
9 -> 1 3
10 -> 1 2 5
79 -> 15120 18480
 166320 -> 159
1441440 -> 287
28.342000 seconds elapsed

version 2

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

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

With the (subroutine) optimization, it's over twenty times faster. <lang rexx>/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/ parse arg bot top inc range xtra /*get optional arguments from CL.*/ top=word(top bot 10,1); bot=word(bot 1,1); inc=word(inc 1,1) /*first range.*/ if range== then range=top /*use TOP for the 2nd range. */ w=length(top)+1; numeric digits max(9,w) /*'nuff digits for // operation*/ m=0 

    do n=bot  to top  by inc;   q=Pdivs(n);  #=words(q);      if q=='∞'  then #=q
    say  right(n,max(20,digits()))  'has'  center(#,4)   "proper divisors: "  q
    end   /*n*/                                 /* [↑]   process 1st range of integers.*/

say; @.= 'and' 

    do r=1  for range;   q=Pdivs(r);  #=words(q);             if #<m     then iterate
    @.#=@.#  @.  r;      m=#
    end   /*r*/                                 /* [↑]   process 2nd range of integers.*/
     __= 'is the highest number of proper divisors in range 1──►'range

say m __", and it's for: " subword(@.m,3) say /* [↓] handle any given extra numbers.*/ 

    do i=1  for words(xtra);                 n=word(xtra,i)
    numeric digits max(9,length(n)+1);       q=Pdivs(n);            #=words(q)
    say  right(n,max(20,w))          'has'   center(#,4)    "proper divisors."
    end   /*i*/                                 /* [↑] support extra specified integers*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ Pdivs: procedure; parse arg x,b; x=abs(x); if x==1 then return 

      odd=x//2;                              if x==0  then return '∞'
      a=1                                       /* [↓]  use all, or only odd #s.    ___*/
          do j=2+odd  by 1+odd  while j*j<x     /*divide by some integers up to    √ X */
          if x//j==0  then do;  a=a j;  b=x%j b /*if ÷, add both divisors to α & ß.    */
                           end
          end   /*j*/                           /* [↑]  %  is the REXX integer division*/
                                                /* [↓]  adjust for a square.        ___*/
      if j*j==x  then  return  a j b            /*Was  X  a square?    If so, add  √ X */
                       return  a   b            /*return the divisors  (both lists).   */</lang>

output   when using the following input:   0   10   1       20000       166320   1441440   11796480000 

                   0 has  ∞   proper divisors:  ∞
                   1 has  0   proper divisors:
                   2 has  1   proper divisors:  1
                   3 has  1   proper divisors:  1
                   4 has  2   proper divisors:  1 2
                   5 has  1   proper divisors:  1
                   6 has  3   proper divisors:  1 2 3
                   7 has  1   proper divisors:  1
                   8 has  3   proper divisors:  1 2 4
                   9 has  2   proper divisors:  1 3
                  10 has  3   proper divisors:  1 2 5

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

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

version 3

When factoring   20,000 integers,   this REXX version is over 12% faster than the REXX version 2.
When factoring 200,000 integers,   this REXX version is over 25% faster.

It accomplishes a faster speed by incorporating the calculation of an   integer square root   of an integer   (without using any floating point arithmetic). <lang rexx>/*REXX program finds proper divisors (and count) of integer ranges; finds the max count.*/ parse arg bot top inc range xtra /*get optional arguments from CL.*/ top=word(top bot 10,1); bot=word(bot 1,1); inc=word(inc 1,1) /*first range.*/ if range== then range=top /*use TOP for the 2nd range. */ w=length(top)+1; numeric digits max(9,w) /*'nuff digits for // operation*/ m=0 

    do n=bot  to top  by inc;    q=Pdivs(n);   #=words(q);  if q=='∞'  then #=q
    say  right(n,max(20,digits()))  'has'  center(#,4)   "proper divisors: "  q
    end   /*n*/                                 /* [↑]   process 1st range of integers.*/

say; @.='and' 

    do r=1  for range;   q=Pdivs(r);  #=words(q);      if #<m  then iterate
    @.#=@.#  @.  r;      m=#
    end   /*r*/                                 /* [↑]   process 2nd range of integers.*/
     __= 'is the highest number of proper divisors in range 1──►'range

say m __", and it's for: " subword(@.m,3) say /* [↓] handle any given extra numbers.*/ 

    do i=1  for words(xtra);                 n=word(xtra,i)
    numeric digits max(9,length(n)+1);       q=Pdivs(n);            #=words(q)
    say  right(n,max(20,w))          'has'   center(#,4)    "proper divisors."
    end   /*i*/                       /* [↑] support extra specified integers*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ Pdivs: procedure; parse arg x,b; x=abs(x); if x==1 then return 

      odd=x//2;                              if x==0  then return '∞'          /*   ___*/
      z=x; r=0; q=1;  do while q<=z; q=q*4; end /*R:  an integer which will be     √ X */
          do while q>1; q=q%4; _=z-r-q; r=r%2; if _>=0  then do;  z=_;  r=r+q;  end;  end
      a=1                                       /* [↓]  use all, or only odd #s.    ___*/
          do j=2+odd  by 1+odd to r-(r*r==x)    /*divide by some integers up to    √ X */
          if x//j==0  then do;  a=a j;  b=x%j b /*if ÷, add both divisors to α & ß.    */
                           end
          end   /*j*/                           /* [↑]  %  is the REXX integer division*/
                                                /* [↓]  adjust for a square.        ___*/
      if j*j==x  then  return  a j b            /*Was  X  a square?    If so, add  √ X */
                       return  a   b            /*return the divisors  (both lists).   */</lang>

output   is identical to the 2nd REXX version.

Ruby

<lang ruby>require "prime"

class Integer 

 def proper_divisors
   return [] if self == 1
   primes = prime_division.flat_map{|prime, freq| [prime] * freq}
   (1...primes.size).each_with_object([1]) do |n, res|
     primes.combination(n).map{|combi| res << combi.inject(:*)}
   end.flatten.uniq
 end

end

(1..10).map{|n| puts "#{n}: #{n.proper_divisors}"}

size, select = (1..20_000).group_by{|n| n.proper_divisors.size}.max select.each do |n| 

 puts "#{n} has #{size} divisors"

end</lang>

Output:
1: []
2: [1]
3: [1]
4: [1, 2]
5: [1]
6: [1, 2, 3]
7: [1]
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120 has 79 divisors
18480 has 79 divisors

An Alternative Approach

<lang ruby>#Determine the integer within a range of integers that has the most proper divisors

  1. Nigel Galloway: December 23rd., 2014

require "prime" n, g = 0 (1..20000).each{|i| e = i.prime_division.inject(1){|n,g| n * (g[1]+1)} 

                   n, g = e, i if e > n}

puts "#{g} has #{n-1} proper divisors"</lang>

Output:

In the range 1..200000 

15120 has 79 proper divisors

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

166320 has 159 proper divisors
1441440 has 287 proper divisors

Scala

Simple proper divisors

<lang Scala>def properDivisors(n: Int) = (1 to n/2).filter(i => n % i == 0) def format(i: Int, divisors: Seq[Int]) = f"$i%5d ${divisors.length}%2d ${divisors mkString " "}"

println(f" n cnt PROPER DIVISORS") val (count, list) = (1 to 20000).foldLeft( (0, List[Int]()) ) { (max, i) => 

   val divisors = properDivisors(i)
   if (i <= 10 || i == 100) println( format(i, divisors) )
   if (max._1 < divisors.length) (divisors.length, List(i))
   else if (max._1 == divisors.length) (divisors.length, max._2 ::: List(i))
   else max

}

list.foreach( number => println(f"$number%5d ${properDivisors(number).length}") )</lang>

Output:
    n   cnt   PROPER DIVISORS
    1     0   
    2     1   1
    3     1   1
    4     2   1 2
    5     1   1
    6     3   1 2 3
    7     1   1
    8     3   1 2 4
    9     2   1 3
   10     3   1 2 5
  100     8   1 2 4 5 10 20 25 50
15120    79
18480    79

Proper divisors for big (long) numbers

<lang Scala>import annotation.tailrec import collection.parallel.mutable.ParSeq

def factorize(n: Long): List[Long] = { 

   @tailrec
   def factors(tuple: (Long, Long, List[Long], Int)): List[Long] = {
       tuple match {
           case (1, _, acc, _)                 => acc
           case (n, k, acc, _) if (n % k == 0) => factors((n / k, k, acc ++ ParSeq(k), Math.sqrt(n / k).toInt))
           case (n, k, acc, sqr) if (k < sqr)  => factors(n, k + 1, acc, sqr)
           case (n, k, acc, sqr) if (k >= sqr) => factors((1, k, acc ++ ParSeq(n), 0))
       }
   }
   factors((n, 2, List[Long](), Math.sqrt(n).toInt))

} def properDivisors(n: Long) = { 

   val factors = factorize(n)
   val products = (1 until factors.length).map(i =>
       factors.combinations(i).map(_.product).toList).flatten
   (1L +: products).filter(_ < n)

}</lang>

Sidef

Translation of: Perl 6

<lang ruby>func propdiv (x) { 

   var divs = []
   divs << 1 if (x > 1)

   for d in (2 ..^ x.isqrt) {
       var y = (x // d)
       if (y*d == x) {
           divs << d
           y == d || (divs << y)
       }
   }

   divs

}

10.times { |i| say "#{i}\t#{propdiv(i).dump}" }

var max = 0 var candidates = [] for i in (1 ..^ 20_000) { 

   var divs = propdiv(i).len
   if (divs > max) {
       candidates = []
       max = divs
   }
   candidates << i if (divs == max)

}

say "max = #{max}, candidates = #{candidates.dump}"</lang>

Output:
1	[]
2	[1]
3	[1]
4	[1, 2]
5	[1]
6	[1, 2, 3]
7	[1]
8	[1, 2, 4]
9	[1, 3]
10	[1, 2, 5]
max = 79, candidates = [15120, 18480]

Swift

Simple function: <lang Swift>func properDivs1(n: Int) -> [Int] { 

   return filter (1 ..< n) { n % $0 == 0 }

}</lang> More efficient function: <lang Swift>import func Darwin.sqrt

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

func properDivs(n: Int) -> [Int] { 

   if n == 1 { return [] }
   
   var result = [Int]()
   
   for div in filter (1 ... sqrt(n), { n % $0 == 0 }) {
       
       result.append(div)

       if n/div != div && n/div != n { result.append(n/div) }
   }
   
   return sorted(result)

}</lang> Rest of the task: <lang Swift>for i in 1...10 { 

   println("\(i): \(properDivs(i))")

}

var (num, max) = (0,0)

for i in 1...20_000 { 

   let count = properDivs(i).count
   if (count > max) { (num, max) = (i, count) }

}

println("\(num): \(max)")</lang>

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

Tcl

Note that if a number, , greater than 1 divides exactly, both and are proper divisors. (The raw answers are not sorted; the pretty-printer code sorts.) <lang tcl>proc properDivisors {n} { 

   if {$n == 1} return
   set divs 1
   for {set i 2} {$i*$i <= $n} {incr i} {

if {!($n % $i)} { lappend divs $i if {$i*$i < $n} { lappend divs [expr {$n / $i}] } } 

   }
   return $divs

}

for {set i 1} {$i <= 10} {incr i} { 

   puts "$i => {[join [lsort -int [properDivisors $i]] ,]}"

} set maxI [set maxC 0] for {set i 1} {$i <= 20000} {incr i} { 

   set c [llength [properDivisors $i]]
   if {$c > $maxC} {

set maxI $i set maxC $c 

   }

} puts "max: $maxI => (...$maxC…)"</lang>

Output:
1 => {}
2 => {1}
3 => {1}
4 => {1,2}
5 => {1}
6 => {1,2,3}
7 => {1}
8 => {1,2,4}
9 => {1,3}
10 => {1,2,5}
max: 15120 => (...79...)

zkl

Translation of: D

<lang zkl>fcn properDivs(n){ [1.. (n + 1)/2 + 1].filter('wrap(x){ n%x==0 and n!=x }) } [1..10].apply(properDivs).println(); [1..20_001].apply('wrap(n){ T(properDivs(n).len(),n) }) 

  .reduce(fcn([(a,_)]ab, [(c,_)]cd){ a>c and ab or cd },T(0,0))
  .println();</lang>
Output:
L(L(),L(1),L(1),L(1,2),L(1),L(1,2,3),L(1),L(1,2,4),L(1,3),L(1,2,5))
L(79,18480)
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