# Proper divisors

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

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

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

Examples

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

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

Show all output here.

## 11l

Translation of: Python
```F proper_divs(n)
R Array(Set((1 .. (n + 1) I/ 2).filter(x -> @n % x == 0 & @n != x)))

print((1..10).map(n -> proper_divs(n)))

V (n, leng) = max(((1..20000).map(n -> (n, proper_divs(n).len))), key' pd -> pd)
print(n‘ ’leng)```
Output:
```[[], , , [1, 2], , [1, 2, 3], , [1, 2, 4], [1, 3], [1, 2, 5]]
15120 79
```

## 360 Assembly

Translation of: Rexx

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

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

## Action!

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

```BYTE FUNC GetDivisors(INT a INT ARRAY divisors)
INT i,max
BYTE count

max=a/2
count=0
FOR i=1 TO max
DO
IF a MOD i=0 THEN
divisors(count)=i
count==+1
FI
OD
RETURN (count)

PROC Main()
DEFINE MAXNUM="20000"
INT i,j,count,max,ind
INT ARRAY divisors(100)
BYTE ARRAY pdc(MAXNUM+1)

FOR i=1 TO 10
DO
count=GetDivisors(i,divisors)
PrintF("%I has %I proper divisors: [",i,count)
FOR j=0 TO count-1
DO
PrintI(divisors(j))
IF j<count-1 THEN
Put(32)
FI
OD
PrintE("]")
OD

PutE() PrintE("Searching for max number of divisors:")

FOR i=1 TO MAXNUM
DO
pdc(i)=1
OD
FOR i=2 TO MAXNUM
DO
FOR j=i+i TO MAXNUM STEP i
DO
pdc(j)==+1
OD
OD

max=0 ind=0
FOR i=1 TO MAXNUM
DO
count=pdc(i)
IF count>max THEN
max=count ind=i
FI
OD
PrintF("%I has %I proper divisors%E",ind,max)
RETURN```
Output:
```1 has 0 proper divisors: []
2 has 1 proper divisors: 
3 has 1 proper divisors: 
4 has 2 proper divisors: [1 2]
5 has 1 proper divisors: 
6 has 3 proper divisors: [1 2 3]
7 has 1 proper divisors: 
8 has 3 proper divisors: [1 2 4]
9 has 2 proper divisors: [1 3]
10 has 3 proper divisors: [1 2 5]

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

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

```generic
type Result_Type (<>) is limited private;
None: Result_Type;
with function One(X: Positive) return Result_Type;
with function Add(X, Y: Result_Type) return Result_Type
is <>;
package Generic_Divisors is

function Process
(N: Positive; First: Positive := 1) return Result_Type is
(if First**2 > N or First = N then None
elsif (N mod First)=0 then
(if First = 1 or First*First = N
else Process(N, First+1));

end Generic_Divisors;
```

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

```with Ada.Text_IO, Ada.Containers.Generic_Array_Sort, Generic_Divisors;

procedure Proper_Divisors is

begin
-- show the proper divisors of the numbers 1 to 10 inclusive.
declare
type Pos_Arr is array(Positive range <>) of Positive;
subtype Single_Pos_Arr is Pos_Arr(1 .. 1);
Empty: Pos_Arr(1 .. 0);

function Arr(P: Positive) return Single_Pos_Arr is ((others => P));

package Divisor_List is new Generic_Divisors
(Result_Type => Pos_Arr, None => Empty, One => Arr, Add =>  "&");

(Positive, Positive, Pos_Arr);
begin
for I in 1 .. 10 loop
declare
List: Pos_Arr := Divisor_List.Process(I);
begin
(Positive'Image(I) & " has" &
Natural'Image(List'Length) & " proper divisors:");
Sort(List);
for Item of List loop
end loop;
end;
end loop;
end;

-- find a number 1 .. 20,000 with the most proper divisors
declare
Number: Positive := 1;
Number_Count: Natural := 0;
Current_Count: Natural;

function Cnt(P: Positive) return Positive is (1);

package Divisor_Count is new Generic_Divisors
(Result_Type => Natural, None => 0, One => Cnt, Add =>  "+");

begin
for Current in 1 .. 20_000 loop
Current_Count := Divisor_Count.Process(Current);
if Current_Count > Number_Count then
Number := Current;
Number_Count := Current_Count;
end if;
end loop;
(Positive'Image(Number) & " has the maximum number of" &
Natural'Image(Number_Count) & " proper divisors.");
end;
end Proper_Divisors;
```
Output:
``` 1 has 0 proper divisors:
2 has 1 proper divisors: 1
3 has 1 proper divisors: 1
4 has 2 proper divisors: 1 2
5 has 1 proper divisors: 1
6 has 3 proper divisors: 1 2 3
7 has 1 proper divisors: 1
8 has 3 proper divisors: 1 2 4
9 has 2 proper divisors: 1 3
10 has 3 proper divisors: 1 2 5

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

## ALGOL 68

### As required by the Task

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32
```# MODE to hold an element of a list of proper divisors            #
MODE DIVISORLIST = STRUCT( INT divisor, REF DIVISORLIST next );

# end of divisor list value                                       #
REF DIVISORLIST nil divisor list = REF DIVISORLIST(NIL);

# resturns a DIVISORLIST containing the proper divisors of n      #
# if n = 1, 0 or -1, we return no divisors                        #
PROC proper divisors = ( INT n )REF DIVISORLIST:
BEGIN
REF DIVISORLIST result   := nil divisor list;
REF DIVISORLIST end list := result;
INT abs n  = ABS n;
IF abs n > 1 THEN
# build the list of divisors backeards, so they are  #
# returned in ascending order                        #
INT root n = ENTIER sqrt( abs n );
FOR d FROM root n BY -1 TO 2 DO
IF abs n MOD d = 0 THEN
# found another divisor                      #
result := HEAP DIVISORLIST
:= DIVISORLIST( d, result );
IF end list IS nil divisor list THEN
# first result                           #
end list := result
FI;
IF d * d /= n THEN
# add the other divisor to the end of    #
# the list                               #
next OF end list := HEAP DIVISORLIST
:= DIVISORLIST( abs n OVER d, nil divisor list );
end list         := next OF end list
FI
FI
OD;
# 1 is always a proper divisor of numbers > 1        #
result := HEAP DIVISORLIST
:= DIVISORLIST( 1, result )
FI;
result
END # proper divisors # ;

# returns the number of divisors in a DIVISORLIST                 #
PROC count divisors = ( REF DIVISORLIST list )INT:
BEGIN
INT result := 0;
REF DIVISORLIST divisors := list;
WHILE divisors ISNT nil divisor list DO
result +:= 1;
divisors := next OF divisors
OD;
result
END # count divisors # ;

# find the proper divisors of 1 : 10                              #
FOR n TO 10 DO
REF DIVISORLIST divisors := proper divisors( n );
print( ( "Proper divisors of: ", whole( n, -2 ), ": " ) );
WHILE divisors ISNT nil divisor list DO
print( ( " ", whole( divisor OF divisors, 0 ) ) );
divisors := next OF divisors
OD;
print( ( newline ) )
OD;

# find the first/only number in 1 : 20 000 with the most divisors  #
INT max number         = 20 000;
INT max divisors      :=      0;
INT has max divisors  :=      0;
INT with max divisors :=      0;
FOR d TO max number DO
INT divisor count = count divisors( proper divisors( d ) );
IF divisor count > max divisors THEN
# found a number with more divisors than the previous max  #
max divisors       := divisor count;
has max divisors   := d;
with max divisors  := 1
ELIF divisor count = max divisors THEN
# found another number with that many divisors             #
with max divisors +:= 1
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
) )```
Output:
```Proper divisors of:  1:
Proper divisors of:  2:  1
Proper divisors of:  3:  1
Proper divisors of:  4:  1 2
Proper divisors of:  5:  1
Proper divisors of:  6:  1 2 3
Proper divisors of:  7:  1
Proper divisors of:  8:  1 2 4
Proper divisors of:  9:  1 3
Proper divisors of: 10:  1 2 5
15120 is the first number upto 20000 with 79 divisors
```

### Faster Proper Divisor Counting

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

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

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

## ALGOL-M

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

```BEGIN

% COMPUTE P MOD Q %
INTEGER FUNCTION MOD (P, Q);
INTEGER P, Q;
BEGIN
MOD := P - Q * (P / Q);
END;

% COUNT, AND OPTIONALLY DISPLAY, PROPER DIVISORS OF N %
INTEGER FUNCTION DIVISORS(N, DISPLAY);
INTEGER N, DISPLAY;
BEGIN
INTEGER I, LIMIT, COUNT, START, DELTA;
IF MOD(N, 2) = 0 THEN
BEGIN
START := 2;
DELTA := 1;
END
ELSE  % ONLY NEED TO CHECK ODD DIVISORS %
BEGIN
START := 3;
DELTA := 2;
END;
% 1 IS A DIVISOR OF ANY NUMBER > 1 %
IF N > 1 THEN COUNT := 1 ELSE COUNT := 0;
IF (DISPLAY <> 0) AND (COUNT <> 0) THEN WRITEON(1);
% CHECK REMAINING POTENTIAL DIVISORS %
I := START;
LIMIT := N / START;
WHILE I <= LIMIT DO
BEGIN
IF MOD(N, I) = 0 THEN
BEGIN
IF DISPLAY <> 0 THEN WRITEON(I);
COUNT := COUNT + 1;
END;
I := I + DELTA;
IF COUNT = 1 THEN LIMIT := N / I;
END;
DIVISORS := COUNT;
END;

COMMENT MAIN PROGRAM BEGINS HERE;
INTEGER I, NDIV, TRUE, FALSE, HIGHDIV, HIGHNUM;
TRUE := -1;
FALSE := 0;

WRITE("PROPER DIVISORS OF FIRST TEN NUMBERS:");
FOR I := 1 STEP 1 UNTIL 10 DO
BEGIN
WRITE(I, " : ");
NDIV := DIVISORS(I, TRUE);
END;

WRITE("SEARCHING FOR NUMBER UP TO 10000 WITH MOST DIVISORS ...");
HIGHDIV := 1;
HIGHNUM := 1;
FOR I := 1 STEP 1 UNTIL 10000 DO
BEGIN
NDIV := DIVISORS(I, FALSE);
IF NDIV > HIGHDIV THEN
BEGIN
HIGHDIV := NDIV;
HIGHNUM := I;
END;
END;
WRITE("THE NUMBER IS:", HIGHNUM);
WRITE("IT HAS", HIGHDIV, " DIVISORS");

END```
Output:
```PROPER DIVISORS OF FIRST TEN NUMBERS:
1 :
2 :      1
3 :      1
4 :      1     2
5 :      1
6 :      1     2     3
7 :      1
8 :      1     2     4
9 :      1     3
10 :      1     2     5
SEARCHING FOR NUMBER UP TO 10000 WITH MOST DIVISORS:
THE NUMBER IS:  7560
IT HAS    63 DIVISORS
```

## AppleScript

### Functional

Translation of: JavaScript
```-- PROPER DIVISORS -----------------------------------------------------------

-- properDivisors :: Int -> [Int]
on properDivisors(n)
if n = 1 then
{1}
else
set realRoot to n ^ (1 / 2)
set intRoot to realRoot as integer
set blnPerfectSquare to intRoot = realRoot

-- isFactor :: Int -> Bool
script isFactor
on |λ|(x)
n mod x = 0
end |λ|
end script

-- Factors up to square root of n,
set lows to filter(isFactor, enumFromTo(1, intRoot))

-- and quotients of these factors beyond the square root,

-- integerQuotient :: Int -> Int
script integerQuotient
on |λ|(x)
(n / x) as integer
end |λ|
end script

-- excluding n itself (last item)
items 1 thru -2 of (lows & map(integerQuotient, ¬
items (1 + (blnPerfectSquare as integer)) thru -1 of reverse of lows))
end if
end properDivisors

-- TEST ----------------------------------------------------------------------
on run
-- numberAndDivisors :: Int -> [Int]
script numberAndDivisors
on |λ|(n)
{num:n, divisors:properDivisors(n)}
end |λ|
end script

-- maxDivisorCount :: Record -> Int -> Record
script maxDivisorCount
on |λ|(a, n)
set intDivisors to length of properDivisors(n)

if intDivisors ≥ divisors of a then
{num:n, divisors:intDivisors}
else
a
end if
end |λ|
end script

{oneToTen:map(numberAndDivisors, ¬
enumFromTo(1, 10)), mostDivisors:foldl(maxDivisorCount, ¬
{num:0, divisors:0}, enumFromTo(1, 20000))} ¬

end run

-- GENERIC FUNCTIONS ---------------------------------------------------------

-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m > n then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo

-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter

-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
```
Output:
```{oneToTen:{{num:1, divisors:{1}}, {num:2, divisors:{1}}, {num:3, divisors:{1}},
{num:4, divisors:{1, 2}}, {num:5, divisors:{1}}, {num:6, divisors:{1, 2, 3}},
{num:7, divisors:{1}}, {num:8, divisors:{1, 2, 4}}, {num:9, divisors:{1, 3}},
{num:10, divisors:{1, 2, 5}}},
mostDivisors:{num:18480, divisors:79}}
```

### Idiomatic

```on properDivisors(n)
set output to {}

if (n > 1) then
set sqrt to n ^ 0.5
set limit to sqrt div 1
if (limit = sqrt) then
set end of output to limit
set limit to limit - 1
end if
repeat with i from limit to 2 by -1
if (n mod i is 0) then
set beginning of output to i
set end of output to n div i
end if
end repeat
set beginning of output to 1
end if

return output
end properDivisors

local output, astid, i, maxPDs, maxPDNums, pdCount
set output to {}
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to ", "
repeat with i from 1 to 10
set end of output to (i as text) & "'s proper divisors:  {" & properDivisors(i) & "}"
end repeat
set maxPDs to 0
set maxPDNums to {}
repeat with i from 1 to 20000
set pdCount to (count properDivisors(i))
if (pdCount > maxPDs) then
set maxPDs to pdCount
set maxPDNums to {i}
else if (pdCount = maxPDs) then
set end of maxPDNums to i
end if
end repeat
set end of output to linefeed & "Largest number of proper divisors for any number from 1 to 20,000:  " & maxPDs
set end of output to "Numbers with this many:  " & maxPDNums
set AppleScript's text item delimiters to linefeed
set output to output as text
set AppleScript's text item delimiters to astid
return output
```
Output:
```"1's proper divisors:  {}
2's proper divisors:  {1}
3's proper divisors:  {1}
4's proper divisors:  {1, 2}
5's proper divisors:  {1}
6's proper divisors:  {1, 2, 3}
7's proper divisors:  {1}
8's proper divisors:  {1, 2, 4}
9's proper divisors:  {1, 3}
10's proper divisors:  {1, 2, 5}

Largest number of proper divisors for any number from 1 to 20,000:  79
Numbers with this many:  15120, 18480"
```

## Arc

```;; Given num, return num and the list of its divisors
(= divisor (fn (num)
(= dlist '())
(when (is 1 num) (= dlist '(1 0)))
(when (is 2 num) (= dlist '(2 1)))
(unless (or (is 1 num) (is 2 num))
(up i 1 (+ 1 (/ num 2))
(if (is 0 (mod num i))
(push i dlist)))
(= dlist (cons num dlist)))
dlist))

;; Find out what number has the most divisors between 2 and 20,000.
;; Print a list of the largest known number's divisors as it is found.
(= div-lists (fn (cnt (o show 0))
(= tlist '()) (= clist tlist)
(when (> show 0) (prn tlist))
(up i 1 cnt
(divisor i)
(when (is 1 show) (prn dlist))
(when (>= (len dlist) (len tlist))
(= tlist dlist)
(when (is show 2) (prn tlist))
(let c (- (len dlist) 1)
(push (list i c) clist))))

(= many-divisors (list ((clist 0) 1)))
(for n 0 (is ((clist n) 1) ((clist 0) 1)) (= n (+ 1 n))
(push ((clist n) 0) many-divisors))
(= many-divisors (rev many-divisors))
(prn "The number with the most divisors under " cnt
" has " (many-divisors 0) " divisors.")
(prn "It is the number "
(if (> 2 (len many-divisors)) (cut (many-divisors) 1)
(many-divisors 1)) ".")
(prn "There are " (- (len many-divisors) 1) " numbers"
" with this trait, and they are "
(map [many-divisors _] (range 1 (- (len many-divisors) 1))))
(prn (map [divisor _] (cut many-divisors 1)))
many-divisors))

(div-lists 10 1)
(div-lists 20000)
;; This took about 10 minutes on my machine.```
Output:
```(1 0)
(2 1)
(3 1)
(4 2 1)
(5 1)
(6 3 2 1)
(7 1)
(8 4 2 1)
(9 3 1)
(10 5 2 1)
The number with the most divisors under 10 has 3 divisors.
It is the number 10.
There are 3 numbers with this trait, and they are (10 8 6)
((10 5 2 1) (8 4 2 1) (6 3 2 1))
'(3 10 8 6)

The number with the most divisors under 20000 has 79 divisors.
It is the number 18480.
There are 2 numbers with this trait, and they are (18480 15120)```

## ARM Assembly

Works with: as version Raspberry Pi
```/* ARM assembly Raspberry PI  */
/*  program proFactor.s   */

/* REMARK 1 : this program use routines in a include file
see task Include a file language arm assembly
for the routine affichageMess conversion10
see at end of this program the instruction include */
/*******************************************/
/* Constantes                              */
/*******************************************/
.equ STDOUT, 1                           @ Linux output console
.equ EXIT,   1                           @ Linux syscall
.equ WRITE,  4                           @ Linux syscall

/*******************************************/
/* Initialized data                        */
/*******************************************/
.data
szMessStartPgm:          .asciz "Program start \n"
szMessEndPgm:            .asciz "Program normal end.\n"
szMessError:             .asciz "\033[31mError Allocation !!!\n"
szCarriageReturn:        .asciz "\n"

/* datas message display */
szMessEntete:            .ascii "Number :"
sNumber:                 .space 12,' '
.asciz " Divisors :"
szMessResult:            .ascii " "
sValue:                  .space 12,' '
.asciz ""
szMessDivNumber:         .ascii "\nnumber divisors :"
sCounter:                .space 12,' '
.asciz "\n"
szMessNumberMax:         .ascii "Number :"
sNumberMax:              .space 12,' '
.ascii " has "
sDivMax:                 .space 12, ' '
.asciz " divisors\n"
/*******************************************/
/* UnInitialized data                      */
/*******************************************/
.bss
/*******************************************/
/*  code section                           */
/*******************************************/
.text
.global main
main:                               @ program start
ldr r0,iAdrszMessStartPgm       @ display start message
bl affichageMess
mov r2,#1
1:
mov r0,r2                       @  number
ldr r1,iAdrsNumber              @ and convert ascii string
bl conversion10
ldr r0,iAdrszMessEntete         @ display result message
bl affichageMess
mov r0,r2                       @  number
mov r1,#1                       @ display flag
bl divisors                     @ display divisors
ldr r1,iAdrsCounter              @ and convert ascii string
bl conversion10
ldr r0,iAdrszMessDivNumber      @ display result message
bl affichageMess
cmp r2,#10
ble 1b

mov r2,#2
mov r3,#0
mov r4,#0
ldr r5,iMaxi
2:
mov r0,r2
mov r1,#0                       @ display flag
bl divisors                     @ display divisors
cmp r0,r3
movgt r3,r0
movgt r4,r2
cmp r2,r5
ble 2b
mov r0,r4
ldr r1,iAdrsNumberMax           @ and convert ascii string
bl conversion10
mov r0,r3
ldr r1,iAdrsDivMax              @ and convert ascii string
bl conversion10
bl affichageMess

ldr r0,iAdrszMessEndPgm         @ display end message
bl affichageMess
b 100f
99:                                 @ display error message
bl affichageMess
100:                                @ standard end of the program
mov r0, #0                      @ return code
mov r7, #EXIT                   @ request to exit program
svc 0                           @ perform system call
iMaxi:                     .int 20000
/******************************************************************/
/*     divisors function                         */
/******************************************************************/
/* r0 contains the number  */
/* r1 contains display flag (<>0: display, 0: no display )
/* r0 return divisors number */
divisors:
push {r1-r8,lr}             @ save  registers
cmp r0,#1                   @ = 1 ?
movle r0,#0
ble 100f
mov r7,r0
mov r8,r1
cmp r8,#0
beq 1f
mov r0,#1                   @ first divisor = 1
ldr r1,iAdrsValue           @ and convert ascii string
bl conversion10
ldr r0,iAdrszMessResult     @ display result message
bl affichageMess
1:                              @ begin loop
lsr r4,r7,#1                @ Maxi
mov r6,r4                   @ first divisor
mov r5,#1                   @ Counter divisors
2:
mov r0,r7                   @ dividende = number
mov r1,r6                   @ divisor
bl division
cmp r3,#0                   @ remainder = 0 ?
bne 3f
cmp r8,#0                   @ display divisor ?
beq 3f
mov r0,r2                   @ divisor
ldr r1,iAdrsValue           @ and convert ascii string
bl conversion10
ldr r0,iAdrszMessResult     @ display result message
bl affichageMess
3:
sub r6,r6,#1                @ decrement divisor
cmp r6,#2                   @ End ?
bge 2b                      @ no loop
mov r0,r5                   @ return divisors number

100:
pop {r1-r8,lr}              @ restaur registers
bx lr                       @ return

/***************************************************/
/*      ROUTINES INCLUDE                 */
/***************************************************/
.include "../affichage.inc"```
Output:
```Program start
Number :1            Divisors :
number divisors :0
Number :2            Divisors : 1            2
number divisors :2
Number :3            Divisors : 1            3
number divisors :2
Number :4            Divisors : 1            2
number divisors :2
Number :5            Divisors : 1
number divisors :1
Number :6            Divisors : 1            2            3
number divisors :3
Number :7            Divisors : 1
number divisors :1
Number :8            Divisors : 1            2            4
number divisors :3
Number :9            Divisors : 1            3
number divisors :2
Number :10           Divisors : 1            2            5
number divisors :3
Number :15120        has 79           divisors
Program normal end.

```

## Arturo

```properDivisors: function [x] ->
(factors x) -- x

loop 1..10 'x ->
print ["proper divisors of" x "=>" properDivisors x]

maxN: 0
maxProperDivisors: 0

loop 1..20000 'x [
pd: size properDivisors x
if maxProperDivisors < pd [
maxN: x
maxProperDivisors: pd
]
]

print ["The number with the most proper divisors (" maxProperDivisors ") is" maxN]
```
Output:
```proper divisors of 1 => []
proper divisors of 2 => 
proper divisors of 3 => 
proper divisors of 4 => [1 2]
proper divisors of 5 => 
proper divisors of 6 => [1 2 3]
proper divisors of 7 => 
proper divisors of 8 => [1 2 4]
proper divisors of 9 => [1 3]
proper divisors of 10 => [1 2 5]
The number with the most proper divisors ( 79 ) is 15120```

## AutoHotkey

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

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

## AWK

```# syntax: GAWK -f PROPER_DIVISORS.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
}
```

output:

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

## BASIC

Works with: QBasic version 1.1
Works with: QuickBasic version 4.5
```FUNCTION CountProperDivisors (number)
IF number < 2 THEN CountProperDivisors = 0
count = 0
FOR i = 1 TO number \ 2
IF number MOD i = 0 THEN count = count + 1
NEXT i
CountProperDivisors = count
END FUNCTION

SUB ListProperDivisors (limit)
IF limit < 1 THEN EXIT SUB
FOR i = 1 TO limit
PRINT USING "## ->"; i;
IF i = 1 THEN PRINT " (None)";
FOR j = 1 TO i \ 2
IF i MOD j = 0 THEN PRINT " "; j;
NEXT j
PRINT
NEXT i
END SUB

most = 1
maxCount = 0

PRINT "The proper divisors of the following numbers are: "; CHR\$(10)
ListProperDivisors (10)

FOR n = 2 TO 20000
'' It is extremely slow in this loop
count = CountProperDivisors(n)
IF count > maxCount THEN
maxCount = count
most = n
END IF
NEXT n

PRINT
PRINT most; " has the most proper divisors, namely "; maxCount
END
```
Output:
```Igual que la entrada de FreeBASIC o PureBasic.
```

### BASIC256

```subroutine ListProperDivisors(limit)
if limit < 1 then return
for i = 1 to limit
print i; " ->";
if i = 1 then
print " (None)"
continue for
end if
for j = 1 to i \ 2
if i mod j = 0 then print " "; j;
next j
print
next i
end subroutine

function CountProperDivisors(number)
if number < 2 then return 0
dim cont = 0
for i = 1 to number \ 2
if number mod i = 0 then cont += 1
next i
return cont
end function

most = 1
maxCount = 0

print "The proper divisors of the following numbers are:"
print
call ListProperDivisors(10)

for n = 2 to 20000
cont = CountProperDivisors(n)
if cont > maxCount then
maxCount = cont
most = n
end if
next n

print
print most; " has the most proper divisors, namely "; maxCount
end```
Output:
```Igual que la entrada de FreeBASIC o PureBasic.
```

### Craft Basic

```let m = 1
let l = 10

if l >= 1 then

for i = 1 to l

if i = 1 then

print i, " : (None)"

else

for j = 1 to int(i / 2)

if i % j = 0 then

print i, " :", j

endif

next j

endif

next i

endif

for n = 2 to 20000

let c = 0

if n >= 2 then

for i = 1 to int(n / 2)

if n % i = 0 then

let c = c + 1

endif

next i

endif

if c > x then

let x = c
let m = n

endif

wait

next n

print m, " has the most proper divisors", comma, " namely ", x
```
Output:
```
1 : (None)
2 : 1
3 : 1
4 : 1 2
5 : 1
6 : 1 2 3
7 : 1
8 : 1 2 4
9 : 1 3
10 : 1 2 5
15120 has the most proper divisors, namely 79

```

### True BASIC

```FUNCTION CountProperDivisors (number)
IF number < 2 THEN LET CountProperDivisors = 0
LET count = 0
FOR i = 1 TO INT(number / 2)
IF MOD(number, i) = 0 THEN LET count = count + 1
NEXT i
LET CountProperDivisors = count
END FUNCTION

SUB ListProperDivisors (limit)
IF limit < 1 THEN EXIT SUB
FOR i = 1 TO limit
PRINT i; " ->";
IF i = 1 THEN PRINT " (None)";
FOR j = 1 TO INT(i / 2)
IF MOD(i, j) = 0 THEN PRINT " "; j;
NEXT j
PRINT
NEXT i
END SUB

LET most = 1
LET maxCount = 0

PRINT "The proper divisors of the following numbers are: "; CHR\$(10)
CALL LISTPROPERDIVISORS (10)

FOR n = 2 TO 20000
LET count = CountProperDivisors(n)
IF count > maxCount THEN
LET maxCount = count
LET most = n
END IF
NEXT n

PRINT
PRINT most; "has the most proper divisors, namely"; maxCount
END
```
Output:
```Igual que la entrada de FreeBASIC o PureBasic.
```

### Yabasic

```sub ListProperDivisors(limit)
if limit < 1 then return : fi
for i = 1 to limit
print i, " ->";
if i = 1 then
print " (None)"
continue //for
endif
for j = 1 to int(i / 2)
if mod(i, j) = 0 then print " ", j; : fi
next j
print
next i
end sub

sub CountProperDivisors(number)
if number < 2 then return 0 : fi
count = 0
for i = 1 to int(number / 2)
if mod(number, i) = 0 then count = count + 1 : fi
next i
return count
end sub

most = 1
maxCount = 0

print "The proper divisors of the following numbers are: \n"
ListProperDivisors(10)

for n = 2 to 20000
count = CountProperDivisors(n)
if count > maxCount then
maxCount = count
most = n
endif
next n

print
print most, " has the most proper divisors, namely ", maxCount
end```
Output:
```Igual que la entrada de FreeBASIC o PureBasic.
```

## BaCon

```FUNCTION ProperDivisor(nr, show)

LOCAL probe, total

FOR probe = 1 TO nr-1
IF MOD(nr, probe) = 0 THEN
IF show THEN PRINT " ", probe;
INCR total
END IF
NEXT

END FUNCTION

FOR x = 1 TO 10
PRINT x, ":";
IF ProperDivisor(x, 1) = 0 THEN PRINT " 0";
PRINT
NEXT

FOR x = 1 TO 20000
DivisorCount = ProperDivisor(x, 0)
IF DivisorCount > MaxDivisors THEN
MaxDivisors = DivisorCount
MagicNumber = x
END IF
NEXT

PRINT "Most proper divisors for number in the range 1-20000: ", MagicNumber, " with ", MaxDivisors, " divisors."
```
Output:
```1: 0
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
Most proper divisors for number in the range 1-20000: 15120 with 79 divisors.
```

## C

### Brute Force

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

```#include <stdio.h>
#include <stdbool.h>

int proper_divisors(const int n, bool print_flag)
{
int count = 0;

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

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

return count;
}

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

int max = 0;
int max_i = 1;

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

printf("%d with %d divisors\n", max_i, max);
return 0;
}
```
Output:
```1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
18480 with 79 divisors
```

### Number Theoretic

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

```#include <stdio.h>
#include <stdbool.h>

int proper_divisors(const int n, bool print_flag)
{
int count = 0;

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

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

return count;
}

int countProperDivisors(int n){
int prod = 1,i,count=0;

while(n%2==0){
count++;
n /= 2;
}

prod *= (1+count);

for(i=3;i*i<=n;i+=2){
count = 0;

while(n%i==0){
count++;
n /= i;
}

prod *= (1+count);
}

if(n>2)
prod *= 2;

return prod - 1;
}

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

int max = 0;
int max_i = 1;

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

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

## C#

```namespace RosettaCode.ProperDivisors
{
using System;
using System.Collections.Generic;
using System.Linq;

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

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

var record = Enumerable.Range(1, 20000).Select(number => new
{
Number = number,
Count = ProperDivisors(number).Count()
}).OrderByDescending(currentRecord => currentRecord.Count).First();
Console.WriteLine("{0}: {1}", record.Number, record.Count);
}
}
}
```
Output:
```1: {}
2: {1}
3: {1}
4: {1, 2}
5: {1}
6: {1, 2, 3}
7: {1}
8: {1, 2, 4}
9: {1, 3}
10: {1, 2, 5}
15120: 79```

## C++

```#include <vector>
#include <iostream>
#include <algorithm>

std::vector<int> properDivisors ( int number ) {
std::vector<int> divisors ;
for ( int i = 1 ; i < number / 2 + 1 ; i++ )
if ( number % i == 0 )
divisors.push_back( i ) ;
return divisors ;
}

int main( ) {
std::vector<int> divisors ;
unsigned int maxdivisors = 0 ;
int corresponding_number = 0 ;
for ( int i = 1 ; i < 11 ; i++ ) {
divisors =  properDivisors ( i ) ;
std::cout << "Proper divisors of " << i << ":\n" ;
for ( int number : divisors ) {
std::cout << number << " " ;
}
std::cout << std::endl ;
divisors.clear( ) ;
}
for ( int i = 11 ; i < 20001 ; i++ ) {
divisors =  properDivisors ( i ) ;
if ( divisors.size( ) > maxdivisors ) {
maxdivisors = divisors.size( ) ;
corresponding_number = i ;
}
divisors.clear( ) ;
}

std::cout << "Most divisors has " << corresponding_number <<
" , it has " << maxdivisors << " divisors!\n" ;
return 0 ;
}
```
Output:
```Proper divisors of 1:

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

## Ceylon

```shared void run() {

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

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

value start = 1;
value end = 20k;

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

print("the number(s) with the most divisors between ``start`` and ``end`` is/are:
``mostDivisors?.item else "nothing"`` with ``mostDivisors?.key else "no"`` divisors");
}
```
Output:
```1 => []
2 => { 1 }
3 => { 1 }
4 => { 1, 2 }
5 => { 1 }
6 => { 1, 2, 3 }
7 => { 1 }
8 => { 1, 2, 4 }
9 => { 1, 3 }
10 => { 1, 2, 5 }
the number(s) with the most divisors between 1 and 20000 is/are:
[15120, 18480] with 79 divisors```

## Clojure

```(ns properdivisors
(:gen-class))

(defn proper-divisors [n]
" Proper divisors of n"
(if (= n 1)
[]
(filter #(= 0 (rem n %)) (range 1 n))))

;; Property divisors of numbers 1 to 20,000 inclusive
(def data (for [n (range 1 (inc 20000))]
[n (proper-divisors n)]))

;; Find Max
(defn maximal-key [k x & xs]
" Normal max-key only finds one key that produces maximum, while this function finds them all "
(reduce (fn [ys x]
(let [c (compare (k x) (k (peek ys)))]
(cond
(pos? c) [x]
(neg? c) ys
:else    (conj ys x))))
[x]
xs))

(println "n\tcnt\tPROPER DIVISORS")
(doseq [n (range 1 11)]
(let [factors (proper-divisors n)]
(println n "\t" (count factors) "\t" factors)))

(def max-data (apply maximal-key (fn [[i pd]] (count pd)) data))

(doseq [[n factors] max-data]
(println n " has " (count factors) " divisors"))
```
Output:
```n	cnt	PROPER DIVISORS
1 	 0 	 []
2 	 1 	 (1)
3 	 1 	 (1)
4 	 2 	 (1 2)
5 	 1 	 (1)
6 	 3 	 (1 2 3)
7 	 1 	 (1)
8 	 3 	 (1 2 4)
9 	 2 	 (1 3)
10 	 3 	 (1 2 5)
15120  has  79  divisors
18480  has  79  divisors
```

## Common Lisp

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

```(defun proper-divisors-recursive (product &optional (results '(1)))
"(int,list)->list::Function to find all proper divisors of a +ve integer."

(defun smallest-divisor (x)
"int->int::Find the smallest divisor of an integer > 1."
(if (evenp x) 2
(do ((lim (truncate (sqrt x)))
(sd 3 (+ sd 2)))
((or (integerp (/ x sd)) (> sd lim)) (if (> sd lim) x sd)))))

(defun pd-rec (fac)
"(int,int)->nil::Recursive function to find proper divisors of a +ve integer"
(when (not (member fac results))
(push fac results)
(let ((hifac (/ fac (smallest-divisor fac))))
(pd-rec hifac)
(pd-rec (/ product hifac)))))

(pd-rec product)
(butlast (sort (copy-list results) #'<)))

(defun task (method &optional (n 1) (most-pds '(0)))
(dotimes (i 19999)
(let ((npds (length (funcall method (incf n))))
(hiest (car most-pds)))
(when (>= npds hiest)
(if (> npds hiest)
(setf most-pds (list npds (list n)))
(setf most-pds (list npds (cons n (second most-pds))))))))
most-pds)

(defun main ()
(format t "Task 1:Proper Divisors of [1,10]:~%")
(dotimes (i 10) (format t "~A:~A~%" (1+ i) (proper-divisors-recursive (1+ i))))
(format t "Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:~%~A~%"
```
Output:
```CL-USER(10): (main)
1:NIL
2:(1)
3:(1)
4:(1 2)
5:(1)
6:(1 2 3)
7:(1)
8:(1 2 4)
9:(1 3)
10:(1 2 5)
Task 2:Count & list of numbers <=20,000 with the most Proper Divisors:
(79 (18480 15120))
NIL```

## Component Pascal

```MODULE RosettaProperDivisor;
IMPORT StdLog;

PROCEDURE Pd*(n: LONGINT;OUT r: ARRAY OF LONGINT):LONGINT;
VAR
i,j: LONGINT;
BEGIN
i := 1;j := 0;
IF n >  1 THEN
WHILE (i < n) DO
IF (n MOD i) = 0 THEN
IF (j < LEN(r)) THEN r[j] := i END; INC(j)
END;
INC(i)
END;
END;
RETURN j
END Pd;

PROCEDURE Do*;
VAR
r: ARRAY 128 OF LONGINT;
i,j,found,max,idxMx: LONGINT;
mx: ARRAY 128 OF LONGINT;
BEGIN
FOR i := 1 TO 10 DO
found := Pd(i,r);
IF found > LEN(r) THEN (* Error. more pd than r can admit *) HALT(1) END;
StdLog.Int(i);StdLog.String("[");StdLog.Int(found);StdLog.String("]:> ");
FOR j := 0 TO found - 1 DO
StdLog.Int(r[j]);StdLog.Char(' ');
END;
StdLog.Ln
END;

max := 0;idxMx := 0;
FOR i := 1 TO 20000 DO
found := Pd(i,r);
IF found > max THEN
idxMx:= 0;mx[idxMx] := i;max := found
ELSIF found = max THEN
INC(idxMx);mx[idxMx] := i
END;
END;
StdLog.String("Found: ");StdLog.Int(idxMx + 1);
StdLog.String(" Numbers with the longest proper divisors [");
StdLog.Int(max);StdLog.String("]: ");StdLog.Ln;
FOR i := 0 TO idxMx DO
StdLog.Int(mx[i]);StdLog.Ln
END
END Do;

END RosettaProperDivisor.

^Q RosettaProperDivisor.Do~```
Output:
``` 1[ 0]:>
2[ 1]:>  1
3[ 1]:>  1
4[ 2]:>  1  2
5[ 1]:>  1
6[ 3]:>  1  2  3
7[ 1]:>  1
8[ 3]:>  1  2  4
9[ 2]:>  1  3
10[ 3]:>  1  2  5
Found:  2 Numbers with the longest proper divisors [ 79]:
15120
18480
```

## D

Translation of: Python

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

```void main() /*@safe*/ {
import std.stdio, std.algorithm, std.range, std.typecons;

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

iota(1, 11).map!properDivs.writeln;
iota(1, 20_001).map!(n => tuple(properDivs(n).count, n)).reduce!max.writeln;
}
```
Output:
```[[], , , [1, 2], , [1, 2, 3], , [1, 2, 4], [1, 3], [1, 2, 5]]
Tuple!(uint, int)(79, 18480)```

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

## Delphi

Translation of: C#
```program ProperDivisors;

{\$APPTYPE CONSOLE}

{\$R *.res}

uses
System.SysUtils,
System.Generics.Collections;

type
TProperDivisors = TArray<Integer>;

function GetProperDivisors(const value: Integer): TProperDivisors;
var
i, count: Integer;
begin
count := 0;

for i := 1 to value div 2 do
begin
if value mod i = 0 then
begin
inc(count);
SetLength(result, count);
Result[count - 1] := i;
end;
end;
end;

procedure Println(values: TProperDivisors);
var
i: Integer;
begin
Write('[');
if Length(values) > 0 then
for i := 0 to High(values) do
Write(Format('%2d', [values[i]]));
Writeln(']');
end;

var
number, max_count, count, max_number: Integer;

begin
for number := 1 to 10 do
begin
write(number, ': ');
Println(GetProperDivisors(number));
end;

max_count := 0;
for number := 1 to 20000 do
begin
count := length(GetProperDivisors(number));
if count > max_count then
begin
max_count := count;
max_number := number;
end;
end;

Write(max_number, ': ', max_count);

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

Version with TParallel.For

```program ProperDivisors;

{\$APPTYPE CONSOLE}

{\$R *.res}

uses
System.SysUtils,
System.SyncObjs;

type
TProperDivisors = array of Integer;

function GetProperDivisors(const value: Integer): TProperDivisors;
var
i, count: Integer;
begin
count := 0;

for i := 1 to value div 2 do
begin
if value mod i = 0 then
begin
inc(count);
SetLength(result, count);
Result[count - 1] := i;
end;
end;
end;

procedure Println(values: TProperDivisors);
var
i: Integer;
begin
Write('[');
if Length(values) > 0 then
for i := 0 to High(values) do
Write(Format('%2d', [values[i]]));
Writeln(']');
end;

var
number, max_count, count, max_number: Integer;

begin
for number := 1 to 10 do
begin
write(number, ': ');
Println(GetProperDivisors(number));
end;

max_count := 0;
TParallel.for (1, 20000,
procedure(I: Int64)
begin
count := length(GetProperDivisors(I));
if count > max_count then
begin
TInterlocked.Exchange(max_count, count);
TInterlocked.Exchange(max_number, I);
end;
end);

Writeln(max_number, ': ', max_count);
end.
```

## Dyalect

Translation of: Swift
```func properDivs(n) {
if n == 1 {
yield break
}
for x in 1..<n {
if n % x == 0 {
yield x
}
}
}

for i in 1..10 {
print("\(i): \(properDivs(i).ToArray())")
}

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

for i in 1..20000 {
let count = properDivs(i).Length()
if count > max {
set (num, max) = (i, count)
}
}

print("\(num): \(max)")```
Output:
```1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79```

## EchoLisp

```(lib 'list) ;; list-delete

;; let n = product p_i^a_i , p_i prime
;; number of divisors = product (a_i + 1) - 1
(define (numdivs n)
(1- (apply * (map (lambda(g) (1+ (length g))) (group (prime-factors n))))))

(remember 'numdivs)

;; prime powers
;; input : a list g of grouped prime factors ( 3 3 3 ..)
;; returns (1 3 9 27 ...)
(define (ppows g (mult 1))
(for/fold (ppows '(1)) ((a g))
(set! mult (* mult a))
(cons mult ppows)))

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

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

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

(define (most-proper N)
(define maxdivs 1)
(define ndivs 0)
(for/fold (most-proper null) ((n (in-range 2 N)))
(set! ndivs (numdivs n))
#:continue (< ndivs maxdivs)
(when (> ndivs maxdivs)
(set!-values (most-proper maxdivs) (values null ndivs)))
(cons (cons n maxdivs) most-proper)))
```
Output:
```(for ((i (in-range 1 11))) (writeln i (divs i)))
1     null
2     (1)
3     (1)
4     (2 1)
5     (1)
6     (2 3 1)
7     (1)
8     (4 2 1)
9     (3 1)
10     (2 5 1)

(most-proper 20000)
→ ((18480 . 79) (15120 . 79))
(most-proper 1_000_000)
→ ((997920 . 239) (982800 . 239) (942480 . 239) (831600 . 239) (720720 . 239))

(lib 'bigint)
(numdivs 95952222101012742144)  → 666 ;; 🎩
```

## Eiffel

```class
APPLICATION

create
make

feature

make
-- Test the feature proper_divisors.
local
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 of 'n'.
do
create Result.make
across
1 |..| (n - 1) as c
loop
if n \\ c.item = 0 then
Result.extend (c.item)
end
end
end

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

## Elixir

Translation of: Erlang
```defmodule Proper do
def divisors(1), do: []
def divisors(n), do: [1 | divisors(2,n,:math.sqrt(n))] |> Enum.sort

defp divisors(k,_n,q) when k>q, do: []
defp divisors(k,n,q) when rem(n,k)>0, do: divisors(k+1,n,q)
defp divisors(k,n,q) when k * k == n, do: [k | divisors(k+1,n,q)]
defp divisors(k,n,q)                , do: [k,div(n,k) | divisors(k+1,n,q)]

def most_divisors(limit) do
{length,nums} = Enum.group_by(1..limit, fn n -> length(divisors(n)) end)
|> Enum.max_by(fn {length,_nums} -> length end)
IO.puts "With #{length}, Number #{inspect nums} has the most divisors"
end
end

Enum.each(1..10, fn n ->
IO.puts "#{n}: #{inspect Proper.divisors(n)}"
end)
Proper.most_divisors(20000)
```
Output:
```1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
With 79, Number [18480, 15120] has the most divisors
```

## Erlang

```-module(properdivs).
-export([divs/1,sumdivs/1,longest/1]).

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

divisors(K,_N,Q) when K > Q -> [];
divisors(K,N,Q) when N rem K =/= 0 ->
divisors(K+1,N,Q);
divisors(K,N,Q) when K * K  == N ->
[K] ++ divisors(K+1,N,Q);
divisors(K,N,Q) ->
[K, N div K] ++ divisors(K+1,N,Q).

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

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

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

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

## F#

```// the simple function with the answer
let propDivs n = [1..n/2] |> List.filter (fun x->n % x = 0)

// to cache the result length; helpful for a long search
let propDivDat n = propDivs n |> fun xs -> n, xs.Length, xs

// UI: always the longest and messiest
let show (n,count,divs) =
let showCount = count |> function | 0-> "no proper divisors" | 1->"1 proper divisor" | _-> sprintf "%d proper divisors" count
let showDiv = divs |> function | []->"" | x::[]->sprintf ": %d" x | _->divs |> Seq.map string |> String.concat "," |> sprintf ": %s"
printfn "%d has %s%s" n showCount showDiv

// generate output
[1..10] |> List.iter (propDivDat >> show)

// use a sequence: we don't really need to hold this data, just iterate over it
Seq.init 20000 ( ((+) 1) >> propDivDat)
|> Seq.fold (fun a b ->match a,b with | (_,c1,_),(_,c2,_) when c2 > c1 -> b | _-> a) (0,0,[])
|> fun (n,count,_) -> (n,count,[]) |> show
```
Output:
```1 has no proper divisors
2 has 1 proper divisor: 1
3 has 1 proper divisor: 1
4 has 2 proper divisors: 1,2
5 has 1 proper divisor: 1
6 has 3 proper divisors: 1,2,3
7 has 1 proper divisor: 1
8 has 3 proper divisors: 1,2,4
9 has 2 proper divisors: 1,3
10 has 3 proper divisors: 1,2,5
15120 has 79 proper divisors
```

## Factor

```USING: formatting io kernel math math.functions
math.primes.factors math.ranges prettyprint sequences ;

: #divisors ( m -- n )
dup sqrt >integer 1 + [1,b] [ divisor? ] with count dup +
1 - ;

10 [1,b] [ dup pprint bl divisors but-last . ] each
20000 [1,b] [ #divisors ] supremum-by dup #divisors
"%d with %d divisors.\n" printf
```
Output:
```1 { }
2 { 1 }
3 { 1 }
4 { 1 2 }
5 { 1 }
6 { 1 2 3 }
7 { 1 }
8 { 1 2 4 }
9 { 1 3 }
10 { 1 2 5 }
15120 with 79 divisors.
```

## Fermat

```Func Divisors(n) =
[d]:=[(1)];                                            {start divisor list with just 1, which is a divisor of everything}
for i = 2 to n\2 do                                    {loop through possible divisors of n}
if Divides(i, n) then [d]:=[d]_[(i)] fi
od;
.;

for n = 1 to 10 do
Divisors(n);
!!(n,'    ',[d);
od;

record:=0;
champ:=1;
for n=2 to 20000 do
Divisors(n);
m:=Cols[d];                                            {this gets the length of the array}
if m > record then
champ:=n;
record:=m;
fi;
od;

!!('The number up to 20,000 with the most divisors was ',champ,' with ',record,' divisors.');```
Output:
```
1    1
2    1
3    1
4    1,  2
5    1
6    1,  2,  3
7    1
8    1,  2,  4
9    1,  3
10    1,  2,  5

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

## Forth

Works with: gforth version 0.7.3
```: .proper-divisors
dup 1 ?do
dup i mod 0= if i . then
loop cr drop
;

: proper-divisors-count
0 swap
dup 1 ?do
dup i mod 0= if swap 1 + swap then
loop drop
;

: rosetta-proper-divisors
cr
11 1 do
i . ." : " i .proper-divisors
loop

1 0
20000 2 do
i proper-divisors-count
2dup < if nip nip i swap else drop then
loop
swap cr . ." has " . ." divisors" cr
;

rosetta-proper-divisors
```
Output:
```1 :
2 : 1
3 : 1
4 : 1 2
5 : 1
6 : 1 2 3
7 : 1
8 : 1 2 4
9 : 1 3
10 : 1 2 5

15120 has 79 divisors
ok```

## Fortran

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

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

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

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

end do

print *,' '
print *,' from 1 to ',limit
print *,maxj,' has max proper divisors: ',maxcnt
end
```
Output:
```
N   divisors
1
2
1
3
1
4
1
2
5
1
6
1
2
3
7
1
8
1
2
4
9
1
3
10
1
2
5

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

## FreeBASIC

```' FreeBASIC v1.05.0 win64

Sub ListProperDivisors(limit As Integer)
If limit < 1 Then Return
For i As Integer = 1 To limit
Print Using "##"; i;
Print " ->";
If i = 1 Then
Print " (None)"
Continue For
End if
For j As Integer = 1 To i \ 2
If i Mod j = 0 Then Print " "; j;
Next j
Print
Next i
End Sub

Function CountProperDivisors(number As Integer) As Integer
If number < 2 Then Return 0
Dim count As Integer = 0
For i As Integer = 1 To number \ 2
If number Mod i = 0 Then count += 1
Next
Return count
End Function

Dim As Integer n, count, most = 1, maxCount = 0

Print "The proper divisors of the following numbers are :"
Print
ListProperDivisors(10)

For n As Integer = 2 To 20000
count = CountProperDivisors(n)
If count > maxCount Then
maxCount = count
most = n
EndIf
Next

Print
Print Str(most); " has the most proper divisors, namely"; maxCount
Print
Print "Press any key to exit the program"
Sleep
End```
Output:
```The proper divisors of the following numbers are :

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

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

## Free Pascal

```Program ProperDivisors;

Uses fgl;

Type
TIntegerList = Specialize TfpgList<longint>;

Var list : TintegerList;

Function GetProperDivisors(x : longint): longint;
{this function will return the number of proper divisors
and put them in the list}

Var i : longint;
Begin
list.clear;
If x = 1 Then {by default 1 has no proper divisors}
GetProperDivisors := 0
Else
Begin
i := 2;
While i * i < x Do
Begin
If (x Mod i) = 0 Then //found a proper divisor
Begin
End;
inc(i);
End;
If i*i=x Then list.add(i); //make sure to capture the sqrt only once
GetProperDivisors := list.count;
End;
End;

Var i,j,count,most : longint;
Begin

list := TIntegerList.Create;
For i := 1 To 10 Do
Begin
write(i:4,' has ', GetProperDivisors(i),' proper divisors:');
For j := 0 To pred(list.count) Do
write(list[j]:3);
writeln();
End;
count := 0; //store highest number of proper divisors
most := 0;  //store number with highest number of proper divisors
For i := 1 To 20000 Do
If GetProperDivisors(i) > count Then
Begin
count := list.count;
most := i;
End;
writeln(most,' has ',count,' proper divisors');
list.free;
End.
```
Output:
```   1 has 0 proper divisors:
2 has 1 proper divisors:  1
3 has 1 proper divisors:  1
4 has 2 proper divisors:  1  2
5 has 1 proper divisors:  1
6 has 3 proper divisors:  1  2  3
7 has 1 proper divisors:  1
8 has 3 proper divisors:  1  2  4
9 has 2 proper divisors:  1  3
10 has 3 proper divisors:  1  2  5
15120 has 79 proper divisors
```

## Frink

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

```for n = 1 to 10
println["\$n\t" + join[" ", properDivisors[n]]]

println[]

d = new dict
for n = 1 to 20000
{
c = length[properDivisors[n]]
}

most = max[keys[d]]
println[d@most + " have \$most factors"]

properDivisors[n] := allFactors[n, true, false, true]```
Output:
```1
2       1
3       1
4       1 2
5       1
6       1 2 3
7       1
8       1 2 4
9       1 3
10      1 2 5

[15120, 18480] have 79 factors
```

## FutureBasic

```local fn ProperDivisors( n as long ) as CFArrayRef
CFMutableArrayRef array = fn MutableArrayWithCapacity(0)
if ( n < 2 ) then exit fn
long i
for i = 1 to n - 1
if ( n mod i == 0 )
end if
next
end fn = array

void local fn DoIt
long n, count, num, max = 0

for n = 1 to 10
printf @"%2ld: %@",n,fn ArrayComponentsJoinedByString( fn ProperDivisors( n ), @" " )
next

for n = 1 to 20000
count = len( fn Properdivisors( n ) )
if ( count > max )
max = count
num = n
end if
next

print: print num;@" has the most proper divisors with ";max
end fn

fn DoIt

HandleEvents```
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 the most proper divisors with 79
```

## GFA Basic

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

2> properdivs:longest(20000).
With 79, Number 15120 has the most divisors
re root of n%
f%(count%)=n%/i%
count%=count%+1
ENDIF
ENDIF
NEXT i%
ENDIF
'
RETURN count%-1
ENDFUNC
```

Output is:

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

## Go

Translation of: Kotlin
```package main

import (
"fmt"
"strconv"
)

func listProperDivisors(limit int) {
if limit < 1 {
return
}
width := len(strconv.Itoa(limit))
for i := 1; i <= limit; i++ {
fmt.Printf("%*d -> ", width, i)
if i == 1 {
fmt.Println("(None)")
continue
}
for j := 1; j <= i/2; j++ {
if i%j == 0 {
fmt.Printf(" %d", j)
}
}
fmt.Println()
}
}

func countProperDivisors(n int) int {
if n < 2 {
return 0
}
count := 0
for i := 1; i <= n/2; i++ {
if n%i == 0 {
count++
}
}
return count
}

func main() {
fmt.Println("The proper divisors of the following numbers are :\n")
listProperDivisors(10)
fmt.Println()
maxCount := 0
most := []int{1}
for n := 2; n <= 20000; n++ {
count := countProperDivisors(n)
if count == maxCount {
most = append(most, n)
} else if count > maxCount {
maxCount = count
most = most[0:1]
most = n
}
}
fmt.Print("The following number(s) <= 20000 have the most proper divisors, ")
fmt.Println("namely", maxCount, "\b\n")
for _, n := range most {
fmt.Println(n)
}
}
```
Output:
```The proper divisors of the following numbers are :

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

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

15120
18480
```

```import Data.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]]
```
Output:
```divisors of 1 to 10:
[]


[1,2]

[1,2,3]

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

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

```import Data.List (maximumBy)
import Data.Ord (comparing)
import Data.Bool (bool)

properDivisors
:: Integral a
=> a -> [a]
properDivisors n =
let root = (floor . sqrt . fromIntegral) n
lows = filter ((0 ==) . rem n) [1 .. root]
in init (lows ++ bool id tail (n == root * root) (reverse (quot n <\$> lows)))

main :: IO ()
main = do
putStrLn "Proper divisors of 1 to 10:"
mapM_ (print . properDivisors) [1 .. 10]
mapM_
putStrLn
[ ""
, "A number in the range 1 to 20,000 with the most proper divisors,"
, "as (number, count of proper divisors):"
, ""
]
print \$
maximumBy (comparing snd) \$
(,) <*> (length . properDivisors) <\$> [1 .. 20000]
```
Output:
```Proper divisors of 1 to 10:
[]


[1,2]

[1,2,3]

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

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

(18480,79)```

and we can also define properDivisors in terms of primeFactors:

```import Data.Numbers.Primes (primeFactors)
import Data.List (group, maximumBy, sort)
import Data.Ord (comparing)

properDivisors :: Int -> [Int]
properDivisors =
init . sort . foldr (
flip ((<*>) . fmap (*)) . scanl (*) 1
)  . group . primeFactors

---------------------------TEST----------------------------
main :: IO ()
main = do
putStrLn \$
fTable "Proper divisors of [1..10]:" show show properDivisors [1 .. 10]
mapM_
putStrLn
[ ""
, "A number in the range 1 to 20,000 with the most proper divisors,"
, "as (number, count of proper divisors):"
, ""
]
print \$
maximumBy (comparing snd) \$
(,) <*> (length . properDivisors) <\$> [1 .. 20000]

--------------------------DISPLAY--------------------------
fTable :: String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String
fTable s xShow fxShow f xs =
let rjust n c = (drop . length) <*> (replicate n c ++)
w = maximum (length . xShow <\$> xs)
in unlines \$
s : fmap (((++) . rjust w ' ' . xShow) <*> ((" -> " ++) . fxShow . f)) xs
```
Output:
```Proper divisors of [1..10]:
1 -> []
2 -> 
3 -> 
4 -> [1,2]
5 -> 
6 -> [1,2,3]
7 -> 
8 -> [1,2,4]
9 -> [1,3]
10 -> [1,2,5]

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

(18480,79)```

## J

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

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

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

Proper divisors of numbers 1 through 10:

```   (,&": ' -- ' ,&": 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
```

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

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

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

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

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

## Java

Works with: Java version 1.5+
```import java.util.Collections;
import java.util.List;

public class Proper{
public static List<Integer> properDivs(int n){
if(n == 1) return divs;
for(int x = 2; x < n; x++){
if(n % x == 0) divs.add(x);
}

Collections.sort(divs);

return divs;
}

public static void main(String[] args){
for(int x = 1; x <= 10; x++){
System.out.println(x + ": " + properDivs(x));
}

int x = 0, count = 0;
for(int n = 1; n <= 20000; n++){
if(properDivs(n).size() > count){
x = n;
count = properDivs(n).size();
}
}
System.out.println(x + ": " + count);
}
}```
Output:
```1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79```

## JavaScript

### ES5

```(function () {

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

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

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

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

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

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

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

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

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

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

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

})();
```
Output:
Number Proper Divisors Count
1 0
2 1 1
3 1 1
4 1, 2 2
5 1 1
6 1, 2, 3 3
7 1 1
8 1, 2, 4 3
9 1, 3 2
10 1, 2, 5 3

Most proper divisors below 20,000:

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

### ES6

```(() => {
'use strict';

// properDivisors :: Int -> [Int]
const properDivisors = n => {
// The integer divisors of n, excluding n itself.
const
rRoot = Math.sqrt(n),
intRoot = Math.floor(rRoot),
blnPerfectSquare = rRoot === intRoot,
lows = enumFromTo(1)(intRoot)
.filter(x => 0 === (n % x));

// For perfect squares, we can drop
// the head of the 'highs' list
return lows.concat(lows
.map(x => n / x)
.reverse()
.slice(blnPerfectSquare | 0)
)
.slice(0, -1); // except n itself
};

// ------------------------TESTS-----------------------
// main :: IO ()
const main = () =>
console.log([
fTable('Proper divisors of [1..10]:')(str)(
JSON.stringify
)(properDivisors)(enumFromTo(1)(10)),
'',
'Example of maximum divisor count in the range [1..20000]:',
'    ' + maximumBy(comparing(snd))(
enumFromTo(1)(20000).map(
n => [n, properDivisors(n).length]
)
).join(' has ') + ' proper divisors.'
].join('\n'));

// -----------------GENERIC FUNCTIONS------------------

// comparing :: (a -> b) -> (a -> a -> Ordering)
const comparing = f =>
x => y => {
const
a = f(x),
b = f(y);
return a < b ? -1 : (a > b ? 1 : 0);
};

// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = m => n =>
Array.from({
length: 1 + n - m
}, (_, i) => m + i);

// fTable :: String -> (a -> String) -> (b -> String)
//                      -> (a -> b) -> [a] -> String
const fTable = s => xShow => fxShow => f => xs => {
// Heading -> x display function ->
//           fx display function ->
//    f -> values -> tabular string
const
ys = xs.map(xShow),
w = Math.max(...ys.map(x => x.length));
return s + '\n' + zipWith(
a => b => a.padStart(w, ' ') + ' -> ' + b
)(ys)(
xs.map(x => fxShow(f(x)))
).join('\n');
};

// maximumBy :: (a -> a -> Ordering) -> [a] -> a
const maximumBy = f => xs =>
0 < xs.length ? (
xs.slice(1)
.reduce((a, x) => 0 < f(x)(a) ? x : a, xs)
) : undefined;

// snd :: (a, b) -> b
const snd = tpl => tpl;

// str :: a -> String
const str = x => x.toString();

// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = p => f => x => {
let v = x;
while (!p(v)) v = f(v);
return v;
};

// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = f => xs => ys => {
const
lng = Math.min(xs.length, xs.length),
as = xs.slice(0, lng),
bs = ys.slice(0, lng);
return Array.from({
length: lng
}, (_, i) => f(as[i])(
bs[i]
));
};

// MAIN ---
return main();
})();
```
Output:
```Proper divisors of [1..10]:
1 -> []
2 -> 
3 -> 
4 -> [1,2]
5 -> 
6 -> [1,2,3]
7 -> 
8 -> [1,2,4]
9 -> [1,3]
10 -> [1,2,5]

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

## jq

Works with: jq version 1.4

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

```def count(stream): reduce stream as \$i (0; . + 1);

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

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

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

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

## Julia

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

```using Primes
function properdivisors(n::T) where {T<:Integer}
0 < n || throw(ArgumentError("number to be factored must be ≥ 0, got \$n"))
1 < n || return T[]
!isprime(n) || return T[one(T)]
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, ".")
```
Output:
```List the proper divisors for 1 through 10.
1 []
2 [1,2]
3 [1,3]
4 [1,2]
5 [1,5]
6 [1,2,3]
7 [1,7]
8 [1,2,4]
9 [1,3]
10 [1,2,5]

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

## Kotlin

```// version 1.0.5-2

fun listProperDivisors(limit: Int) {
if (limit < 1) return
for(i in 1..limit) {
if (i == 1) {
println("(None)")
continue
}
(1..i/2).filter{ i % it == 0 }.forEach { print(" \$it") }
println()
}
}

fun countProperDivisors(n: Int): Int {
if (n < 2) return 0
return (1..n/2).count { (n % it) == 0 }
}

fun main(args: Array<String>) {
println("The proper divisors of the following numbers are :\n")
listProperDivisors(10)
println()
var count: Int
var maxCount = 0
val most: MutableList<Int> = mutableListOf(1)
for (n in 2..20000) {
count = countProperDivisors(n)
if (count == maxCount)
else if (count > maxCount) {
maxCount = count
most.clear()
}
}
println("The following number(s) have the most proper divisors, namely " + maxCount + "\n")
for (n in most) println(n)
}
```
Output:
```The proper divisors of the following numbers are :

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

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

15120
18480
```

## langur

Translation of: Go
Works with: langur version 0.10
```val .getproper = f(.x) for[=[]] .i of .x \ 2 { if .x div .i: _for ~= [.i] }
val .cntproper = f(.x) for[=0] .i of .x \ 2 { if .x div .i: _for += 1 }

val .listproper = f(.x) {
if .x < 1: return null
for .i of .x {
writeln \$"\.i:2; -> ", .getproper(.i)
}
writeln()
}

writeln "The proper divisors of the following numbers are :"
.listproper(10)

var .max = 0
var .most = []
for .n in 2 .. 20_000 {
val .cnt = .cntproper(.n)
if .cnt == .max {
.most = more .most, .n
} else if .cnt > .max {
.max, .most = .cnt, [.n]
}
}

writeln \$"The following number(s) <= 20000 have the most proper divisors (\.max;)"
writeln .most```
Output:
```The proper divisors of the following numbers are :
1 -> []
2 -> 
3 -> 
4 -> [1, 2]
5 -> 
6 -> [1, 2, 3]
7 -> 
8 -> [1, 2, 4]
9 -> [1, 3]
10 -> [1, 2, 5]

The following number(s) <= 20000 have the most proper divisors (79)
[15120, 18480]
```

## Lua

```-- Return a table of the proper divisors of n
function propDivs (n)
if n < 2 then return {} end
local divs, sqr = {1}, math.sqrt(n)
for d = 2, sqr do
if n % d == 0 then
table.insert(divs, d)
if d ~= sqr then table.insert(divs, n/d) end
end
end
table.sort(divs)
return divs
end

-- Show n followed by all values in t
function show (n, t)
io.write(n .. ":\t")
for _, v in pairs(t) do io.write(v .. " ") end
print()
end

-- Main procedure
local mostDivs, numDivs, answer = 0
for i = 1, 10 do show(i, propDivs(i)) end
for i = 1, 20000 do
numDivs = #propDivs(i)
if numDivs > mostDivs then
mostDivs = numDivs
end
end
print(answer .. " has " .. mostDivs .. " proper divisors.")
```
Output:
```1:
2:      1
3:      1
4:      1 2
5:      1
6:      1 2 3
7:      1
8:      1 2 4
9:      1 3
10:     1 2 5
15120 has 79 proper divisors.```

## Mathematica / Wolfram Language

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

```ProperDivisors[n_Integer /; n > 0] := Most@Divisors@n;
```

Proper divisors of n from 1 to 10:

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

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

```Fold[
Last[SortBy[{#1, {#2, Length@ProperDivisors[#2]}}, Last]] &,
{0, 0},
Range]
```
Output:
`{18480, 79}`

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

```Last@SortBy[
Table[
{n, Length@ProperDivisors[n]},
{n, 1, 20000}],
Last]
```
Output:
`{15120, 79}`

## MATLAB

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

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

maxI =

15120

maxL

maxL =

79
```

## Modula-2

```MODULE ProperDivisors;
FROM FormatString IMPORT FormatString;

PROCEDURE WriteInt(n : INTEGER);
VAR buf : ARRAY[0..15] OF CHAR;
BEGIN
FormatString("%i", buf, n);
WriteString(buf)
END WriteInt;

PROCEDURE proper_divisors(n : INTEGER; print_flag : BOOLEAN) : INTEGER;
VAR count,i : INTEGER;
BEGIN
count := 0;
FOR i:=1 TO n-1 DO
IF n MOD i = 0 THEN
INC(count);
IF print_flag THEN
WriteInt(i);
WriteString(" ")
END
END
END;
IF print_flag THEN WriteLn END;
RETURN count;
END proper_divisors;

VAR
buf : ARRAY[0..63] OF CHAR;
i,max,max_i,v : INTEGER;
BEGIN
FOR i:=1 TO 10 DO
WriteInt(i);
WriteString(": ");
proper_divisors(i, TRUE)
END;

max := 0;
max_i := 1;

FOR i:=1 TO 20000 DO
v := proper_divisors(i, FALSE);
IF v>= max THEN
max := v;
max_i := i
END
END;

FormatString("%i with %i divisors\n", buf, max_i, max);
WriteString(buf);

END ProperDivisors.
```

## Nim

Translation of: C
```import strformat

proc properDivisors(n: int) =
var count = 0
for i in 1..<n:
if n mod i == 0:
inc count
write(stdout, fmt"{i} ")
write(stdout, "\n")

proc countProperDivisors(n: int): int =
var nn = n
var prod = 1
var count = 0
while nn mod 2 == 0:
inc count
nn = nn div 2
prod *= (1 + count)
for i in countup(3, n, 2):
count = 0
while nn mod i == 0:
inc count
nn = nn div i
prod *= (1 + count)
if nn > 2:
prod *= 2
prod - 1

for i in 1..10:
write(stdout, fmt"{i:2}: ")
properDivisors(i)

var max = 0
var maxI = 1

for i in 1..20000:
var v = countProperDivisors(i)
if v >= max:
max = v
maxI = i

echo fmt"{maxI} with {max} divisors"
```
Output:
``` 1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
18480 with 79 divisors
```

## Oberon-2

```MODULE ProperDivisors;
IMPORT
Out;

CONST
initialSize = 128;
TYPE
Result* = POINTER TO ResultDesc;
ResultDesc = RECORD
found-: LONGINT; (* number of slots in pd *)
pd-: POINTER TO ARRAY OF LONGINT;
cap: LONGINT;   (* Capacity *)
END;

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

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

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);

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

## Objeck

```use Collection;

class Proper{
function : Main(args : String[]) ~ Nil {
for(x := 1; x <= 10; x++;) {
Print(x, ProperDivs(x));
};

x := 0;
count := 0;

for(n := 1; n <= 20000; n++;) {
if(ProperDivs(n)->Size() > count) {
x := n;
count := ProperDivs(n)->Size();
};
};
"{\$x}: {\$count}"->PrintLine();
}

function : ProperDivs(n : Int) ~ IntVector {
divs := IntVector->New();

if(n = 1) {
return divs;
};

for(x := 2; x < n; x++;) {
if(n % x = 0) {
};
};
divs->Sort();

return divs;
}

function : Print(x : Int, result : IntVector) ~ Nil {
"{\$x}: "->Print();
result->ToArray()->ToString()->PrintLine();
}
}```

Output:

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

## Oforth

```Integer method: properDivs  self 2 / seq filter(#[ self swap mod 0 == ]) }

10 seq apply(#[ dup print " : " print properDivs println ])
20000 seq map(#[ dup properDivs size Pair new ]) reduce(#maxKey) println```
Output:
```1 : []
2 : 
3 : 
4 : [1, 2]
5 : 
6 : [1, 2, 3]
7 : 
8 : [1, 2, 4]
9 : [1, 3]
10 : [1, 2, 5]
[79, 15120]
```

## PARI/GP

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

## Pascal

Works with: Free Pascal

Using prime factorisation

```{\$IFDEF FPC}{\$MODE DELPHI}{\$ELSE}{\$APPTYPE CONSOLE}{\$ENDIF}
uses
sysutils;
const
MAXPROPERDIVS = 1920;
type
tRes = array[0..MAXPROPERDIVS] of LongWord;
tPot = record
potPrim,
potMax :LongWord;
end;

tprimeFac = record
pfPrims : array[1..10] of tPot;
pfCnt,
pfNum   : LongWord;
end;
tSmallPrimes = array[0..6541] of longWord;

var
SmallPrimes: tSmallPrimes;

procedure InitSmallPrimes;
var
pr,testPr,j,maxprimidx: Longword;
isPrime : boolean;
Begin
maxprimidx := 0;
SmallPrimes := 2;
pr := 3;
repeat
isprime := true;
j := 0;
repeat
testPr := SmallPrimes[j];
IF testPr*testPr > pr then
break;
If pr mod testPr = 0 then
Begin
isprime := false;
break;
end;
inc(j);
until false;

if isprime then
Begin
inc(maxprimidx);
SmallPrimes[maxprimidx]:= pr;
end;
inc(pr,2);
until pr > 1 shl 16 -1;
end;

procedure PrimeFacOut(primeDecomp:tprimeFac);
var
i : LongWord;
begin
with primeDecomp do
Begin
write(pfNum,' = ');
For i := 1 to pfCnt-1 do
with pfPrims[i] do
If potMax = 1 then
write(potPrim,'*')
else
write(potPrim,'^',potMax,'*');
with pfPrims[pfCnt] do
If potMax = 1 then
write(potPrim)
else
write(potPrim,'^',potMax);
end;
end;

procedure PrimeDecomposition(n:LongWord;var res:tprimeFac);
var
i,pr,cnt,quot{to minimize divisions} : LongWord;
Begin
res.pfNum := n;
res.pfCnt:= 0;
i := 0;
cnt := 0;
repeat
pr := SmallPrimes[i];
IF pr*pr>n then
Break;

quot := n div pr;
IF pr*quot = n then
with res do
Begin
inc(pfCnt);
with pfPrims[pfCnt] do
Begin
potPrim := pr;
potMax := 0;
repeat
n := quot;
quot := quot div pr;
inc(potMax);
until pr*quot <> n;
end;
end;
inc(i);
until false;
//a big prime left over?
IF n <> 1 then
with res do
Begin
inc(pfCnt);
with pfPrims[pfCnt] do
Begin
potPrim := n;
potMax := 1;
end;
end;
end;

function CntProperDivs(const primeDecomp:tprimeFac):LongWord;
//count of proper divisors
var
i: LongWord;
begin
result := 1;
with primeDecomp do
For i := 1 to pfCnt do
result := result*(pfPrims[i].potMax+1);
//remove
dec(result);
end;

function findProperdivs(n:LongWord;var res:TRes):LongWord;
//simple trial division to get a sorted list of all proper divisors
var
i,j: LongWord;
Begin
result := 0;
i := 1;
j := n;
while j>i do
begin
j := n DIV i;
IF i*j = n then
Begin
//smaller factor part at the beginning upwards
res[result]:= i;
IF i <> j then
//bigger factor at the end downwards
res[MAXPROPERDIVS-result]:= j
else
//n is square number
res[MAXPROPERDIVS-result]:= 0;
inc(result);
end;
inc(i);
end;

If result>0 then
Begin
//move close together
i := result;
j := MAXPROPERDIVS-result+1;
result := 2*result-1;
repeat
res[i] := res[j];
inc(j);
inc(i);
until i > result;

if res[result-1] = 0 then
dec(result);
end;
end;

procedure AllFacsOut(n: Longword);
var
res:TRes;
i,k,j:LongInt;
Begin
j := findProperdivs(n,res);
write(n:5,' : ');
For k := 0 to j-2 do write(res[k],',');
IF j>=1 then
write(res[j-1]);
writeln;
end;

var
primeDecomp: tprimeFac;
rs : tRes;
i,j,max,maxcnt: LongWord;
BEGIN
InitSmallPrimes;
For i := 1 to 10 do
AllFacsOut(i);
writeln;
max    := 0;
maxCnt := 0;
For i := 1 to 20*1000 do
Begin
PrimeDecomposition(i,primeDecomp);
j := CntProperDivs(primeDecomp);
IF j> maxCnt then
Begin
maxcnt := j;
max := i;
end;
end;
PrimeDecomposition(max,primeDecomp);
j := CntProperDivs(primeDecomp);

PrimeFacOut(primeDecomp);writeln('  ',j:10,' factors'); writeln;
//https://en.wikipedia.org/wiki/Highly_composite_number <= HCN
//http://wwwhomes.uni-bielefeld.de/achim/highly.txt the first 1200 HCN
max := 3491888400;
PrimeDecomposition(max,primeDecomp);
j := CntProperDivs(primeDecomp);
PrimeFacOut(primeDecomp);writeln('  ',j:10,' factors'); writeln;
END.
```
Output:
```
1 :
2 : 1
3 : 1
4 : 1,2
5 : 1
6 : 1,2,3
7 : 1
8 : 1,2,4
9 : 1,3
10 : 1,2,5

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

real    0m0.004s```

## Perl

### Using a module for divisors

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

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

## Phix

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

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

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

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

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

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

## PHP

```<?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";
```

Outputs:

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

## Picat

```go =>
println(11111=proper_divisors(11111)),
nl,
foreach(N in 1..10)
println(N=proper_divisors(N))
end,
nl,

find_most_divisors(20_000),
nl.

% Proper divisors of number N
proper_divisors(N) = Divisors =>
Div1 = [ I : I in 1..ceiling(sqrt(N)), N mod I == 0],
Divisors = (Div1 ++ [N div I : I in Div1]).sort_remove_dups().delete(N).

% Find the number(s) with the most proper divisors below Limit
find_most_divisors(Limit) =>
MaxN = [],
MaxNumDivisors = [],
MaxLen = 1,

foreach(N in 1..Limit, not prime(N))
D = proper_divisors(N),
Len = D.len,
% Get all numbers with most proper divisors
if Len = MaxLen then
MaxN := MaxN ++ [N],
MaxNumDivisors := MaxNumDivisors ++ [[N=D]]
elseif Len > MaxLen then
MaxLen := Len,
MaxN := [N],
MaxNumDivisors := [N=D]
end
end,

println(maxN=MaxN),
println(maxLen=MaxLen),
nl.```
Output:
```11111 = [1,41,271]

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

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

### Larger tests

Some larger tests of most number of divisors:

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

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

## PicoLisp

```# Generate all proper divisors.
(de propdiv (N)
'((X) (=0 (% N X)))
(range 1 N) )) )

# Obtaining the values from 1 to 10 inclusive.
(mapcar propdiv (range 1 10))
# Output:
# (NIL (1) (1) (1 2) (1) (1 2 3) (1) (1 2 4) (1 3) (1 2 5))```

### Brute-force

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

### Factorization

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

Output:

```15120 79
0.081 sec
```

## PL/I

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

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

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

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

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

## PowerShell

### version 1

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

### version 2

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

### version 3

```function eratosthenes (\$n) {
if(\$n -gt 1){
\$prime = @(0..\$n| foreach{\$true})
\$m = [Math]::Floor([Math]::Sqrt(\$n))
function multiple(\$i) {
for(\$j = \$i*\$i; \$j -le \$n; \$j += \$i) {
\$prime[\$j] = \$false
}
}
multiple 2
for(\$i = 3; \$i -le \$m; \$i += 2) {
if(\$prime[\$i]) {multiple \$i}
}
2
for(\$i = 3; \$i -le \$n; \$i += 2) {
if(\$prime[\$i]) {\$i}
}

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

Output:

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

## Prolog

Works with: SWI-Prolog 7

Taking a cue from an SO answer:

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

proper_divisor(N, D) :-
divisor(N, D),
D =\= N.

%

proper_divisors(N, Ds) :-
setof(D, proper_divisor(N, D), Ds).

%

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

%

proper_divisor_count(N, Count) :-
proper_divisors(N, Ds),
length(Ds, Count).

find_most_proper_divisors_in_range(Low, High, Result) :-
aggregate_all( max(Count, N),
( between(Low, High, N),
proper_divisor_count(N, Count) ),
max(MaxCount, Num) ),
Result = (num(Num)-divisor_count(MaxCount)).
```

Output:

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

?- find_most_proper_divisors_in_range(1,20000,Result).
Result = num(15120)-divisor_count(79).
```

## PureBasic

```EnableExplicit

Procedure ListProperDivisors(Number, List Lst())
If Number < 2 : ProcedureReturn : EndIf
Protected i
For i = 1 To Number / 2
If Number % i = 0
Lst() = i
EndIf
Next
EndProcedure

Procedure.i CountProperDivisors(Number)
If Number < 2 : ProcedureReturn 0 : EndIf
Protected i, count = 0
For i = 1 To Number / 2
If Number % i = 0
count + 1
EndIf
Next
ProcedureReturn count
EndProcedure

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

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

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

## Python

### Python: Literal

A very literal interpretation

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

### Python: From prime factors

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

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

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

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

```from math import 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))
```
Output:
```[set(), {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}, {1, 3}, {1, 2, 5}]
15120 79```

### Python: Functional

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

Works with: Python version 3.7
```'''Proper divisors'''

from itertools import accumulate, chain, groupby, product
from functools import reduce
from math import floor, sqrt
from operator import mul

# properDivisors :: Int -> [Int]
def properDivisors(n):
'''The ordered divisors of n, excluding n itself.
'''
def go(a, group):
return [x * y for x, y in product(
a,
accumulate(chain(, group), mul)
)]
return sorted(
reduce(go, [
list(g) for _, g in groupby(primeFactors(n))
], )
)[:-1] if 1 < n else []

# --------------------------TEST---------------------------
# main :: IO ()
def main():
'''Tests'''

print(
fTable('Proper divisors of [1..10]:')(str)(str)(
properDivisors
)(range(1, 1 + 10))
)

print('\nExample of maximum divisor count in the range [1..20000]:')
print(
max(
[(n, len(properDivisors(n))) for n in range(1, 1 + 20000)],
key=snd
)
)

# -------------------------DISPLAY-------------------------

# fTable :: String -> (a -> String) ->
# (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
'''
def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
return s + '\n' + '\n'.join(map(
lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)

# -------------------------GENERIC-------------------------

# primeFactors :: Int -> [Int]
def primeFactors(n):
'''A list of the prime factors of n.
'''
def f(qr):
r = qr
return step(r), 1 + r

def step(x):
return 1 + (x << 2) - ((x >> 1) << 1)

def go(x):
root = floor(sqrt(x))

def p(qr):
q = qr
return root < q or 0 == (x % q)

q = until(p)(f)(
(2 if 0 == x % 2 else 3, 1)
)
return [x] if q > root else [q] + go(x // q)

return go(n)

# snd :: (a, b) -> b
def snd(tpl):
'''Second member of a pair.'''
return tpl

# until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
'''The result of repeatedly applying f until p holds.
The initial seed value is x.
'''
def go(f, x):
v = x
while not p(v):
v = f(v)
return v
return lambda f: lambda x: go(f, x)

# MAIN ---
if __name__ == '__main__':
main()
```
Output:
```Proper divisors of [1..10]:
1 -> []
2 -> 
3 -> 
4 -> [1, 2]
5 -> 
6 -> [1, 2, 3]
7 -> 
8 -> [1, 2, 4]
9 -> [1, 3]
10 -> [1, 2, 5]

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

### Python: The Simple Way

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

```def pd(num):
factors = []
for divisor in range(1,1+num//2):
if num % divisor == 0: factors.append(divisor)
return factors

def pdc(num):
count = 0
for divisor in range(1,1+num//2):
if num % divisor == 0: count += 1
return count

def fmtres(title, lmt, best, bestc):
return "The " + title + " number up to and including " + str(lmt) + " with the highest number of proper divisors is " + str(best) + ", which has " + str(bestc)

def showcount(limit):
best, bestc, bh, bhc = 0, 0, 0, 0
for i in range(limit+1):
divc = pdc(i)
if divc > bestc: bestc, best = divc, i
if divc >= bhc: bhc, bh = divc, i
if best == bh:
print(fmtres("only", limit, best, bestc))
else:
print(fmtres("lowest", limit, best, bestc))
print(fmtres("highest", limit, bh, bhc))
print()

lmt = 10
for i in range(1, lmt + 1):
divs = pd(i)
if len(divs) == 0:
print("There are no proper divisors of", i)
elif len(divs) == 1:
print(divs, "is the only proper divisor of", i)
else:
print(divs, "are the proper divisors of", i)
print()
showcount(20000)
showcount(25000)
```
Output:
```There are no proper divisors of 1
1 is the only proper divisor of 2
1 is the only proper divisor of 3
[1, 2] are the proper divisors of 4
1 is the only proper divisor of 5
[1, 2, 3] are the proper divisors of 6
1 is the only proper divisor of 7
[1, 2, 4] are the proper divisors of 8
[1, 3] are the proper divisors of 9
[1, 2, 5] are the proper divisors of 10

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

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

## Quackery

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

```  [ factors -1 split drop ]      is properdivisors ( n --> [ )

10 times [ i^ 1+ properdivisors echo cr ]

0 0
20000 times
[ i^ 1+ properdivisors size
2dup < iff
[ dip [ 2drop i^ 1+ ] ]
else drop ]
swap echo say " has "
echo say " proper divisors." cr```
Output:
```[ ]
[ 1 ]
[ 1 ]
[ 1 2 ]
[ 1 ]
[ 1 2 3 ]
[ 1 ]
[ 1 2 4 ]
[ 1 3 ]
[ 1 2 5 ]
15120 has 79 proper divisors.
```

## R

### Package solution

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

### Filter solution

```#Task 1
properDivisors <- function(N) Filter(function(x) N %% x == 0, seq_len(N %/% 2))

Vectorize(properDivisors)(1:10)

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

[]
 1

[]
 1

[]
 1 2

[]
 1

[]
 1 2 3

[]
 1

[]
 1 2 4

[]
 1 3

[]
 1 2 5

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

## Racket

### Short version

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

### Long version

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

```#lang racket/base
(provide fold-divisors ; name as per "Abundant..."
proper-divisors)

(define (fold-divisors v n k0 kons)
(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])
[c (in-value (count-proper-divisors i))]
#:when [> c C])
(values c i)))
(printf "~a has ~a proper divisors\n" I C))
```

The output is the same as the short version above.

## Raku

(formerly Perl 6)

Works with: rakudo version 2018.10

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

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

```sub propdiv (\x) {
my @l = 1 if x > 1;
(2 .. x.sqrt.floor).map: -> \d {
unless x % d { @l.push: d; my \y = x div d; @l.push: y if y != d }
}
@l
}

put "\$_ [{propdiv(\$_)}]" for 1..10;

my @candidates;
loop (my int \$c = 30; \$c <= 20_000; \$c += 30) {
#(30, *+30 …^ * > 500_000).race.map: -> \$c {
my \mx = +propdiv(\$c);
next if mx < @candidates - 1;
@candidates[mx].push: \$c
}

say "max = {@candidates - 1}, candidates = {@candidates.tail}";
```
Output:
```1 []
2 
3 
4 [1 2]
5 
6 [1 2 3]
7 
8 [1 2 4]
9 [1 3]
10 [1 2 5]
max = 79, candidates = 15120 18480```

## REXX

### version 1

```/*REXX*/

Call time 'R'
Do x=1 To 10
Say x '->' proper_divisors(x)
End

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

### version 2

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

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

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

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

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

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

### version 3

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

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

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

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

### version 4

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

For larger numbers,   it is about   7%   faster.

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

## Ring

```# Project : Proper divisors

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

Output:

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

## RPL

Works with: HP version 49
```≪ DIVIS REVLIST TAIL REVLIST       @ or DIVIS 1 OVER SIZE 1 - SUB
≫ 'PDIVIS' STO

≪ 0 → max n
≪ 0
1 max FOR j
j PDIVIS SIZE
IF DUP2 < THEN SWAP j 'n' STO END
DROP
NEXT
DROP n DUP PDIVIS SIZE
```
```≪ n PDIVIS ≫ 'n' 1 10 1 SEQ
```
Output:
```3: {{} {1} {1} {1 2} {1} {1 2 3} {1} {1 2 4} {1 3} {1 2 5}}
2: 15120
1: 39
```

## Ruby

```require "prime"

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

### An Alternative Approach

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

In the range 1..200000

```15120 has 79 proper divisors
```

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

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

## Rust

```trait ProperDivisors {
fn proper_divisors(&self) -> Option<Vec<u64>>;
}

impl ProperDivisors for u64 {
fn proper_divisors(&self) -> Option<Vec<u64>> {
if self.le(&1) {
return None;
}
let mut divisors: Vec<u64> = Vec::new();

for i in 1..*self {
if *self % i == 0 {
divisors.push(i);
}
}
Option::from(divisors)
}
}

fn main() {
for i in 1..11 {
println!("Proper divisors of {:2}: {:?}", i,
i.proper_divisors().unwrap_or(vec![]));
}

let mut most_idx: u64 = 0;
let mut most_divisors: Vec<u64> = Vec::new();
for i in 1..20_001 {
let divs = i.proper_divisors().unwrap_or(vec![]);
if divs.len() > most_divisors.len() {
most_divisors = divs;
most_idx = i;
}
}
println!("In 1 to 20000, {} has the most proper divisors at {}", most_idx,
most_divisors.len());
}
```
Output:
```Proper divisors of  1: []
Proper divisors of  2: 
Proper divisors of  3: 
Proper divisors of  4: [1, 2]
Proper divisors of  5: 
Proper divisors of  6: [1, 2, 3]
Proper divisors of  7: 
Proper divisors of  8: [1, 2, 4]
Proper divisors of  9: [1, 3]
Proper divisors of 10: [1, 2, 5]
In 1 to 20000, 15120 has the most proper divisors at 79
```

## S-BASIC

```\$lines

\$constant false = 0
\$constant true = FFFFH

rem - compute p mod q
function mod(p, q = integer) = integer
end = p - q * (p/q)

rem - count, and optionally display, proper divisors of n
function divisors(n, display = integer) = integer
var i, limit, count, start, delta = integer
if mod(n, 2) = 0 then
begin
start = 2
delta = 1
end
else
begin
start = 3
delta = 2
end
if n < 2 then count = 0 else count = 1
if display and (count = 1) then print using "#####"; 1;
i = start
limit = n / start
while i <= limit do
begin
if mod(n, i) = 0 then
begin
if display then print using "#####"; i;
count = count + 1
end
i = i + delta
if count = 1 then limit = n / i
end
if display then print
end = count

rem - main program begins here
var i, ndiv, highdiv, highnum = integer

print "Proper divisors of first 10 numbers:"
for i = 1 to 10
print using "### : "; i;
ndiv = divisors(i, true)
next i

print "Searching for number with most divisors ..."
highdiv = 1
highnum = 1
for i = 1 to 20000
ndiv = divisors(i, false)
if ndiv > highdiv then
begin
highdiv = ndiv
highnum = i
end
next i
print "Searched up to"; i
print highnum; " has the most divisors: "; highdiv

end
```
Output:
```Proper divisors of first 10 numbers:
1 :
2 :     1
3 :     1
4 :     1     2
5 :     1
6 :     1     2    3
7 :     1
8 :     1     2    4
9 :     1     3
10 :     1     2    5
Searching for number with most divisors ...
Searched up to 20000
15120 has the most divisors:  79
```

## Scala

### Simple proper divisors

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

println(f"    n   cnt   PROPER DIVISORS")
val (count, list) = (1 to 20000).foldLeft( (0, List[Int]()) ) { (max, i) =>
val divisors = properDivisors(i)
if (i <= 10 || i == 100) println( format(i, divisors) )
if (max._1 < divisors.length) (divisors.length, List(i))
else if (max._1 == divisors.length) (divisors.length, max._2 ::: List(i))
else max
}

list.foreach( number => println(f"\$number%5d    \${properDivisors(number).length}") )
```
Output:
```    n   cnt   PROPER DIVISORS
1     0
2     1   1
3     1   1
4     2   1 2
5     1   1
6     3   1 2 3
7     1   1
8     3   1 2 4
9     2   1 3
10     3   1 2 5
100     8   1 2 4 5 10 20 25 50
15120    79
18480    79```

### Proper divisors for integers for big integers

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

```import scala.annotation.tailrec

def factorize(x: BigInt): List[BigInt] = {
@tailrec
def foo(x: BigInt, a: BigInt = 2, list: List[BigInt] = Nil): List[BigInt] = a * a > x match {
case false if x % a == 0 => foo(x / a, a, a :: list)
case false => foo(x, a + 1, list)
case true => x :: list
}

foo(x)
}

def properDivisors(n: BigInt): List[BigInt] = {
val factors = factorize(n)
val products = (1 until factors.length).flatMap(i => factors.combinations(i).map(_.product).toList).toList
(BigInt(1) :: products).filter(_ < n)
}
```

## Seed7

```\$ include "seed7_05.s7i";

const proc: writeProperDivisors (in integer: n) is func
local
var integer: i is 0;
begin
for i range 1 to n div 2 do
if n rem i = 0 then
write(i <& " ");
end if;
end for;
writeln;
end func;

const func integer: countProperDivisors (in integer: n) is func
result
var integer: count is 0;
local
var integer: i is 0;
begin
for i range 1 to n div 2 step succ(n rem 2) do
if n rem i = 0 then
incr(count);
end if;
end for;
end func;

const proc: main is func
local
var integer: i is 0;
var integer: v is 0;
var integer: max is 0;
var integer: max_i is 1;
begin
for i range 1 to 10 do
write(i <& ": ");
writeProperDivisors(i);
end for;
for i range 1 to 20000 do
v := countProperDivisors(i);
if v > max then
max := v;
max_i := i;
end if;
end for;
writeln(max_i <& " with " <& max <& " divisors");
end func;```
Output:
```1:
2: 1
3: 1
4: 1 2
5: 1
6: 1 2 3
7: 1
8: 1 2 4
9: 1 3
10: 1 2 5
15120 with 79 divisors
```

## Sidef

Translation of: Raku
```func propdiv (n) {
n.divisors.first(-1)
}

{|i| printf("%2d: %s\n", i, propdiv(i)) } << 1..10

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

## Swift

Simple function:

```func properDivs1(n: Int) -> [Int] {

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

More efficient function:

```import func Darwin.sqrt

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

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

if n == 1 { return [] }

var result = [Int]()

for div in filter (1 ... sqrt(n), { n % \$0 == 0 }) {

result.append(div)

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

return sorted(result)

}
```

```for i in 1...10 {
println("\(i): \(properDivs(i))")
}

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

for i in 1...20_000 {

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

println("\(num): \(max)")
```
Output:
```1: []
2: 
3: 
4: [1, 2]
5: 
6: [1, 2, 3]
7: 
8: [1, 2, 4]
9: [1, 3]
10: [1, 2, 5]
15120: 79
```

## tbas

```dim _proper_divisors(100)

sub proper_divisors(n)
dim i
dim _proper_divisors_count = 0
if n <> 1 then
for i = 1 to (n \ 2)
if n %% i = 0 then
_proper_divisors_count = _proper_divisors_count + 1
_proper_divisors(_proper_divisors_count) = i
end if
next
end if
return _proper_divisors_count
end sub

sub show_proper_divisors(n, tabbed)
dim cnt = proper_divisors(n)
print str\$(n) + ":"; tab(4);"(" + str\$(cnt) + " items) ";
dim j
for j = 1 to cnt
if tabbed then
print str\$(_proper_divisors(j)),
else
print str\$(_proper_divisors(j));
end if
if (j < cnt) then print ",";
next
print
end sub

dim i
for i = 1 to 10
show_proper_divisors(i, false)
next

dim c
dim maxindex = 0
dim maxlength = 0
for t = 1 to 20000
c = proper_divisors(t)
if c > maxlength then
maxindex = t
maxlength = c
end if
next

print "A maximum at ";
show_proper_divisors(maxindex, false)```
```>tbas proper_divisors.bas
1:  (0 items)
2:  (1 items) 1
3:  (1 items) 1
4:  (2 items) 1,2
5:  (1 items) 1
6:  (3 items) 1,2,3
7:  (1 items) 1
8:  (3 items) 1,2,4
9:  (2 items) 1,3
10: (3 items) 1,2,5
A maximum at 15120:(79 items) 1,2,3,4,5,6,7,8,9,10,12,14,15,16,18,20,21,24,27,28,30,
35,36,40,42,45,48,54,56,60,63,70,72,80,84,90,105,108,112,120,126,135,
140,144,168,180,189,210,216,240,252,270,280,315,336,360,378,420,432,
504,540,560,630,720,756,840,945,1008,1080,1260,1512,1680,1890,2160,
2520,3024,3780,5040,7560
```

## Tcl

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

```proc properDivisors {n} {
if {\$n == 1} return
set divs 1
for {set i 2} {\$i*\$i <= \$n} {incr i} {
if {!(\$n % \$i)} {
lappend divs \$i
if {\$i*\$i < \$n} {
lappend divs [expr {\$n / \$i}]
}
}
}
return \$divs
}

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

## uBasic/4tH

Translation of: True BASIC
```LET m = 1
LET c = 0

PRINT "The proper divisors of the following numbers are:\n"
PROC _ListProperDivisors (10)

FOR n = 2 TO 20000
LET d = FUNC(_CountProperDivisors(n))
IF d > c THEN
LET c = d
LET m = n
ENDIF
NEXT

PRINT
PRINT m; " has the most proper divisors, namely "; c
END

_CountProperDivisors
PARAM (1)
LOCAL (2)

IF a@ < 2 THEN RETURN (0)
LET c@ = 0
FOR b@ = 1 TO a@ / 2
IF a@ % b@ = 0 THEN LET c@ = c@ + 1
NEXT
RETURN (c@)

_ListProperDivisors
PARAM (1)
LOCAL (2)

IF a@ < 1 THEN RETURN
FOR b@ = 1 TO a@
PRINT b@; " ->";
IF b@ = 1 THEN PRINT " (None)";
FOR c@ = 1 TO b@ / 2
IF b@ % c@ = 0 THEN PRINT " "; c@;
NEXT
PRINT
NEXT
RETURN
```
Output:
```The proper divisors of the following numbers are:

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

15120 has the most proper divisors, namely 79

0 OK, 0:415```

## VBA

```Public Sub Proper_Divisor()
Dim t() As Long, i As Long, l As Long, j As Long, c As Long
For i = 1 To 10
Debug.Print "Proper divisor of " & i & " : " & Join(S(i), ", ")
Next
For i = 2 To 20000
l = UBound(S(i)) + 1
If l > c Then c = l: j = i
Next
Debug.Print "Number in the range 1 to 20,000 with the most proper divisors is : " & j
Debug.Print j & " count " & c & " proper divisors"
End Sub

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

## Visual Basic .NET

Translation of: C#
```Module Module1

Function ProperDivisors(number As Integer) As IEnumerable(Of Integer)
Return Enumerable.Range(1, number / 2).Where(Function(divisor As Integer) number Mod divisor = 0)
End Function

Sub Main()
For Each number In Enumerable.Range(1, 10)
Console.WriteLine("{0}: {{{1}}}", number, String.Join(", ", ProperDivisors(number)))
Next

Dim record = Enumerable.Range(1, 20000).Select(Function(number) New With {.Number = number, .Count = ProperDivisors(number).Count()}).OrderByDescending(Function(currentRecord) currentRecord.Count).First()
Console.WriteLine("{0}: {1}", record.Number, record.Count)
End Sub

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

## Wren

Library: Wren-fmt
Library: Wren-math
```import "/fmt" for Fmt
import "/math" for Int

for (i in 1..10) System.print("%(Fmt.d(2, i)) -> %(Int.properDivisors(i))")

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

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

## XPL0

```func PropDiv(N, Show);  \Count and optionally show proper divisors of N
int  N, Show, D, C;
[C:= 0;
if N > 1 then
[D:= 1;
repeat  if rem(N/D) = 0 then
[C:= C+1;
if Show then
[if D > 1 then ChOut(0, ^ );
IntOut(0, D);
];
];
D:= D+1;
until   D >= N;
];
return C;
];

int N, SN, Cnt, Max;
[for N:= 1 to 10 do
[ChOut(0, ^[);  PropDiv(N, true);  ChOut(0, ^]);
ChOut(0, ^ );
];
CrLf(0);

Max:= 0;
for N:= 1 to 20000 do
[Cnt:= PropDiv(N, false);
if Cnt > Max then
[Max:= Cnt;  SN:= N];
];
IntOut(0, SN);  ChOut(0, ^ ); IntOut(0, Max);  CrLf(0);
]```
Output:
```[]   [1 2]  [1 2 3]  [1 2 4] [1 3] [1 2 5]
15120 79
```

## zkl

Translation of: D

This is the simple version :

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

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

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