# Abundant, deficient and perfect number classifications

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

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

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

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


Example

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

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

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

Show the results here.

## 11l

Translation of: Kotlin
F sum_proper_divisors(n)
R I n < 2 {0} E sum((1 .. n I/ 2).filter(it -> (@n % it) == 0))

V deficient = 0
V perfect = 0
V abundant = 0

L(n) 1..20000
V sp = sum_proper_divisors(n)
I sp < n
deficient++
E I sp == n
perfect++
E I sp > n
abundant++

print(‘Deficient = ’deficient)
print(‘Perfect   = ’perfect)
print(‘Abundant  = ’abundant)
Output:
Deficient = 15043
Perfect   = 4
Abundant  = 4953


## 360 Assembly

Translation of: VBScript

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

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


## 8086 Assembly

LIMIT:	equ	20000
cpu	8086
org	100h
mov	ax,data		; Set DS and ES to point right after the
mov	cl,4		; program, so we can store the array there
shr	ax,cl
mov	dx,cs
inc	ax
mov	ds,ax
mov	es,ax
mov	ax,1		; Set each element to 1 at the beginning
xor	di,di
mov	cx,LIMIT+1
rep	stosw
mov	[2],cx		; Except the value for 1, which is 0
mov 	bp,LIMIT/2	; BP = limit / 2 - keep values ready in regs
mov	di,LIMIT	; DI = limit
oloop:	inc	ax		; Let AX be the outer loop counter (divisor)
cmp	ax,bp		; Are we there yet?
ja	clsfy		; If so, stop
mov	dx,ax		; Let DX be the inner loop counter (number)
cmp	dx,di		; Are we there yet?
ja	oloop		; Loop
mov	bx,dx		; Each entry is 2 bytes wide
shl	bx,1
jmp	iloop
clsfy:	xor	bp,bp		; BP = deficient number counter
xor	dx,dx		; DX = perfect number counter
xor	cx,cx 		; CX = abundant number counter
xor	bx,bx		; BX = current number under consideration
mov	si,2		; SI = pointer to divsum of current number
cloop:	inc	bx		; Next number
cmp	bx,di		; Are we done yet?
ja	done		; If so, stop
lodsw			; Otherwise, get divsum of current number
cmp	ax,bx		; Compare to current number
jb	defic		; If smaller, the number is deficient
je	prfct		; If equal, the number is perfect
inc	cx		; Otherwise, the number is abundant
jmp	cloop
defic:	inc	bp
jmp	cloop
prfct:	inc	dx
jmp	cloop
done:	mov	ax,cs		; Set DS and ES back to the code segment
mov	ds,ax
mov	es,ax
mov	di,dx		; Move the perfect numbers to DI
mov	dx,sdef		; Print "Deficient"
call	prstr
mov	ax,bp		; Print amount of deficient numbers
call	prnum
mov	dx,sper		; Print "Perfect"
call	prstr
mov	ax,di		; Print amount of perfect numbers
call	prnum
mov	dx,sabn		; Print "Abundant"
call 	prstr
mov	ax,cx		; Print amount of abundant numbers
prnum:	mov	bx,snum		; Print number in AX
pdgt:	xor	dx,dx
div	word [ten]	; Extract digit
dec	bx		; Move pointer
mov	[bx],dl		; Store digit
test	ax,ax		; Any more digits?
jnz	pdgt
mov	dx,bx		; Print string
prstr:	mov	ah,9
int	21h
ret
ten:	dw	10		; Divisor for number output routine
sdef:	db	'Deficient: $' sper: db 'Perfect:$'
sabn:	db	'Abundant: $' db '.....' snum: db 13,10,'$'
data:	equ	 Output: Deficient: 15043 Perfect: 4 Abundant: 4953 ## AArch64 Assembly Works with: as version Raspberry Pi 3B version Buster 64 bits or android 64 bits with application Termux /* ARM assembly AARCH64 Raspberry PI 3B or android 64 bits */ /* program numberClassif64.s */ /*******************************************/ /* Constantes file */ /*******************************************/ /* for this file see task include a file in language AArch64 assembly*/ .include "../includeConstantesARM64.inc" .equ NBDIVISORS, 1000 /*******************************************/ /* Initialized data */ /*******************************************/ .data szMessStartPgm: .asciz "Program 64 bits start \n" szMessEndPgm: .asciz "Program normal end.\n" szMessErrorArea: .asciz "\033[31mError : area divisors too small.\n" szMessError: .asciz "\033[31mError !!!\n" szMessErrGen: .asciz "Error end program.\n" szMessNbPrem: .asciz "This number is prime !!!.\n" szMessOverflow: .asciz "Overflow function isPrime.\n" szCarriageReturn: .asciz "\n" /* datas message display */ szMessResult: .asciz "Number déficients : @ perfects : @ abundants : @ \n" /*******************************************/ /* UnInitialized data */ /*******************************************/ .bss .align 4 sZoneConv: .skip 24 tbZoneDecom: .skip 8 * NBDIVISORS // facteur 8 octets /*******************************************/ /* code section */ /*******************************************/ .text .global main main: // program start ldr x0,qAdrszMessStartPgm // display start message bl affichageMess mov x4,#1 mov x3,#0 mov x6,#0 mov x7,#0 mov x8,#0 ldr x9,iNBMAX 1: mov x0,x4 // number //================================= ldr x1,qAdrtbZoneDecom bl decompFact // create area of divisors cmp x0,#0 // error ? blt 2f lsl x5,x4,#1 // number * 2 cmp x5,x1 // compare number and sum cinc x7,x7,eq // perfect cinc x6,x6,gt // deficient cinc x8,x8,lt // abundant 2: add x4,x4,#1 cmp x4,x9 ble 1b //================================ mov x0,x6 // deficient ldr x1,qAdrsZoneConv bl conversion10 // convert ascii string ldr x0,qAdrszMessResult ldr x1,qAdrsZoneConv bl strInsertAtCharInc // and put in message mov x5,x0 mov x0,x7 // perfect ldr x1,qAdrsZoneConv bl conversion10 // convert ascii string mov x0,x5 ldr x1,qAdrsZoneConv bl strInsertAtCharInc // and put in message mov x5,x0 mov x0,x8 // abundant ldr x1,qAdrsZoneConv bl conversion10 // convert ascii string mov x0,x5 ldr x1,qAdrsZoneConv bl strInsertAtCharInc // and put in message bl affichageMess ldr x0,qAdrszMessEndPgm // display end message bl affichageMess b 100f 99: // display error message ldr x0,qAdrszMessError bl affichageMess 100: // standard end of the program mov x0, #0 // return code mov x8, #EXIT // request to exit program svc 0 // perform system call qAdrszMessStartPgm: .quad szMessStartPgm qAdrszMessEndPgm: .quad szMessEndPgm qAdrszMessError: .quad szMessError qAdrszCarriageReturn: .quad szCarriageReturn qAdrtbZoneDecom: .quad tbZoneDecom qAdrszMessResult: .quad szMessResult qAdrsZoneConv: .quad sZoneConv iNBMAX: .quad 20000 /******************************************************************/ /* decomposition en facteur */ /******************************************************************/ /* x0 contient le nombre à decomposer */ /* x1 contains factor area address */ decompFact: stp x3,lr,[sp,-16]! // save registres stp x4,x5,[sp,-16]! // save registres stp x6,x7,[sp,-16]! // save registres stp x8,x9,[sp,-16]! // save registres stp x10,x11,[sp,-16]! // save registres mov x5,x1 mov x1,x0 cmp x0,1 beq 100f mov x8,x0 // save number bl isPrime // prime ? cmp x0,#1 beq 98f // yes is prime mov x1,#1 str x1,[x5] // first factor mov x12,#1 // divisors sum mov x4,#1 // indice divisors table mov x1,#2 // first divisor mov x6,#0 // previous divisor mov x7,#0 // number of same divisors 2: mov x0,x8 // dividende udiv x2,x0,x1 // x1 divisor x2 quotient x3 remainder msub x3,x2,x1,x0 cmp x3,#0 bne 5f // if remainder <> zero -> no divisor mov x8,x2 // else quotient -> new dividende cmp x1,x6 // same divisor ? beq 4f // yes mov x7,x4 // number factors in table mov x9,#0 // indice 21: ldr x10,[x5,x9,lsl #3 ] // load one factor mul x10,x1,x10 // multiply str x10,[x5,x7,lsl #3] // and store in the table add x12,x12,x10 add x7,x7,#1 // and increment counter add x9,x9,#1 cmp x9,x4 blt 21b mov x4,x7 mov x6,x1 // new divisor b 7f 4: // same divisor sub x9,x4,#1 mov x7,x4 41: ldr x10,[x5,x9,lsl #3 ] cmp x10,x1 sub x13,x9,1 csel x9,x13,x9,ne bne 41b sub x9,x4,x9 42: ldr x10,[x5,x9,lsl #3 ] mul x10,x1,x10 str x10,[x5,x7,lsl #3] // and store in the table add x12,x12,x10 add x7,x7,#1 // and increment counter add x9,x9,#1 cmp x9,x4 blt 42b mov x4,x7 b 7f // and loop /* not divisor -> increment next divisor */ 5: cmp x1,#2 // if divisor = 2 -> add 1 add x13,x1,#1 // add 1 add x14,x1,#2 // else add 2 csel x1,x13,x14,eq b 2b /* divisor -> test if new dividende is prime */ 7: mov x3,x1 // save divisor cmp x8,#1 // dividende = 1 ? -> end beq 10f mov x0,x8 // new dividende is prime ? mov x1,#0 bl isPrime // the new dividende is prime ? cmp x0,#1 bne 10f // the new dividende is not prime cmp x8,x6 // else dividende is same divisor ? beq 9f // yes mov x7,x4 // number factors in table mov x9,#0 // indice 71: ldr x10,[x5,x9,lsl #3 ] // load one factor mul x10,x8,x10 // multiply str x10,[x5,x7,lsl #3] // and store in the table add x12,x12,x10 add x7,x7,#1 // and increment counter add x9,x9,#1 cmp x9,x4 blt 71b mov x4,x7 mov x7,#0 b 11f 9: sub x9,x4,#1 mov x7,x4 91: ldr x10,[x5,x9,lsl #3 ] cmp x10,x8 sub x13,x9,#1 csel x9,x13,x9,ne bne 91b sub x9,x4,x9 92: ldr x10,[x5,x9,lsl #3 ] mul x10,x8,x10 str x10,[x5,x7,lsl #3] // and store in the table add x12,x12,x10 add x7,x7,#1 // and increment counter add x9,x9,#1 cmp x9,x4 blt 92b mov x4,x7 b 11f 10: mov x1,x3 // current divisor = new divisor cmp x1,x8 // current divisor > new dividende ? ble 2b // no -> loop /* end decomposition */ 11: mov x0,x4 // return number of table items mov x1,x12 // return sum mov x3,#0 str x3,[x5,x4,lsl #3] // store zéro in last table item b 100f 98: //ldr x0,qAdrszMessNbPrem //bl affichageMess add x1,x8,1 mov x0,#0 // return code b 100f 99: ldr x0,qAdrszMessError bl affichageMess mov x0,#-1 // error code b 100f 100: ldp x10,x11,[sp],16 // restaur des 2 registres ldp x8,x9,[sp],16 // restaur des 2 registres ldp x6,x7,[sp],16 // restaur des 2 registres ldp x4,x5,[sp],16 // restaur des 2 registres ldp x3,lr,[sp],16 // restaur des 2 registres ret // retour adresse lr x30 qAdrszMessErrGen: .quad szMessErrGen qAdrszMessNbPrem: .quad szMessNbPrem /***************************************************/ /* Verification si un nombre est premier */ /***************************************************/ /* x0 contient le nombre à verifier */ /* x0 retourne 1 si premier 0 sinon */ isPrime: stp x1,lr,[sp,-16]! // save registres stp x2,x3,[sp,-16]! // save registres mov x2,x0 sub x1,x0,#1 cmp x2,0 beq 99f // retourne zéro cmp x2,2 // pour 1 et 2 retourne 1 ble 2f mov x0,#2 bl moduloPux64 bcs 100f // erreur overflow cmp x0,#1 bne 99f // Pas premier cmp x2,3 beq 2f mov x0,#3 bl moduloPux64 blt 100f // erreur overflow cmp x0,#1 bne 99f cmp x2,5 beq 2f mov x0,#5 bl moduloPux64 bcs 100f // erreur overflow cmp x0,#1 bne 99f // Pas premier cmp x2,7 beq 2f mov x0,#7 bl moduloPux64 bcs 100f // erreur overflow cmp x0,#1 bne 99f // Pas premier cmp x2,11 beq 2f mov x0,#11 bl moduloPux64 bcs 100f // erreur overflow cmp x0,#1 bne 99f // Pas premier cmp x2,13 beq 2f mov x0,#13 bl moduloPux64 bcs 100f // erreur overflow cmp x0,#1 bne 99f // Pas premier 2: cmn x0,0 // carry à zero pas d'erreur mov x0,1 // premier b 100f 99: cmn x0,0 // carry à zero pas d'erreur mov x0,#0 // Pas premier 100: ldp x2,x3,[sp],16 // restaur des 2 registres ldp x1,lr,[sp],16 // restaur des 2 registres ret // retour adresse lr x30 /**************************************************************/ /********************************************************/ /* Calcul modulo de b puissance e modulo m */ /* Exemple 4 puissance 13 modulo 497 = 445 */ /********************************************************/ /* x0 nombre */ /* x1 exposant */ /* x2 modulo */ moduloPux64: stp x1,lr,[sp,-16]! // save registres stp x3,x4,[sp,-16]! // save registres stp x5,x6,[sp,-16]! // save registres stp x7,x8,[sp,-16]! // save registres stp x9,x10,[sp,-16]! // save registres cbz x0,100f cbz x1,100f mov x8,x0 mov x7,x1 mov x6,1 // resultat udiv x4,x8,x2 msub x9,x4,x2,x8 // contient le reste 1: tst x7,1 beq 2f mul x4,x9,x6 umulh x5,x9,x6 //cbnz x5,99f mov x6,x4 mov x0,x6 mov x1,x5 bl divisionReg128U cbnz x1,99f // overflow mov x6,x3 2: mul x8,x9,x9 umulh x5,x9,x9 mov x0,x8 mov x1,x5 bl divisionReg128U cbnz x1,99f // overflow mov x9,x3 lsr x7,x7,1 cbnz x7,1b mov x0,x6 // result cmn x0,0 // carry à zero pas d'erreur b 100f 99: ldr x0,qAdrszMessOverflow bl affichageMess cmp x0,0 // carry à un car erreur mov x0,-1 // code erreur 100: ldp x9,x10,[sp],16 // restaur des 2 registres ldp x7,x8,[sp],16 // restaur des 2 registres ldp x5,x6,[sp],16 // restaur des 2 registres ldp x3,x4,[sp],16 // restaur des 2 registres ldp x1,lr,[sp],16 // restaur des 2 registres ret // retour adresse lr x30 qAdrszMessOverflow: .quad szMessOverflow /***************************************************/ /* division d un nombre de 128 bits par un nombre de 64 bits */ /***************************************************/ /* x0 contient partie basse dividende */ /* x1 contient partie haute dividente */ /* x2 contient le diviseur */ /* x0 retourne partie basse quotient */ /* x1 retourne partie haute quotient */ /* x3 retourne le reste */ divisionReg128U: stp x6,lr,[sp,-16]! // save registres stp x4,x5,[sp,-16]! // save registres mov x5,#0 // raz du reste R mov x3,#128 // compteur de boucle mov x4,#0 // dernier bit 1: lsl x5,x5,#1 // on decale le reste de 1 tst x1,1<<63 // test du bit le plus à gauche lsl x1,x1,#1 // on decale la partie haute du quotient de 1 beq 2f orr x5,x5,#1 // et on le pousse dans le reste R 2: tst x0,1<<63 lsl x0,x0,#1 // puis on decale la partie basse beq 3f orr x1,x1,#1 // et on pousse le bit de gauche dans la partie haute 3: orr x0,x0,x4 // position du dernier bit du quotient mov x4,#0 // raz du bit cmp x5,x2 blt 4f sub x5,x5,x2 // on enleve le diviseur du reste mov x4,#1 // dernier bit à 1 4: // et boucle subs x3,x3,#1 bgt 1b lsl x1,x1,#1 // on decale le quotient de 1 tst x0,1<<63 lsl x0,x0,#1 // puis on decale la partie basse beq 5f orr x1,x1,#1 5: orr x0,x0,x4 // position du dernier bit du quotient mov x3,x5 100: ldp x4,x5,[sp],16 // restaur des 2 registres ldp x6,lr,[sp],16 // restaur des 2 registres ret // retour adresse lr x30 /********************************************************/ /* File Include fonctions */ /********************************************************/ /* for this file see task include a file in language AArch64 assembly */ .include "../includeARM64.inc" Output: Program 64 bits start Number déficients : 15043 perfects : 4 abundants : 4953 Program normal end.  ## ABC PUT 0 IN deficient PUT 0 IN perfect PUT 0 IN abundant HOW TO FIND PROPER DIVISOR SUMS UP TO limit: SHARE p PUT {} IN p FOR i IN {0..limit}: PUT 0 IN p[i] FOR i IN {1..floor (limit/2)}: PUT i+i IN j WHILE j <= limit: PUT p[j]+i IN p[j] PUT j+i IN j HOW TO CLASSIFY n: SHARE deficient, perfect, abundant, p SELECT: p[n] < n: PUT deficient+1 IN deficient p[n] = n: PUT perfect+1 IN perfect p[n] > n: PUT abundant+1 IN abundant PUT 20000 IN limit FIND PROPER DIVISOR SUMS UP TO limit FOR n IN {1..limit}: CLASSIFY n WRITE deficient, "deficient"/ WRITE perfect, "perfect"/ WRITE abundant, "abundant"/ Output: 15043 deficient 4 perfect 4953 abundant ## Action! Because of the memory limitation on the non-expanded Atari 8-bit computer the array containing Proper Divisor Sums is generated and used twice for the first and the second half of numbers separately. PROC FillSumOfDivisors(CARD ARRAY pds CARD size,maxNum,offset) CARD i,j FOR i=0 TO size-1 DO pds(i)=1 OD FOR i=2 TO maxNum DO FOR j=i+i TO maxNum STEP i DO IF j>=offset THEN pds(j-offset)==+i FI OD OD RETURN PROC Main() DEFINE MAXNUM="20000" DEFINE HALFNUM="10000" CARD ARRAY pds(HALFNUM+1) CARD def,perf,abud,i,sum,offset BYTE CRSINH=02F0 ;Controls visibility of cursor

CRSINH=1 ;hide cursor
Put(125) PutE() ;clear the screen

def=1 perf=0 abud=0
FillSumOfDivisors(pds,HALFNUM+1,HALFNUM,0)
FOR i=2 TO HALFNUM
DO
sum=pds(i)
IF sum<i THEN def==+1
ELSEIF sum=i THEN perf==+1
ELSE abud==+1 FI
OD

offset=HALFNUM
FillSumOfDivisors(pds,HALFNUM+1,MAXNUM,offset)
FOR i=HALFNUM+1 TO MAXNUM
DO
sum=pds(i-offset)
IF sum<i THEN def==+1
ELSEIF sum=i THEN perf==+1
ELSE abud==+1 FI
OD

PrintF("  Numbers: %I%E",MAXNUM)
PrintF("Deficient: %I%E",def)
PrintF("  Perfect: %I%E",perf)
PrintF("  Abudant: %I%E",abud)
RETURN
Output:
Please wait...
Numbers: 20000
Deficient: 15043
Perfect: 4
Abudant: 4953


This solution uses the package Generic_Divisors from the Proper Divisors task [[1]].

with Ada.Text_IO, Generic_Divisors;

function Same(P: Positive) return Positive is (P);
package Divisor_Sum is new Generic_Divisors
(Result_Type => Natural, None => 0, One => Same, Add =>  "+");

type Class_Type is (Deficient, Perfect, Abundant);

function Class(D_Sum, N: Natural) return Class_Type is
(if D_Sum < N then Deficient
elsif D_Sum = N then Perfect
else Abundant);

Cls: Class_Type;
Results: array (Class_Type) of Natural := (others => 0);

begin
for N in 1 .. 20_000 loop
Cls := Class(Divisor_Sum.Process(N), N);
Results(Cls) := Results(Cls)+1;
end loop;
for Class in Results'Range loop
CIO.Put(Class, 12);
NIO.Put(Results(Class), 8);
end loop;
NIO.Put(Results(Deficient)+Results(Perfect)+Results(Abundant), 8);

Output:
DEFICIENT      15043
PERFECT            4
ABUNDANT        4953
--------------------
Sum            20000
====================

## ALGOL 68

BEGIN # classify the numbers 1 : 20 000 as abudant, deficient or perfect #
INT abundant count    := 0;
INT deficient count   := 0;
INT perfect count     := 0;
INT max number         = 20 000;
# construct a table of the proper divisor sums                 #
[ 1 : max number ]INT pds;
pds[ 1 ] := 0;
FOR i FROM 2 TO UPB pds DO pds[ i ] := 1 OD;
FOR i FROM 2 TO UPB pds DO
FOR j FROM i + i BY i TO UPB pds DO pds[ j ] +:= i OD
OD;
# classify the numbers                                         #
FOR n TO max number DO
INT pd sum = pds[ n ];
IF     pd sum < n THEN
deficient count +:= 1
ELIF   pd sum = n THEN
perfect count   +:= 1
ELSE # pd sum > n #
abundant count  +:= 1
FI
OD;
print( ( "abundant  ", whole(  abundant count, 0 ), newline ) );
print( ( "deficient ", whole( deficient count, 0 ), newline ) );
print( ( "perfect   ", whole(   perfect count, 0 ), newline ) )
END
Output:
abundant  4953
deficient 15043
perfect   4


## ALGOL W

begin % count abundant, perfect and deficient numbers up to 20 000        %
integer MAX_NUMBER;
MAX_NUMBER := 20000;
begin
integer array pds ( 1 :: MAX_NUMBER );
integer aCount, dCount, pCount, dSum;
% construct a table of proper divisor sums                        %
pds( 1 ) := 0;
for i := 2 until MAX_NUMBER do pds( i ) := 1;
for i := 2 until MAX_NUMBER do begin
for j := i + i step i until MAX_NUMBER do pds( j ) := pds( j ) + i
end for_i ;
aCount := dCount := pCOunt := 0;
for i := 1 until 20000 do begin
dSum := pds( i );
if      dSum > i then aCount := aCount + 1
else if dSum < i then dCount := dCOunt + 1
else %  dSum = i    % pCount := pCount + 1
end for_i ;
write( "Abundant  numbers up to 20 000: ", aCount );
write( "Perfect   numbers up to 20 000: ", pCount );
write( "Deficient numbers up to 20 000: ", dCount )
end
end.
Output:
Abundant  numbers up to 20 000:           4953
Perfect   numbers up to 20 000:              4
Deficient numbers up to 20 000:          15043


## AppleScript

on aliquotSum(n)
if (n < 2) then return 0
set sum to 1
set sqrt to n ^ 0.5
set limit to sqrt div 1
if (limit = sqrt) then
set sum to sum + limit
set limit to limit - 1
end if
repeat with i from 2 to limit
if (n mod i is 0) then set sum to sum + i + n div i
end repeat

return sum
end aliquotSum

set {deficient, perfect, abundant} to {0, 0, 0}
repeat with n from 1 to 20000
set s to aliquotSum(n)
if (s < n) then
set deficient to deficient + 1
else if (s > n) then
set abundant to abundant + 1
else
set perfect to perfect + 1
end if
end repeat

return {deficient:deficient, perfect:perfect, abundant:abundant}


Output:
{deficient:15043, perfect:4, abundant:4953}


## ARM Assembly

Works with: as version Raspberry Pi
or android 32 bits with application Termux
/* ARM assembly Raspberry PI  */
/* program numberClassif.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 */
/* for constantes see task include a file in arm assembly */
/************************************/
/* Constantes                       */
/************************************/
.include "../constantes.inc"

.equ NBDIVISORS,             1000

/*******************************************/
/* Initialized data                        */
/*******************************************/
.data
szMessStartPgm:          .asciz "Program start \n"
szMessEndPgm:            .asciz "Program normal end.\n"
szMessErrorArea:         .asciz "\033[31mError : area divisors too small.\n"
szMessError:             .asciz "\033[31mError  !!!\n"
szMessErrGen:            .asciz "Error end program.\n"
szMessNbPrem:            .asciz "This number is prime !!!.\n"
szMessResultFact:        .asciz "@ "

szCarriageReturn:        .asciz "\n"

/* datas message display */
szMessResult:            .asciz "Number déficients : @ perfects : @ abundants : @ \n"

/*******************************************/
/* UnInitialized data                      */
/*******************************************/
.bss
.align 4
sZoneConv:               .skip 24
tbZoneDecom:             .skip 4 * NBDIVISORS       // facteur 4 octets
/*******************************************/
/*  code section                           */
/*******************************************/
.text
.global main
main:                               @ program start
ldr r0,iAdrszMessStartPgm       @ display start message
bl affichageMess

mov r4,#1
mov r3,#0
mov r6,#0
mov r7,#0
mov r8,#0
ldr r9,iNBMAX
1:
mov r0,r4                       @  number
//=================================
bl decompFact                @ create area of divisors
cmp r0,#0                    @ error ?
blt 2f
lsl r5,r4,#1                 @ number * 2
cmp r5,r1                    @ compare number and sum

2:
cmp r4,r9
ble 1b

//================================

mov r0,r6                        @ deficient
bl conversion10                  @ convert ascii string
bl strInsertAtCharInc               @ and put in message
mov r5,r0
mov r0,r7                        @ perfect
bl conversion10                  @ convert ascii string
mov r0,r5
bl strInsertAtCharInc               @ and put in message
mov r5,r0
mov r0,r8                        @ abundant
bl conversion10                  @ convert ascii string
mov r0,r5
bl strInsertAtCharInc               @ and put in message
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

iNBMAX:                    .int 20000

/******************************************************************/
/*     factor decomposition                                               */
/******************************************************************/
/* r0 contains number */
/* r1 contains address of divisors area */
/* r0 return divisors items in table */
/* r1 return the sum of divisors  */
decompFact:
push {r3-r12,lr}              @ save  registers
cmp r0,#1
moveq r1,#1
beq 100f
mov r5,r1
mov r8,r0                    @ save number
bl isPrime                   @ prime ?
cmp r0,#1
beq 98f                      @ yes is prime
mov r1,#1
str r1,[r5]                  @ first factor
mov r12,#1                   @ divisors sum
mov r10,#1                   @ indice divisors table
mov r9,#2                    @ first divisor
mov r6,#0                    @ previous divisor
mov r7,#0                    @ number of same divisors

/*  division loop  */
2:
mov r0,r8                    @ dividende
mov r1,r9                    @ divisor
bl division                  @ r2 quotient r3 remainder
cmp r3,#0
beq 3f                       @ if remainder  zero  ->  divisor

/* not divisor -> increment next divisor */
cmp r9,#2                    @ if divisor = 2 -> add 1
b 2b

/* divisor   compute the new factors of number */
3:
mov r8,r2                    @ else quotient -> new dividende
cmp r9,r6                    @ same divisor ?
beq 4f                       @ yes

mov r1,r10                   @ number factors in table
mov r2,r9                    @ divisor
mov r3,r12                   @ somme
mov r4,#0
bl computeFactors
mov r10,r1
mov r12,r0
mov r6,r9                    @ new divisor
b 7f

4:                               @ same divisor
sub r7,r10,#1
5:                              @ search in table the first use of divisor
ldr r3,[r5,r7,lsl #2 ]
cmp r3,r9
subne r7,#1
bne 5b
@ and compute new factors after factors
sub r4,r10,r7                @ start indice
mov r0,r5
mov r1,r10
mov r2,r9                    @ divisor
mov r3,r12
bl computeFactors
mov r12,r0
mov r10,r1

/* divisor -> test if new dividende is prime */
7:
cmp r8,#1                    @ dividende = 1 ? -> end
beq 10f
mov r0,r8                    @ new dividende is prime ?
mov r1,#0
bl isPrime                   @ the new dividende is prime ?
cmp r0,#1
bne 10f                      @ the new dividende is not prime

cmp r8,r6                    @ else dividende is same divisor ?
beq 8f                       @ yes

mov r0,r5
mov r1,r10
mov r2,r8
mov r3,r12
mov r4,#0
bl computeFactors
mov r12,r0
mov r10,r1
mov r7,#0
b 11f
8:
sub r7,r10,#1
9:
ldr r3,[r5,r7,lsl #2 ]
cmp r3,r8
subne r7,#1
bne 9b

mov r0,r5
mov r1,r10
sub r4,r10,r7
mov r2,r8
mov r3,r12
bl computeFactors
mov r12,r0
mov r10,r1

b 11f

10:
cmp r9,r8                    @ current divisor  > new dividende ?
ble 2b                       @ no -> loop

/* end decomposition */
11:
mov r0,r10                  @ return number of table items
mov r1,r12                  @ return sum
mov r3,#0
str r3,[r5,r10,lsl #2]      @ store zéro in last table item
b 100f

98:                             @ prime number
//bl   affichageMess
mov r0,#0                   @ return code
b 100f
99:
bl   affichageMess
mov r0,#-1                  @ error code
b 100f
100:
pop {r3-r12,lr}             @ restaur registers
bx lr

/*   r0 table factors address */
/*   r1 number factors in table */
/*   r2 new divisor */
/*   r3 sum  */
/*   r4 start indice */
/*   r0 return sum */
/*   r1 return number factors in table */
computeFactors:
push {r2-r6,lr}              @ save registers
mov r6,r1                    @ number factors in table
1:
ldr r5,[r0,r4,lsl #2 ]       @ load one factor
mul r5,r2,r5                 @ multiply
str r5,[r0,r1,lsl #2]        @ and store in the table

add r1,r1,#1                 @ and increment counter
cmp r4,r6
blt 1b
mov r0,r3
100:                             @ fin standard de la fonction
pop {r2-r6,lr}               @ restaur des registres
bx lr                        @ retour de la fonction en utilisant lr
/***************************************************/
/*   check if a number is prime              */
/***************************************************/
/* r0 contains the number            */
/* r0 return 1 if prime  0 else */
@2147483647
@4294967297
@131071
isPrime:
push {r1-r6,lr}    @ save registers
cmp r0,#0
beq 90f
cmp r0,#17
bhi 1f
cmp r0,#3
bls 80f            @ for 1,2,3 return prime
cmp r0,#5
beq 80f            @ for 5 return prime
cmp r0,#7
beq 80f            @ for 7 return prime
cmp r0,#11
beq 80f            @ for 11 return prime
cmp r0,#13
beq 80f            @ for 13 return prime
cmp r0,#17
beq 80f            @ for 17 return prime
1:
tst r0,#1          @ even ?
beq 90f            @ yes -> not prime
mov r2,r0          @ save number
sub r1,r0,#1       @ exposant n - 1
mov r0,#3          @ base
bl moduloPuR32     @ compute base power n - 1 modulo n
cmp r0,#1
bne 90f            @ if <> 1  -> not prime

mov r0,#5
bl moduloPuR32
cmp r0,#1
bne 90f

mov r0,#7
bl moduloPuR32
cmp r0,#1
bne 90f

mov r0,#11
bl moduloPuR32
cmp r0,#1
bne 90f

mov r0,#13
bl moduloPuR32
cmp r0,#1
bne 90f

mov r0,#17
bl moduloPuR32
cmp r0,#1
bne 90f
80:
mov r0,#1        @ is prime
b 100f
90:
mov r0,#0        @ no prime
100:                 @ fin standard de la fonction
pop {r1-r6,lr}   @ restaur des registres
bx lr            @ retour de la fonction en utilisant lr
/********************************************************/
/*   Calcul modulo de b puissance e modulo m  */
/*    Exemple 4 puissance 13 modulo 497 = 445         */
/*                                             */
/********************************************************/
/* r0  nombre  */
/* r1 exposant */
/* r2 modulo   */
/* r0 return result  */
moduloPuR32:
push {r1-r7,lr}    @ save registers
cmp r0,#0          @ verif <> zero
beq 100f
cmp r2,#0          @ verif <> zero
beq 100f           @ TODO: v鲩fier les cas d erreur
1:
mov r4,r2          @ save modulo
mov r5,r1          @ save exposant
mov r6,r0          @ save base
mov r3,#1          @ start result

mov r1,#0          @ division de r0,r1 par r2
bl division32R
mov r6,r2          @ base <- remainder
2:
tst r5,#1          @  exposant even or odd
beq 3f
umull r0,r1,r6,r3
mov r2,r4
bl division32R
mov r3,r2          @ result <- remainder
3:
umull r0,r1,r6,r6
mov r2,r4
bl division32R
mov r6,r2          @ base <- remainder

lsr r5,#1          @ left shift 1 bit
cmp r5,#0          @ end ?
bne 2b
mov r0,r3
100:                   @ fin standard de la fonction
pop {r1-r7,lr}     @ restaur des registres
bx lr              @ retour de la fonction en utilisant lr

/***************************************************/
/*   division number 64 bits in 2 registers by number 32 bits */
/***************************************************/
/* r0 contains lower part dividende   */
/* r1 contains upper part dividende   */
/* r2 contains divisor   */
/* r0 return lower part quotient    */
/* r1 return upper part quotient    */
/* r2 return remainder               */
division32R:
push {r3-r9,lr}    @ save registers
mov r6,#0          @ init upper upper part remainder  !!
mov r7,r1          @ init upper part remainder with upper part dividende
mov r8,r0          @ init lower part remainder with lower part dividende
mov r9,#0          @ upper part quotient
mov r4,#0          @ lower part quotient
mov r5,#32         @ bits number
1:                     @ begin loop
lsl r6,#1          @ shift upper upper part remainder
lsls r7,#1         @ shift upper  part remainder
orrcs r6,#1
lsls r8,#1         @ shift lower  part remainder
orrcs r7,#1
lsls r4,#1         @ shift lower part quotient
lsl r9,#1          @ shift upper part quotient
orrcs r9,#1
@ divisor sustract  upper  part remainder
subs r7,r2
sbcs  r6,#0        @ and substract carry
bmi 2f             @ n駡tive ?

@ positive or equal
orr r4,#1          @ 1 -> right bit quotient
b 3f
2:                     @ negative
orr r4,#0          @ 0 -> right bit quotient
adds r7,r2         @ and restaur remainder
3:
subs r5,#1         @ decrement bit size
bgt 1b             @ end ?
mov r0,r4          @ lower part quotient
mov r1,r9          @ upper part quotient
mov r2,r7          @ remainder
100:                   @ function end
pop {r3-r9,lr}     @ restaur registers
bx lr

/***************************************************/
/*      ROUTINES INCLUDE                 */
/***************************************************/
.include "../affichage.inc"
Output:
Program start
Number déficients : 15043       perfects : 4           abundants : 4953
Program normal end.


## Arturo

properDivisors: function [n]->
(factors n) -- n

abundant: new 0 deficient: new 0 perfect: new 0

loop 1..20000 'x [
s: sum properDivisors x

case [s]
when? [<x] -> inc 'deficient
when? [>x] -> inc 'abundant
else       -> inc 'perfect
]

print ["Found" abundant "abundant,"
deficient "deficient and"
perfect "perfect numbers."]

Output:
Found 4953 abundant, 15043 deficient and 4 perfect numbers.

## AutoHotkey

Loop
{
m := A_index
; getting factors=====================
loop % floor(sqrt(m))
{
if ( mod(m, A_index) == "0" )
{
if ( A_index ** 2 == m )
{
list .= A_index . ":"
sum := sum + A_index
continue
}
if ( A_index != 1 )
{
list .= A_index . ":" . m//A_index . ":"
sum := sum + A_index + m//A_index
}
if ( A_index == "1" )
{
list .= A_index . ":"
sum := sum + A_index
}
}
}
; Factors obtained above===============
if ( sum == m ) && ( sum != 1 )
{
result := "perfect"
perfect++
}
if ( sum > m )
{
result := "Abundant"
Abundant++
}
if ( sum < m ) or ( m == "1" )
{
result := "Deficient"
Deficient++
}
if ( m == 20000 )
{
MsgBox % "number: " . m . "nFactors:n" . list . "nSum of Factors: " . Sum . "nResult: " . result . "n_______________________nTotals up to: " . m . "nPerfect: " . perfect . "nAbundant: " . Abundant . "nDeficient: " . Deficient
ExitApp
}
list := ""
sum := 0
}

esc::ExitApp

Output:
number: 20000
Factors:
1:2:10000:4:5000:5:4000:8:2500:10:2000:16:1250:20:1000:25:800:32:625:40:500:50:400:80:250:100:200:125:160:
Sum of Factors: 29203
Result: Abundant
_______________________
Totals up to: 20000
Perfect: 4
Abundant: 4953
Deficient: 15043


## AWK

works with GNU Awk 3.1.5 and with BusyBox v1.21.1

#!/bin/gawk -f
function sumprop(num,   i,sum,root) {
if (num == 1) return 0
sum=1
root=sqrt(num)
for ( i=2; i < root; i++) {
if (num % i == 0 )
{
sum = sum + i + num/i
}
}
if (num % root == 0)
{
sum = sum + root
}
return sum
}

BEGIN{
limit = 20000
abundant = 0
defiecient =0
perfect = 0

for (j=1; j < limit+1; j++)
{
sump = sumprop(j)
if (sump < j) deficient = deficient + 1
if (sump == j) perfect = perfect + 1
if (sump > j) abundant = abundant + 1
}
print "For 1 through " limit
print "Perfect: " perfect
print "Abundant: " abundant
print "Deficient: " deficient
}

Output:
For 1 through 20000
Perfect: 4
Abundant: 4953
Deficient: 15043


## Batch File

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

@echo off
setlocal enabledelayedexpansion

:_main

for /l %%i in (1,1,20000) do (

echo Processing %%i

call:_P %%i
set Pn=!errorlevel!
if !Pn! lss %%i set /a deficient+=1
if !Pn!==%%i set /a perfect+=1
if !Pn! gtr %%i set /a abundant+=1
cls
)

echo Deficient - %deficient% ^| Perfect - %perfect% ^| Abundant - %abundant%
pause>nul

:_P
setlocal enabledelayedexpansion
set sumdivisers=0

set /a upperlimit=%1-1

for /l %%i in (1,1,%upperlimit%) do (
set /a isdiviser=%1 %% %%i
if !isdiviser!==0 set /a sumdivisers+=%%i
)

exit /b %sumdivisers%

## BASIC

Works with: Chipmunk Basic
Works with: GW-BASIC
Works with: PC-BASIC version any
Works with: QBasic
10 DEFINT A-Z: LM=20000
20 DIM P(LM)
30 FOR I=1 TO LM: P(I)=-32767: NEXT
40 FOR I=1 TO LM/2: FOR J=I+I TO LM STEP I: P(J)=P(J)+I: NEXT: NEXT
50 FOR I=1 TO LM
60 X=I-32767
70 IF P(I)<X THEN D=D+1 ELSE IF P(I)=X THEN P=P+1 ELSE A=A+1
80 NEXT
90 PRINT "DEFICIENT:";D
100 PRINT "PERFECT:";P
110 PRINT "ABUNDANT:";A

Output:
DEFICIENT: 15043
PERFECT: 4
ABUNDANT: 4953

### BASIC256

deficient   = 0
perfect = 0
abundant = 0

for n = 1 to 20000
sum = SumProperDivisors(n)
begin case
case sum < n
deficient += 1
case  sum = n
perfect += 1
else
abundant += 1
end case
next

print "The classification of the numbers from 1 to 20,000 is as follows :"
print
print "Deficient = "; deficient
print "Perfect   = "; perfect
print "Abundant  = "; abundant
end

function SumProperDivisors(number)
if number < 2 then return 0
sum = 0
for i = 1 to number \ 2
if number mod i = 0 then sum += i
next i
return sum
end function
Output:
Same as FreeBASIC entry.

### Chipmunk Basic

Works with: Chipmunk Basic version 3.6.4
100 cls
110 defic = 0
120 perfe = 0
130 abund = 0
140 for n = 1 to 20000
150   sump = SumProperDivisors(n)
160     if sump < n then
170     defic = defic+1
180   else
190     if sump = n then
200       perfe = perfe+1
210     else
220       if sump > n then abund = abund+1
230     endif
240   endif
250 next
260 print "The classification of the numbers from 1 to 20,000 is as follows :"
270 print
280 print "Deficient = ";defic
290 print "Perfect   = ";perfe
300 print "Abundant  = ";abund
310 end
320 function SumProperDivisors(number)
330 if number < 2 then SumProperDivisors = 0
340 sum = 0
350 for i = 1 to number/2
360   if number mod i = 0 then sum = sum+i
370 next i
380 SumProperDivisors = sum
390 end function

Output:
Same as FreeBASIC entry.

### Gambas

Translation of: FreeBASIC
Public Sub Main()

Dim sum As Integer, deficient As Integer, perfect As Integer, abundant As Integer

For n As Integer = 1 To 20000
sum = SumProperDivisors(n)
If sum < n Then
deficient += 1
Else If sum = n Then
perfect += 1
Else
abundant += 1
Endif
Next

Print "The classification of the numbers from 1 to 20,000 is as follows : \n"
Print "Deficient = "; deficient
Print "Perfect   = "; perfect
Print "Abundant  = "; abundant

End

Function SumProperDivisors(number As Integer) As Integer

If number < 2 Then Return 0
Dim sum As Integer = 0
For i As Integer = 1 To number \ 2
If number Mod i = 0 Then sum += i
Next
Return sum

End Function

Output:
Same as FreeBASIC entry.

### GW-BASIC

Works with: PC-BASIC version any

The BASIC solution works without any changes.

### QBasic

The BASIC solution works without any changes.

### Run BASIC

function sumProperDivisors(num)
if num > 1 then
sum = 1
root = sqr(num)
for i = 2 to root
if num mod i = 0 then
sum = sum + i
if (i*i) <> num then sum = sum + num / i
end if
next i
end if
sumProperDivisors = sum
end function

deficient = 0
perfect = 0
abundant = 0

print "The classification of the numbers from 1 to 20,000 is as follows :"

for n = 1 to 20000
sump = sumProperDivisors(n)
if sump < n then
deficient = deficient +1
else
if sump = n then
perfect = perfect +1
else
if sump > n then abundant = abundant +1
end if
end if
next n

print "Deficient = "; deficient
print "Perfect   = "; perfect
print "Abundant  = "; abundant

### True BASIC

LET lm = 20000
DIM s(0)
MAT REDIM s(lm)

FOR i = 1 TO lm
LET s(i) = -32767
NEXT i
FOR i = 1 TO lm/2
FOR j = i+i TO lm STEP i
LET s(j) = s(j) +i
NEXT j
NEXT i

FOR i = 1 TO lm
LET x = i - 32767
IF s(i) < x THEN
LET d = d +1
ELSE
IF s(i) = x THEN
LET p = p +1
ELSE
LET a = a +1
END IF
END IF
NEXT i

PRINT "The classification of the numbers from 1 to 20,000 is as follows :"
PRINT
PRINT "Deficient ="; d
PRINT "Perfect   ="; p
PRINT "Abundant  ="; a
END

Output:
Similar to FreeBASIC entry.

## BCPL

get "libhdr"
manifest $( maximum = 20000$)

let calcpdivs(p, max) be
$( for i=0 to max do p!i := 0 for i=1 to max/2$(  let j = i+i
while 0 < j <= max
$( p!j := p!j + i j := j + i$)
$)$)

let classify(p, n, def, per, ab) be
$( let z = 0<=p!n<n -> def, p!n=n -> per, ab !z := !z + 1$)

let start() be
$( let p = getvec(maximum) let def, per, ab = 0, 0, 0 calcpdivs(p, maximum) for i=1 to maximum do classify(p, i, @def, @per, @ab) writef("Deficient numbers: %N*N", def) writef("Perfect numbers: %N*N", per) writef("Abundant numbers: %N*N", ab) freevec(p)$)
Output:
Deficient numbers: 15043
Perfect numbers: 4
Abundant numbers: 4953

## Befunge

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

p0"2":*8*>::2/\:2/\28*:*:**+>::28*:*:*/\28*:*:*%%#v_\:28*:*:*%v>00p:0\0\-1v
++\1-:1#^_$:28*:*:*/\28*vv_^#<<<!%*:*:*82:-1\-1\<<<\+**:*:*82<+>*:*:**\2-!#+ v"There are "0\g00+1%*:*:<>28*:*:*/\28*:*:*/:0\28*:*:**+-:!00g^^82!:g01\p01< >:#,_\." ,tneicifed">:#,_\." dna ,tcefrep">:#,_\.55+".srebmun tnadnuba">:#,_@  Output: There are 15043 deficient, 4 perfect, and 4953 abundant numbers. ## Bracmat Two solutions are given. The first solution first decomposes the current number into a multiset of prime factors and then constructs the proper divisors. The second solution finds proper divisors by checking all candidates from 1 up to the square root of the given number. The first solution is a few times faster, because establishing the prime factors of a small enough number (less than 2^32 or less than 2^64, depending on the bitness of Bracmat) is fast. ( clk$:?t0
& ( multiples
=   prime multiplicity
.     !arg:(?prime.?multiplicity)
& !multiplicity:0
& 1
|   !prime^!multiplicity*(.!multiplicity)
+ multiples$(!prime.-1+!multiplicity) ) & ( P = primeFactors prime exp poly S . !arg^1/67:?primeFactors & ( !primeFactors:?^1/67&0 | 1:?poly & whl ' ( !primeFactors:%?prime^?exp*?primeFactors & !poly*multiples$(!prime.67*!exp):?poly
)
& -1+!poly+1:?poly
& 1:?S
& (   !poly
:   ?
+ (#%@?s*?&!S+!s:?S&~)
+ ?
| 1/2*!S
)
)
)
& 0:?deficient:?perfect:?abundant
& 0:?n
&   whl
' ( 1+!n:~>20000:?n
&   P$!n : ( <!n&1+!deficient:?deficient | !n&1+!perfect:?perfect | >!n&1+!abundant:?abundant ) ) & out$(deficient !deficient perfect !perfect abundant !abundant)
& clk$:?t1 & out$(flt$(!t1+-1*!t0,2) sec) & clk$:?t2
& ( P
=   f h S
.   0:?f
& 0:?S
&   whl
' ( 1+!f:?f
& !f^2:~>!n
& (   !arg*!f^-1:~/:?g
& !S+!f:?S
& ( !g:~!f&!S+!g:?S
|
)
|
)
)
& 1/2*!S
)
& 0:?deficient:?perfect:?abundant
& 0:?n
&   whl
' ( 1+!n:~>20000:?n
&   P$!n : ( <!n&1+!deficient:?deficient | !n&1+!perfect:?perfect | >!n&1+!abundant:?abundant ) ) & out$(deficient !deficient perfect !perfect abundant !abundant)
& clk$:?t3 & out$(flt$(!t3+-1*!t2,2) sec) ); Output: deficient 15043 perfect 4 abundant 4953 4,27*10E0 sec deficient 15043 perfect 4 abundant 4953 1,63*10E1 sec ## C #include<stdio.h> #define de 0 #define pe 1 #define ab 2 int main(){ int sum = 0, i, j; int try_max = 0; //1 is deficient by default and can add it deficient list int count_list[3] = {1,0,0}; for(i=2; i <= 20000; i++){ //Set maximum to check for proper division try_max = i/2; //1 is in all proper division number sum = 1; for(j=2; j<try_max; j++){ //Check for proper division if (i % j) continue; //Pass if not proper division //Set new maximum for divisibility check try_max = i/j; //Add j to sum sum += j; if (j != try_max) sum += try_max; } //Categorize summation if (sum < i){ count_list[de]++; continue; } if (sum > i){ count_list[ab]++; continue; } count_list[pe]++; } printf("\nThere are %d deficient," ,count_list[de]); printf(" %d perfect," ,count_list[pe]); printf(" %d abundant numbers between 1 and 20000.\n" ,count_list[ab]); return 0; }  Output: There are 15043 deficient, 4 perfect, 4953 abundant numbers between 1 and 20000.  ## C# Three algorithms presented, the first is fast, but can be a memory hog when tabulating to larger limits. The second is slower, but doesn't have any memory issue. The third is quite a bit slower, but the code may be easier to follow. First method: Initializes a large queue, uses a double nested loop to populate it, and a third loop to interrogate the queue. Second method: Uses a double nested loop with the inner loop only reaching to sqrt(i), as it adds both divisors at once, later correcting the sum when the divisor is a perfect square. Third method: Uses a loop with a inner Enumerable.Range reaching to i / 2, only adding one divisor at a time. using System; using System.Linq; public class Program { public static void Main() { int abundant, deficient, perfect; var sw = System.Diagnostics.Stopwatch.StartNew(); ClassifyNumbers.UsingSieve(20000, out abundant, out deficient, out perfect); sw.Stop(); Console.WriteLine($"Abundant: {abundant}, Deficient: {deficient}, Perfect: {perfect}  {sw.Elapsed.TotalMilliseconds} ms");
sw.Restart();
ClassifyNumbers.UsingOptiDivision(20000, out abundant, out deficient, out perfect);
Console.WriteLine($"Abundant: {abundant}, Deficient: {deficient}, Perfect: {perfect} {sw.Elapsed.TotalMilliseconds} ms"); sw.Restart(); ClassifyNumbers.UsingDivision(20000, out abundant, out deficient, out perfect); Console.WriteLine($"Abundant: {abundant}, Deficient: {deficient}, Perfect: {perfect}  {sw.Elapsed.TotalMilliseconds} ms");
}
}

public static class ClassifyNumbers
{
//Fastest way, but uses memory
public static void UsingSieve(int bound, out int abundant, out int deficient, out int perfect) {
abundant = perfect = 0;
//For very large bounds, this array can get big.
int[] sum = new int[bound + 1];
for (int divisor = 1; divisor <= bound >> 1; divisor++)
for (int i = divisor << 1; i <= bound; i += divisor)
sum[i] += divisor;
for (int i = 1; i <= bound; i++) {
if (sum[i] > i) abundant++;
else if (sum[i] == i) perfect++;
}
deficient = bound - abundant - perfect;
}

//Slower, optimized, but doesn't use storage
public static void UsingOptiDivision(int bound, out int abundant, out int deficient, out int perfect) {
abundant = perfect = 0; int sum = 0;
for (int i = 2, d, r = 1; i <= bound; i++) {
if ((d = r * r - i) < 0) r++;
for (int x = 2; x < r; x++) if (i % x == 0) sum += x + i / x;
if (d == 0) sum += r;
switch (sum.CompareTo(i)) { case 0: perfect++; break; case 1: abundant++; break; }
sum = 1;
}
deficient = bound - abundant - perfect;
}

//Much slower, doesn't use storage and is un-optimized
public static void UsingDivision(int bound, out int abundant, out int deficient, out int perfect) {
abundant = perfect = 0;
for (int i = 2; i <= bound; i++) {
int sum = Enumerable.Range(1, (i + 1) / 2)
.Where(div => i % div == 0).Sum();
switch (sum.CompareTo(i)) {
case 0: perfect++; break;
case 1: abundant++; break;
}
}
deficient = bound - abundant - perfect;
}
}

Output @ Tio.run:

We see the second method is about 10 times slower than the first method, and the third method more than 120 times slower than the second method.

Abundant: 4953, Deficient: 15043, Perfect: 4  0.7277 ms
Abundant: 4953, Deficient: 15043, Perfect: 4  7.3458 ms
Abundant: 4953, Deficient: 15043, Perfect: 4  1048.9541 ms


## C++

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

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

int main( ) {
std::vector<int> deficients , perfects , abundants , divisors ;
for ( int n = 1 ; n < 20001 ; n++ ) {
divisors = findProperDivisors( n ) ;
int sum = std::accumulate( divisors.begin( ) , divisors.end( ) , 0 ) ;
if ( sum < n ) {
deficients.push_back( n ) ;
}
if ( sum == n ) {
perfects.push_back( n ) ;
}
if ( sum > n ) {
abundants.push_back( n ) ;
}
}
std::cout << "Deficient : " << deficients.size( ) << std::endl ;
std::cout << "Perfect   : " << perfects.size( ) << std::endl ;
std::cout << "Abundant  : " << abundants.size( ) << std::endl ;
return 0 ;
}

Output:
Deficient : 15043
Perfect   : 4
Abundant  : 4953


## Ceylon

shared void run() {

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

function classify(Integer int) => sum {0, *divisors(int)} <=> int;

value counts = (1..20k).map(classify).frequencies();

print("deficient: counts[smaller] else "none"");
print("perfect:   counts[equal] else "none"");
print("abundant:  counts[larger] else "none"");
}

Output:
deficient: 15043
perfect:   4
abundant:  4953

## Clojure

(defn pad-class
[n]
(let [divs (filter #(zero? (mod n %)) (range 1 n))
divs-sum (reduce + divs)]
(cond
(< divs-sum n) :deficient
(= divs-sum n) :perfect
(> divs-sum n) :abundant)))

(defn count-classes
[n]
{:perfect (count (filter #(= % :perfect) classes))
:abundant (count (filter #(= % :abundant) classes))
:deficient (count (filter #(= % :deficient) classes))}))


Example:

(count-classes 20000)
;=> {:perfect 4,
;    :abundant 4953,
;    :deficient 15043}


## CLU

% Generate proper divisors from 1 to max
proper_divisors = proc (max: int) returns (array[int])
divs: array[int] := array[int]$fill(1, max, 0) for i: int in int$from_to(1, max/2) do
for j: int in int$from_to_by(i*2, max, i) do divs[j] := divs[j] + i end end return(divs) end proper_divisors % Classify all the numbers for which we have divisors classify = proc (divs: array[int]) returns (int, int, int) def, per, ab: int def, per, ab := 0, 0, 0 for i: int in array[int]$indexes(divs) do
if     divs[i]<i then def := def + 1
elseif divs[i]=i then per := per + 1
elseif divs[i]>i then ab := ab + 1
end
end
return(def, per, ab)
end classify

% Find amount of deficient, perfect, and abundant numbers up to 20000
start_up = proc ()
max = 20000

po: stream := stream$primary_output() def, per, ab: int := classify(proper_divisors(max)) stream$putl(po, "Deficient: " || int$unparse(def)) stream$putl(po, "Perfect:   " || int$unparse(per)) stream$putl(po, "Abundant:  " || int$unparse(ab)) end start_up Output: Deficient: 15043 Perfect: 4 Abundant: 4953 ## Common Lisp (defun number-class (n) (let ((divisor-sum (sum-divisors n))) (cond ((< divisor-sum n) :deficient) ((= divisor-sum n) :perfect) ((> divisor-sum n) :abundant)))) (defun sum-divisors (n) (loop :for i :from 1 :to (/ n 2) :when (zerop (mod n i)) :sum i)) (defun classification () (loop :for n :from 1 :to 20000 :for class := (number-class n) :count (eq class :deficient) :into deficient :count (eq class :perfect) :into perfect :count (eq class :abundant) :into abundant :finally (return (values deficient perfect abundant))))  Output: CL-USER> (classification) 15043 4 4953 ## Cowgol include "cowgol.coh"; const MAXIMUM := 20000; var p: uint16[MAXIMUM+1]; var i: uint16; var j: uint16; MemZero(&p as [uint8], @bytesof p); i := 1; while i <= MAXIMUM/2 loop j := i+i; while j <= MAXIMUM loop p[j] := p[j]+i; j := j+i; end loop; i := i+1; end loop; var def: uint16 := 0; var per: uint16 := 0; var ab: uint16 := 0; i := 1; while i <= MAXIMUM loop if p[i]<i then def := def + 1; elseif p[i]==i then per := per + 1; else ab := ab + 1; end if; i := i + 1; end loop; print_i16(def); print(" deficient numbers.\n"); print_i16(per); print(" perfect numbers.\n"); print_i16(ab); print(" abundant numbers.\n"); Output: 15043 deficient numbers. 4 perfect numbers. 4953 abundant numbers. ## D void main() /*@safe*/ { import std.stdio, std.algorithm, std.range; static immutable properDivs = (in uint n) pure nothrow @safe /*@nogc*/ => iota(1, (n + 1) / 2 + 1).filter!(x => n % x == 0 && n != x); enum Class { deficient, perfect, abundant } static Class classify(in uint n) pure nothrow @safe /*@nogc*/ { immutable p = properDivs(n).sum; with (Class) return (p < n) ? deficient : ((p == n) ? perfect : abundant); } enum rangeMax = 20_000; //iota(1, 1 + rangeMax).map!classify.hashGroup.writeln; iota(1, 1 + rangeMax).map!classify.array.sort().group.writeln; }  Output: [Tuple!(Class, uint)(deficient, 15043), Tuple!(Class, uint)(perfect, 4), Tuple!(Class, uint)(abundant, 4953)] ## Delphi See #Pascal. ## Draco /* Fill a given array such that for each N, * P[n] is the sum of proper divisors of N */ proc nonrec propdivs([*] word p) void: word i, j, max; max := dim(p,1)-1; for i from 0 upto max do p[i] := 0 od; for i from 1 upto max/2 do for j from i*2 by i upto max do p[j] := p[j] + i od od corp proc nonrec main() void: word MAX = 20000; word def, per, ab, i; /* Find all required proper divisor sums */ [MAX+1] word p; propdivs(p); def := 0; per := 0; ab := 0; /* Check each number */ for i from 1 upto MAX do if p[i]<i then def := def + 1 elif p[i]=i then per := per + 1 elif p[i]>i then ab := ab + 1 fi od; writeln("Deficient: ", def:5); writeln("Perfect: ", per:5); writeln("Abundant: ", ab:5) corp Output: Deficient: 15043 Perfect: 4 Abundant: 4953 ## Dyalect Translation of: C# func sieve(bound) { var (a, d, p) = (0, 0, 0) var sum = Array.Empty(bound + 1, 0) for divisor in 1..(bound / 2) { var i = divisor + divisor while i <= bound { sum[i] += divisor i += divisor } } for i in 1..bound { if sum[i] < i { d += 1 } else if sum[i] > i { a += 1 } else { p += 1 } } (abundant: a, deficient: d, perfect: p) } func Iterator.Where(fn) { for x in this { if fn(x) { yield x } } } func Iterator.Sum() { var sum = 0 for x in this { sum += x } sum } func division(bound) { var (a, d, p) = (0, 0, 0) for i in 1..20000 { var sum = ( 1 .. ((i + 1) / 2) ) .Where(div => div != i && i % div == 0) .Sum() if sum < i { d += 1 } else if sum > i { a += 1 } else { p += 1 } } (abundant: a, deficient: d, perfect: p) } func out(res) { print("Abundant: \(res.abundant), Deficient: \(res.deficient), Perfect: \(res.perfect)"); } out( sieve(20000) ) out( division(20000) ) Output: Abundant: 4953, Deficient: 15043, Perfect: 4 Abundant: 4953, Deficient: 15043, Perfect: 4 ## EasyLang Translation of: AWK func sumprop num . if num < 2 return 0 . i = 2 sum = 1 root = sqrt num while i < root if num mod i = 0 sum += i + num / i . i += 1 . if num mod root = 0 sum += root . return sum . for j = 1 to 20000 sump = sumprop j if sump < j deficient += 1 elif sump = j perfect += 1 else abundant += 1 . . print "Perfect: " & perfect print "Abundant: " & abundant print "Deficient: " & deficient ## EchoLisp (lib 'math) ;; sum-divisors function (define-syntax-rule (++ a) (set! a (1+ a))) (define (abondance (N 20000)) (define-values (delta abondant deficient perfect) '(0 0 0 0)) (for ((n (in-range 1 (1+ N)))) (set! delta (- (sum-divisors n) n)) (cond ((< delta 0) (++ deficient)) ((> delta 0) (++ abondant)) (else (writeln 'perfect→ n) (++ perfect)))) (printf "In range 1.. %d" N) (for-each (lambda(x) (writeln x (eval x))) '(abondant deficient perfect))) (abondance) perfect→ 6 perfect→ 28 perfect→ 496 perfect→ 8128 In range 1.. 20000 abondant 4953 deficient 15043 perfect 4  ## Ela Translation of: Haskell open monad io number list divisors n = filter ((0 ==) << (n mod)) [1 .. (n div 2)] classOf n = compare (sum$ divisors n) n

do
let classes = map classOf [1 .. 20000]
let printRes w c = putStrLn $w ++ (show << length$ filter (== c) classes)
printRes "deficient: " LT
printRes "perfect:   " EQ
printRes "abundant:  " GT
Output:
deficient: 15043
perfect:   4
abundant:  4953

## Elena

Translation of: C#

ELENA 6.x :

import extensions;

classifyNumbers(int bound, ref int abundant, ref int deficient, ref int perfect)
{
int a := 0;
int d := 0;
int p := 0;
int[] sum := new int[](bound + 1);

for(int divisor := 1; divisor <= bound / 2; divisor += 1)
{
for(int i := divisor + divisor; i <= bound; i += divisor)
{
sum[i] := sum[i] + divisor
}
};

for(int i := 1; i <= bound; i += 1)
{
int t := sum[i];

if (sum[i]<i)
{
d += 1
}
else
{
if (sum[i]>i)
{
a += 1
}
else
{
p += 1
}
}
};

abundant := a;
deficient := d;
perfect := p
}

public program()
{
int abundant := 0;
int deficient := 0;
int perfect := 0;
classifyNumbers(20000, ref abundant, ref deficient, ref perfect);
console.printLine("Abundant: ",abundant,", Deficient: ",deficient,", Perfect: ",perfect)
}
Output:
Abundant: 4953, Deficient: 15043, Perfect: 4


## Elixir

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

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

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

Output:
Deficient: 15043   Perfect: 4   Abundant: 4953


## Erlang

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

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

divisors(1,N) ->
divisors(2,N,math:sqrt(N),[1]).

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

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

class(Limit) -> class(0,0,0,sumdivs(2),2,Limit).

class(D,P,A,_Sum,Acc,L) when Acc > L +1->
io:format("Deficient: ~w, Perfect: ~w, Abundant: ~w~n", [D,P,A]);

class(D,P,A,Sum,Acc,L) when Acc < Sum ->
class(D,P,A+1,sumdivs(Acc+1),Acc+1,L);
class(D,P,A,Sum,Acc,L) when Acc == Sum ->
class(D,P+1,A,sumdivs(Acc+1),Acc+1,L);
class(D,P,A,Sum,Acc,L) when Acc > Sum  ->
class(D+1,P,A,sumdivs(Acc+1),Acc+1,L).

Output:
24> c(properdivs).
{ok,properdivs}
25> properdivs:class(20000).
Deficient: 15043, Perfect: 4, Abundant: 4953
ok


The above divisors method was slightly rewritten to satisfy the observation below but preserve the different programming style. Now has comparable performance.

### Erlang 2

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

-module(proper_divisors).
-export([classify_range/2]).

classify_range(Start, Stop) ->
lists:foldl(fun (X, A) ->
Class = classify(X),
A#{Class => maps:get(Class, A, 0)+1} end,
#{},
lists:seq(Start, Stop)).

classify(N) ->
SumPD = lists:sum(proper_divisors(N)),
if
SumPD  <  N -> deficient;
SumPD =:= N -> perfect;
SumPD  >  N -> abundant
end.

proper_divisors(1) -> [];
proper_divisors(N) when N > 1, is_integer(N) ->
proper_divisors(2, math:sqrt(N), N, [1]).

proper_divisors(I, L, _, A) when I > L -> lists:sort(A);
proper_divisors(I, L, N, A) when N rem I =/= 0 ->
proper_divisors(I+1, L, N, A);
proper_divisors(I, L, N, A) when I * I =:= N ->
proper_divisors(I+1, L, N, [I|A]);
proper_divisors(I, L, N, A) ->
proper_divisors(I+1, L, N, [N div I, I|A]).

Output:
8>proper_divisors:classify_range(1,20000).
#{abundant => 4953,deficient => 15043,perfect => 4}


## F#

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


An immutable solution.

let deficient, perfect, abundant = 0,1,2

let classify n = ([1..n/2] |> List.filter (fun x->n % x = 0) |> List.sum) |> function
| x when x<n -> deficient | x when x>n -> abundant | _ -> perfect

let incClass xs n =
let cn = n |> classify
xs |> List.mapi (fun i x->if i=cn then x + 1 else x)

[1..20000]
|> List.fold incClass [0;0;0]
|> List.zip [ "deficient"; "perfect"; "abundant" ]
|> List.iter (fun (label, count) -> printfn "%s: %d" label count)


## Factor

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

Output:
Deficient: 15043
Perfect: 4
Abundant: 4953


## Forth

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

Output:
Deficient : 15043
Perfect   : 5
Abundant  : 4953

## Fortran

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

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

Output:

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

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


## FreeBASIC

' FreeBASIC v1.05.0 win64

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

Dim As Integer sum, deficient, perfect, abundant

For n As Integer = 1 To 20000
sum = SumProperDivisors(n)
If sum < n Then
deficient += 1
ElseIf sum = n Then
perfect += 1
Else
abundant += 1
EndIf
Next

Print "The classification of the numbers from 1 to 20,000 is as follows : "
Print
Print "Deficient = "; deficient
Print "Perfect   = "; perfect
Print "Abundant  = "; abundant
Print
Print "Press any key to exit the program"
Sleep
End
Output:
The classification of the numbers from 1 to 20,000 is as follows :

Deficient =  15043
Perfect   =  4
Abundant  =  4953


## Frink

d = new	dict
for n =	1 to 20000
{
s = sum[allFactors[n, true, false, true], 0]
rel = s <=> n
d.increment[rel, 1]
}

println["Deficient: " + d@(-1)]
println["Perfect:   " + d@0]
println["Abundant:  " + d@1]
Output:
Deficient: 15043
Perfect:   4
Abundant:  4953


## FutureBasic

local fn SumProperDivisors( number as long ) as long
long i, result, sum = 0

if number < 2 then exit fn = 0
for i = 1 to number / 2
if number mod i == 0 then sum += i
next
result = sum
end fn = result

void local fn NumberCategories( limit as long )
long i, sum, deficient = 0, perfect = 0, abundant = 0

for i = 1 to limit
sum = fn SumProperDivisors(i)
if sum  < i then deficient++ : continue
if sum == i then   perfect++ : continue
abundant++
next
printf @"\nClassification of integers from 1 to %ld is:\n", limit
printf @"Deficient = %ld\nPerfect   = %ld\nAbundant  = %ld", deficient, perfect, abundant
printf @"-----------------\nTotal     = %ld\n", deficient + perfect + abundant
end fn

CFTimeInterval t
t = fn CACurrentMediaTime
fn NumberCategories( 20000 )
printf @"Compute time: %.3f ms",(fn CACurrentMediaTime-t)*1000

HandleEvents
Output:
Classification of integers from 1 to 20000 is:

Deficient = 15043
Perfect   = 4
Abundant  = 4953
-----------------
Total     = 20000

Compute time: 1761.443 ms


## GFA Basic

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


Output is:

Number deficient 15043
Number perfect   4
Number abundant  4953


## Go

package main

import "fmt"

func pfacSum(i int) int {
sum := 0
for p := 1; p <= i/2; p++ {
if i%p == 0 {
sum += p
}
}
return sum
}

func main() {
var d, a, p = 0, 0, 0
for i := 1; i <= 20000; i++ {
j := pfacSum(i)
if j < i {
d++
} else if j == i {
p++
} else {
a++
}
}
fmt.Printf("There are %d deficient numbers between 1 and 20000\n", d)
fmt.Printf("There are %d abundant numbers  between 1 and 20000\n", a)
fmt.Printf("There are %d perfect numbers between 1 and 20000\n", p)
}

Output:
There are 15043 deficient numbers between 1 and 20000
There are 4953 abundant numbers  between 1 and 20000
There are 4 perfect numbers between 1 and 20000


## Groovy

##### Solution:

Uses the "factorize" closure from Factors of an integer

def dpaCalc = { factors ->
def n = factors.pop()
def fSum = factors.sum()
fSum < n
? 'deficient'
: fSum > n
? 'abundant'
: 'perfect'
}

(1..20000).inject([deficient:0, perfect:0, abundant:0]) { map, n ->
map[dpaCalc(factorize(n))]++
map
}
.each { e -> println e }

Output:
deficient=15043
perfect=4
abundant=4953

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

classOf :: (Integral a) => a -> Ordering
classOf n = compare (sum $divisors n) n main :: IO () main = do let classes = map classOf [1 .. 20000 :: Int] printRes w c = putStrLn$ w ++ (show . length $filter (== c) classes) printRes "deficient: " LT printRes "perfect: " EQ printRes "abundant: " GT  Output: deficient: 15043 perfect: 4 abundant: 4953 Or, a little faster and more directly, as a single fold: import Data.Numbers.Primes (primeFactors) import Data.List (group, sort) deficientPerfectAbundantCountsUpTo :: Int -> (Int, Int, Int) deficientPerfectAbundantCountsUpTo = foldr go (0, 0, 0) . enumFromTo 1 where go x (deficient, perfect, abundant) | divisorSum < x = (succ deficient, perfect, abundant) | divisorSum > x = (deficient, perfect, succ abundant) | otherwise = (deficient, succ perfect, abundant) where divisorSum = sum$ properDivisors x

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

main :: IO ()
main = print $deficientPerfectAbundantCountsUpTo 20000  Output: (15043,4,4953) ## J factors=: [: /:~@, */&>@{@((^ i.@>:)&.>/)@q:~&__ properDivisors=: factors -. ]  We can subtract the sum of a number's proper divisors from itself to classify the number:  (- +/@properDivisors&>) 1+i.10 1 1 2 1 4 0 6 1 5 2  Except, we are only concerned with the sign of this difference:  *(- +/@properDivisors&>) 1+i.30 1 1 1 1 1 0 1 1 1 1 1 _1 1 1 1 1 1 _1 1 _1 1 1 1 _1 1 1 1 0 1 _1  Also, we do not care about the individual classification but only about how many numbers fall in each category:  #/.~ *(- +/@properDivisors&>) 1+i.20000 15043 4 4953  So: 15043 deficient, 4 perfect and 4953 abundant numbers in this range. How do we know which is which? We look at the unique values (which are arranged by their first appearance, scanning the list left to right):  ~. *(- +/@properDivisors&>) 1+i.20000 1 0 _1  The sign of the difference is negative for the abundant case - where the sum is greater than the number. And we rely on order being preserved in sequences (this happens to be a fundamental property of computer memory, also). ## Java Works with: Java version 8 import java.util.stream.LongStream; public class NumberClassifications { public static void main(String[] args) { int deficient = 0; int perfect = 0; int abundant = 0; for (long i = 1; i <= 20_000; i++) { long sum = properDivsSum(i); if (sum < i) deficient++; else if (sum == i) perfect++; else abundant++; } System.out.println("Deficient: " + deficient); System.out.println("Perfect: " + perfect); System.out.println("Abundant: " + abundant); } public static long properDivsSum(long n) { return LongStream.rangeClosed(1, (n + 1) / 2).filter(i -> n != i && n % i == 0).sum(); } }  Deficient: 15043 Perfect: 4 Abundant: 4953 ## JavaScript ### ES5 for (var dpa=[1,0,0], n=2; n<=20000; n+=1) { for (var ds=0, d=1, e=n/2+1; d<e; d+=1) if (n%d==0) ds+=d dpa[ds<n ? 0 : ds==n ? 1 : 2]+=1 } document.write('Deficient:',dpa[0], ', Perfect:',dpa[1], ', Abundant:',dpa[2], '<br>' )  Or: for (var dpa=[1,0,0], n=2; n<=20000; n+=1) { for (var ds=1, d=2, e=Math.sqrt(n); d<e; d+=1) if (n%d==0) ds+=d+n/d if (n%e==0) ds+=e dpa[ds<n ? 0 : ds==n ? 1 : 2]+=1 } document.write('Deficient:',dpa[0], ', Perfect:',dpa[1], ', Abundant:',dpa[2], '<br>' )  Or: function primes(t) { var ps = {2:true, 3:true} next: for (var n=5, i=2; n<=t; n+=i, i=6-i) { var s = Math.sqrt( n ) for ( var p in ps ) { if ( p > s ) break if ( n % p ) continue continue next } ps[n] = true } return ps } function factorize(f, t) { var cs = {}, ps = primes(t) for (var n=f; n<=t; n++) if (!ps[n]) cs[n] = factors(n) return cs function factors(n) { for ( var p in ps ) if ( n % p == 0 ) break var ts = {} ts[p] = 1 if ( ps[n /= p] ) { if ( !ts[n]++ ) ts[n]=1 } else { var fs = cs[n] if ( !fs ) fs = cs[n] = factors(n) for ( var e in fs ) ts[e] = fs[e] + (e==p) } return ts } } function pContrib(p, e) { for (var pc=1, n=1, i=1; i<=e; i+=1) pc+=n*=p; return pc } for (var dpa=[1,0,0], t=20000, cs=factorize(2,t), n=2; n<=t; n+=1) { var ds=1, fs=cs[n] if (fs) { for (var p in fs) ds *= pContrib(p, fs[p]) ds -= n } dpa[ds<n ? 0 : ds==n ? 1 : 2]+=1 } document.write('Deficient:',dpa[0], ', Perfect:',dpa[1], ', Abundant:',dpa[2], '<br>' )  Output: Deficient:15043, Perfect:4, Abundant:4953 ### ES6 Translation of: Haskell (() => { 'use strict'; const // divisors :: (Integral a) => a -> [a] divisors = n => range(1, Math.floor(n / 2)) .filter(x => n % x === 0), // classOf :: (Integral a) => a -> Ordering classOf = n => compare(divisors(n) .reduce((a, b) => a + b, 0), n), classTypes = { deficient: -1, perfect: 0, abundant: 1 }; // GENERIC FUNCTIONS const // compare :: Ord a => a -> a -> Ordering compare = (a, b) => a < b ? -1 : (a > b ? 1 : 0), // range :: Int -> Int -> [Int] range = (m, n) => Array.from({ length: Math.floor(n - m) + 1 }, (_, i) => m + i); // TEST // classes :: [Ordering] const classes = range(1, 20000) .map(classOf); return Object.keys(classTypes) .map(k => k + ": " + classes .filter(x => x === classTypes[k]) .length.toString()) .join('\n'); })();  Output: deficient: 15043 perfect: 4 abundant: 4953 Translation of: Lua // classify the numbers 1 : 20 000 as abudant, deficient or perfect "use strict" let abundantCount = 0 let deficientCount = 0 let perfectCount = 0 const maxNumber = 20000 // construct a table of the proper divisor sums let pds = [] pds[ 1 ] = 0 for( let i = 2; i <= maxNumber; i ++ ){ pds[ i ] = 1 } for( let i = 2; i <= maxNumber; i ++ ) { for( let j = i + i; j <= maxNumber; j += i ){ pds[ j ] += i } } // classify the numbers for( let n = 1; n <= maxNumber; n ++ ) { if( pds[ n ] < n ) { deficientCount ++ } else if( pds[ n ] == n ) { perfectCount ++ } else // pds[ n ] > n { abundantCount ++ } } console.log( "abundant " + abundantCount.toString() ) console.log( "deficient " + deficientCount.toString() ) console.log( "perfect " + perfectCount.toString() )  Output: abundant 4953 deficient 15043 perfect 4  ## jq Works with: jq version 1.4 The definition of proper_divisors is taken from Proper_divisors#jq: # 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;

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

def classify:
. as $n | sum(proper_divisors) | if . <$n then "deficient" elif . == $n then "perfect" else "abundant" end; reduce (range(1; 20001) | classify) as$c ({}; .[$c] += 1 ) Output: $ jq -n -c -f AbundantDeficientPerfect.jq
{"deficient":15043,"perfect":4,"abundant":4953}


## Jsish

From Javascript ES5 entry.

/* Classify Deficient, Perfect and Abdundant integers */
function classifyDPA(stop:number, start:number=0, step:number=1):array {
var dpa = [1, 0, 0];
for (var n=start; n<=stop; n+=step) {
for (var ds=0, d=1, e=n/2+1; d<e; d+=1) if (n%d == 0) ds += d;
dpa[ds < n ? 0 : ds==n ? 1 : 2] += 1;
}
return dpa;
}

var dpa = classifyDPA(20000, 2);
printf('Deficient: %d, Perfect: %d, Abundant: %d\n', dpa[0], dpa[1], dpa[2]);

Output:
prompt$jsish classifyDPA.jsi Deficient: 15043, Perfect: 4, Abundant: 4953 ## Julia This post was created with Julia version 0.3.6. The code uses no exotic features and should work for a wide range of Julia versions. The Math A natural number can be written as a product of powers of its prime factors, ${\displaystyle \prod _{i}p_{i}^{a_{i}}}$. Handily Julia has the factor function, which provides these parameters. The sum of n's divisors (n inclusive) is ${\displaystyle \prod _{i}{\frac {p_{i}^{a_{i}+1}-1}{p_{i}-1}}=\prod _{i}p_{i}^{a_{i}}+p_{i}^{a_{i}-1}+\cdots +p_{i}+1}$. Functions divisorsum calculates the sum of aliquot divisors. It uses pcontrib to calculate the contribution of each prime factor. function pcontrib(p::Int64, a::Int64) n = one(p) pcon = one(p) for i in 1:a n *= p pcon += n end return pcon end function divisorsum(n::Int64) dsum = one(n) for (p, a) in factor(n) dsum *= pcontrib(p, a) end dsum -= n end  Perhaps pcontrib could be made more efficient by caching results to avoid repeated calculations. Main Use a three element array, iclass, rather than three separate variables to tally the classifications. Take advantage of the fact that the sign of divisorsum(n) - n depends upon its class to increment iclass. 1 is a difficult case, it is deficient by convention, so I manually add its contribution and start the accumulation with 2. All primes are deficient, so I test for those and tally accordingly, bypassing divisorsum. const L = 2*10^4 iclasslabel = ["Deficient", "Perfect", "Abundant"] iclass = zeros(Int64, 3) iclass[1] = one(Int64) #by convention 1 is deficient for n in 2:L if isprime(n) iclass[1] += 1 else iclass[sign(divisorsum(n)-n)+2] += 1 end end println("Classification of integers from 1 to ", L) for i in 1:3 println(" ", iclasslabel[i], ", ", iclass[i]) end  Output:    Classification of integers from 1 to 20000 Deficient, 15043 Perfect, 4 Abundant, 4953  ### Using Primes versions >= 0.5.4 Recent revisions of the Primes package include a divisors() which returns divisors of n including 1 and n. using Primes """ Return tuple of (perfect, abundant, deficient) counts from 1 up to nmax """ function per_abu_def_classify(nmax::Int) results = [0, 0, 0] for n in 1:nmax results[sign(sum(divisors(n)) - 2 * n) + 2] += 1 end return (perfect, abundant, deficient) = results end let MAX = 20_000 NPE, NAB, NDE = per_abu_def_classify(MAX) println("$NPE perfect, $NAB abundant, and$NDE deficient numbers in 1:$MAX.") end Output: 4 perfect, 4953 abundant, and 15043 deficient numbers in 1:20000. ## K Works with: Kona /Classification of numbers into abundant, perfect and deficient / numclass.k /return 0,1 or -1 if perfect or abundant or deficient respectively numclass: {s:(+/&~x!'!1+x)-x; :[s>x;:1;:[s<x;:-1;:0]]} /classify numbers from 1 to 20000 into respective groups c: =numclass' 1+!20000 /print statistics 0: ,"Deficient = ",$(#c[0])
0: ,"Perfect   = ", $(#c[1]) 0: ,"Abundant = ",$(#c[2])


Works with: ngn/k
/Classification of numbers into abundant, perfect and deficient
/ numclass.k

/return 0,1 or -1 if perfect or abundant or deficient respectively
numclass: {s:(+/&~(!1+x)!\:x)-x; $[s>x;:1;$[s<x;:-1;:0]]}
/classify numbers from 1 to 20000 into respective groups
c: =numclass' 1+!20000
/print statistics
0: ,"Deficient = ", $(#c[-1]) 0: ,"Perfect = ",$(#c[0])
0: ,"Abundant  = ", $(#c[1])  (indentation optional, used to emphasize lines which are not comment lines) Output: Deficient = 15043 Perfect = 4 Abundant = 4953  ## Kotlin Translation of: FreeBASIC // version 1.1 fun sumProperDivisors(n: Int) = if (n < 2) 0 else (1..n / 2).filter { (n % it) == 0 }.sum() fun main(args: Array<String>) { var sum: Int var deficient = 0 var perfect = 0 var abundant = 0 for (n in 1..20000) { sum = sumProperDivisors(n) when { sum < n -> deficient++ sum == n -> perfect++ sum > n -> abundant++ } } println("The classification of the numbers from 1 to 20,000 is as follows:\n") println("Deficient =$deficient")
println("Perfect   = $perfect") println("Abundant =$abundant")
}

Output:
The classification of the numbers from 1 to 20,000 is as follows:

Deficient = 15043
Perfect   = 4
Abundant  = 4953


## Liberty BASIC

print "ROSETTA CODE - Abundant, deficient and perfect number classifications"
print
for x=1 to 20000
x$=NumberClassification$(x)
select case x$case "deficient": de=de+1 case "perfect": pe=pe+1: print x; " is a perfect number" case "abundant": ab=ab+1 end select select case x case 2000: print "Checking the number classifications of 20,000 integers..." case 4000: print "Please be patient." case 7000: print "7,000" case 10000: print "10,000" case 12000: print "12,000" case 14000: print "14,000" case 16000: print "16,000" case 18000: print "18,000" case 19000: print "Almost done..." end select next x print "Deficient numbers = "; de print "Perfect numbers = "; pe print "Abundant numbers = "; ab print "TOTAL = "; pe+de+ab [Quit] print "Program complete." end function NumberClassification$(n)
x=ProperDivisorCount(n)
for y=1 to x
PDtotal=PDtotal+ProperDivisor(y)
next y
if PDtotal=n then NumberClassification$="perfect": exit function if PDtotal<n then NumberClassification$="deficient": exit function
if PDtotal>n then NumberClassification$="abundant": exit function end function function ProperDivisorCount(n) n=abs(int(n)): if n=0 or n>20000 then exit function dim ProperDivisor(100) for y=2 to n if (n mod y)=0 then ProperDivisorCount=ProperDivisorCount+1 ProperDivisor(ProperDivisorCount)=n/y end if next y end function Output: ROSETTA CODE - Abundant, deficient and perfect number classifications 6 is a perfect number 28 is a perfect number 496 is a perfect number Checking the number classifications of 20,000 integers... Please be patient. 7,000 8128 is a perfect number 10,000 12,000 14,000 16,000 18,000 Almost done... Deficient numbers = 15043 Perfect numbers = 4 Abundant numbers = 4953 TOTAL = 20000 Program complete.  ## Lua ### Summing the factors using modulo/division function sumDivs (n) if n < 2 then return 0 end local sum, sr = 1, math.sqrt(n) for d = 2, sr do if n % d == 0 then sum = sum + d if d ~= sr then sum = sum + n / d end end end return sum end local a, d, p, Pn = 0, 0, 0 for n = 1, 20000 do Pn = sumDivs(n) if Pn > n then a = a + 1 end if Pn < n then d = d + 1 end if Pn == n then p = p + 1 end end print("Abundant:", a) print("Deficient:", d) print("Perfect:", p)  Output: Abundant: 4953 Deficient: 15043 Perfect: 4 ### Summing the factors using a table Translation of: ALGOL 68 do -- classify the numbers 1 : 20 000 as abudant, deficient or perfect local abundantCount = 0 local deficientCount = 0 local perfectCount = 0 local maxNumber = 20000 -- construct a table of the proper divisor sums local pds = {} pds[ 1 ] = 0 for i = 2, maxNumber do pds[ i ] = 1 end for i = 2, maxNumber do for j = i + i, maxNumber, i do pds[ j ] = pds[ j ] + i end end -- classify the numbers for n = 1, maxNumber do local pdSum = pds[ n ] if pdSum < n then deficientCount = deficientCount + 1 elseif pdSum == n then perfectCount = perfectCount + 1 else -- pdSum > n abundantCount = abundantCount + 1 end end io.write( "abundant ", abundantCount, "\n" ) io.write( "deficient ", deficientCount, "\n" ) io.write( "perfect ", perfectCount, "\n" ) end  Output: abundant 4953 deficient 15043 perfect 4  ## MAD  NORMAL MODE IS INTEGER DIMENSION P(20000) MAX = 20000 THROUGH INIT, FOR I=1, 1, I.G.MAX INIT P(I) = 0 THROUGH CALC, FOR I=1, 1, I.G.MAX/2 THROUGH CALC, FOR J=I+I, I, J.G.MAX CALC P(J) = P(J)+I DEF = 0 PER = 0 AB = 0 THROUGH CLSFY, FOR N=1, 1, N.G.MAX WHENEVER P(N).L.N, DEF = DEF+1 WHENEVER P(N).E.N, PER = PER+1 CLSFY WHENEVER P(N).G.N, AB = AB+1 PRINT FORMAT FDEF,DEF PRINT FORMAT FPER,PER PRINT FORMAT FAB,AB VECTOR VALUES FDEF =$I5,S1,9HDEFICIENT*$VECTOR VALUES FPER =$I5,S1,7HPERFECT*$VECTOR VALUES FAB =$I5,S1,8HABUNDANT*$END OF PROGRAM Output: 15043 DEFICIENT 4 PERFECT 4953 ABUNDANT ## Maple  classify_number := proc(n::posint); if evalb(NumberTheory:-SumOfDivisors(n) < 2*n) then return "Deficient"; elif evalb(NumberTheory:-SumOfDivisors(n) = 2*n) then return "Perfect"; else return "Abundant"; end if; end proc: classify_sequence := proc(k::posint) local num_list; num_list := map(classify_number, [seq(1..k)]); return Statistics:-Tally(num_list) end proc: Output: ["Perfect" = 4, "Abundant" = 4953, "Deficient" = 15043] ## Mathematica / Wolfram Language classify[n_Integer] := Sign[Total[Most@Divisors@n] - n] StringJoin[ Flatten[Tally[ Table[classify[n], {n, 20000}]] /. {-1 -> "deficient: ", 0 -> " perfect: ", 1 -> " abundant: "}] /. n_Integer :> ToString[n]]  Output: deficient: 15043 perfect: 4 abundant: 4953 ## MatLab abundant=0; deficient=0; perfect=0; p=[]; for N=2:20000 K=1:ceil(N/2); D=K(~(rem(N, K))); sD=sum(D); if sD<N deficient=deficient+1; elseif sD==N perfect=perfect+1; else abundant=abundant+1; end end disp(table([deficient;perfect;abundant],'RowNames',{'Deficient','Perfect','Abundant'},'VariableNames',{'Quantities'}))  Output:  Quantities __________ Deficient 15042 Perfect 4 Abundant 4953  ## Maxima /* Given a number it returns wether it is perfect, deficient or abundant */ number_class(n):=if divsum(n)-n=n then "perfect" else if divsum(n)-n<n then "deficient" else if divsum(n)-n>n then "abundant"$

/* Function that displays the number of each kind below n */
classification_count(n):=block(makelist(number_class(i),i,1,n),
[[length(sublist(%%,lambda([x],x="deficient")))," deficient"],[length(sublist(%%,lambda([x],x="perfect")))," perfect"],[length(sublist(%%,lambda([x],x="abundant")))," abundant"]])$/* Test case */ classification_count(20000);  Output: [[15043," deficient"],[4," perfect"],[4953," abundant"]]  ## MiniScript Translation of: Lua – Summing the factors using a table // classify the numbers 1 : 20 000 as abudant, deficient or perfect abundantCount = 0 deficientCount = 0 perfectCount = 0 maxNumber = 20000 // construct a table of the proper divisor sums pds = [0] * ( maxNumber + 1 ) pds[ 1 ] = 0 for i in range( 2, maxNumber ) pds[ i ] = 1 end for for i in range( 2, maxNumber ) for j in range( i + i, maxNumber, i ) pds[ j ] = pds[ j ] + i end for end for // classify the numbers for n in range( 1, maxNumber ) pdSum = pds[ n ] if pdSum < n then deficientCount = deficientCount + 1 else if pdSum == n then perfectCount = perfectCount + 1 else // pdSum > n abundantCount = abundantCount + 1 end if end for print "abundant " + abundantCount print "deficient " + deficientCount print "perfect " + perfectCount  Output: abundant 4953 deficient 15043 perfect 4  ## ML ### mLite fun proper (number, count, limit, remainder, results) where (count > limit) = rev results | (number, count, limit, remainder, results) = proper (number, count + 1, limit, number rem (count+1), if remainder = 0 then count :: results else results) | number = (proper (number, 1, number div 2, 0, [])) ; fun is_abundant number = number < (fold (op +, 0)  proper number); fun is_deficient number = number > (fold (op +, 0)  proper number); fun is_perfect number = number = (fold (op +, 0)  proper number); val one_to_20000 = iota 20000; print "Abundant numbers between 1 and 20000: "; println  fold (op +, 0)  map ((fn n = if n then 1 else 0) o is_abundant) one_to_20000; print "Deficient numbers between 1 and 20000: "; println  fold (op +, 0)  map ((fn n = if n then 1 else 0) o is_deficient) one_to_20000; print "Perfect numbers between 1 and 20000: "; println  fold (op +, 0)  map ((fn n = if n then 1 else 0) o is_perfect) one_to_20000;  Output Abundant numbers between 1 and 20000: 4953 Deficient numbers between 1 and 20000: 15043 Perfect numbers between 1 and 20000: 4  ## Modula-2 MODULE ADP; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,WriteLn,ReadChar; PROCEDURE ProperDivisorSum(n : INTEGER) : INTEGER; VAR i,sum : INTEGER; BEGIN sum := 0; IF n<2 THEN RETURN 0 END; FOR i:=1 TO (n DIV 2) DO IF n MOD i = 0 THEN INC(sum,i) END END; RETURN sum END ProperDivisorSum; VAR buf : ARRAY[0..63] OF CHAR; n : INTEGER; d,p,a : INTEGER = 0; sum : INTEGER; BEGIN FOR n:=1 TO 20000 DO sum := ProperDivisorSum(n); IF sum<n THEN INC(d) ELSIF sum=n THEN INC(p) ELSIF sum>n THEN INC(a) END END; WriteString("The classification of the numbers from 1 to 20,000 is as follows:"); WriteLn; FormatString("Deficient = %i\n", buf, d); WriteString(buf); FormatString("Perfect = %i\n", buf, p); WriteString(buf); FormatString("Abundant = %i\n", buf, a); WriteString(buf); ReadChar END ADP.  ## NewLisp ;;; The list (1 .. n-1) of integers is generated ;;; then each non-divisor of n is replaced by 0 ;;; finally all these numbers are summed. ;;; fn defines an anonymous function inline. (define (sum-divisors n) (apply + (map (fn (x) (if (> (% n x) 0) 0 x)) (sequence 1 (- n 1))))) ; ;;; Returns the symbols -, p or + for deficient, perfect or abundant numbers respectively. (define (number-type n) (let (sum (sum-divisors n)) (if (< sum n) '- (= sum n) 'p true '+))) ; ;;; Tallies the types from 2 to n. (define (count-types n) (count '(- p +) (map number-type (sequence 2 n)))) ; ;;; Running: (println (count-types 20000))  Output: (15042 4 4953)  ## Nim proc sumProperDivisors(number: int) : int = if number < 2 : return 0 for i in 1 .. number div 2 : if number mod i == 0 : result += i var sum : int deficient = 0 perfect = 0 abundant = 0 for n in 1 .. 20000 : sum = sumProperDivisors(n) if sum < n : inc(deficient) elif sum == n : inc(perfect) else : inc(abundant) echo "The classification of the numbers between 1 and 20,000 is as follows :\n" echo " Deficient = " , deficient echo " Perfect = " , perfect echo " Abundant = " , abundant  Output: The classification of the numbers between 1 and 20,000 is as follows : Deficient = 15043 Perfect = 4 Abundant = 4953  ## Oforth import: mapping Integer method: properDivs -- [] self 2 / seq filter( #[ self swap mod 0 == ] ) ; : numberClasses | i deficient perfect s | 0 0 ->deficient ->perfect 0 20000 loop: i [ 0 #+ i properDivs apply ->s s i < ifTrue: [ deficient 1+ ->deficient continue ] s i == ifTrue: [ perfect 1+ ->perfect continue ] 1+ ] "Deficients :" . deficient .cr "Perfects :" . perfect .cr "Abundant :" . .cr ; Output: numberClasses Deficients : 15043 Perfects : 4 Abundant : 4953  ## PARI/GP classify(k)= { my(v=[0,0,0],t); for(n=1,k, t=sigma(n,-1); if(t<2,v[1]++,t>2,v[3]++,v[2]++) ); v; } classify(20000) Output: %1 = [15043, 4, 4953] ## Pascal ### Free Pascal Library: PrimTrial search for "UNIT for prime decomposition". program KindOfN; //[deficient,perfect,abundant] {$IFDEF FPC}
{$MODE DELPHI}{$OPTIMIZATION ON,ALL}{$COPERATORS ON}{$CODEALIGN proc=16}
{$ENDIF} {$IFDEF WINDOWS} {$APPTYPE CONSOLE}{$ENDIF}
uses
sysutils,PrimeDecomp // limited to 1.2e11
{$IFDEF WINDOWS},Windows{$ENDIF}
;
//alternative copy and paste PrimeDecomp.inc for TIO.RUN
{$I PrimeDecomp.inc} type tKindIdx = 0..2;//[deficient,perfect,abundant]; tKind = array[tKindIdx] of Uint64; procedure GetKind(Limit:Uint64); var pPrimeDecomp :tpPrimeFac; SumOfKind : tKind; n: NativeUInt; c: NativeInt; T0:Int64; Begin writeln('Limit: ',LIMIT); T0 := GetTickCount64; fillchar(SumOfKind,SizeOf(SumOfKind),#0); n := 1; Init_Sieve(n); repeat pPrimeDecomp:= GetNextPrimeDecomp; c := pPrimeDecomp^.pfSumOfDivs-2*n; c := ORD(c>0)-ORD(c<0)+1;//sgn(c)+1 inc(SumOfKind[c]); inc(n); until n > LIMIT; T0 := GetTickCount64-T0; writeln('deficient: ',SumOfKind[0]); writeln('abundant: ',SumOfKind[2]); writeln('perfect: ',SumOfKind[1]); writeln('runtime ',T0/1000:0:3,' s'); writeln; end; Begin InitSmallPrimes; //using PrimeDecomp.inc GetKind(20000); GetKind(10*1000*1000); GetKind(524*1000*1000); end.  @TIO.RUN: Limit: 20000 deficient: 15043 abundant: 4953 perfect: 4 runtime 0.003 s Limit: 1000000 deficient: 752451 abundant: 247545 perfect: 4 runtime 0.052 s Limit: 524000000 deficient: 394250308 abundant: 129749687 perfect: 5 runtime 32.987 s Real time: 33.203 s User time: 32.881 s Sys. time: 0.048 s CPU share: 99.17 %  ## Perl ### Using a module Library: ntheory Use the <=> operator to return a comparison of -1, 0, or 1, which classifies the results. 1 is classified as a deficient number, 6 is a perfect number, 12 is an abundant number. As per task spec, also showing the totals for the first 20,000 numbers. use ntheory qw/divisor_sum/; my @type = <Perfect Abundant Deficient>; say join "\n", map { sprintf "%2d %s",$_, $type[divisor_sum($_)-$_ <=>$_] } 1..12;
my %h;
$h{divisor_sum($_)-$_ <=>$_}++ for 1..20000;
say "Perfect: $h{0} Deficient:$h{-1}    Abundant: $h{1}";  Output:  1 Deficient 2 Deficient 3 Deficient 4 Deficient 5 Deficient 6 Perfect 7 Deficient 8 Deficient 9 Deficient 10 Deficient 11 Deficient 12 Abundant Perfect: 4 Deficient: 15043 Abundant: 4953 ### Not using a module Everything as above, but done more slowly with div_sum providing sum of proper divisors. sub div_sum { my($n) = @_;
my $sum = 0; map {$sum += $_ unless$n % $_ } 1 ..$n-1;
$sum; } my @type = <Perfect Abundant Deficient>; say join "\n", map { sprintf "%2d %s",$_, $type[div_sum($_) <=> $_] } 1..12; my %h;$h{div_sum($_) <=>$_}++ for 1..20000;
say "Perfect: $h{0} Deficient:$h{-1}    Abundant: $h{1}";  ## Phix integer deficient=0, perfect=0, abundant=0, N for i=1 to 20000 do N = sum(factors(i))+(i!=1) if N=i then perfect += 1 elsif N<i then deficient += 1 else abundant += 1 end if end for printf(1,"deficient:%d, perfect:%d, abundant:%d\n",{deficient, perfect, abundant})  Output: deficient:15043, perfect:4, abundant:4953  ## Picat go => Classes = new_map([deficient=0,perfect=0,abundant=0]), foreach(N in 1..20_000) C = classify(N), Classes.put(C,Classes.get(C)+1) end, println(Classes), nl. % Classify a number N classify(N) = Class => S = sum_divisors(N), if S < N then Class1 = deficient elseif S = N then Class1 = perfect elseif S > N then Class1 = abundant end, Class = Class1. % Alternative (slightly slower) approach. classify2(N,S) = C, S < N => C = deficient. classify2(N,S) = C, S == N => C = perfect. classify2(N,S) = C, S > N => C = abundant. % Sum of divisors sum_divisors(N) = Sum => sum_divisors(2,N,cond(N>1,1,0),Sum). % Part 0: base case sum_divisors(I,N,Sum0,Sum), I > floor(sqrt(N)) => Sum = Sum0. % Part 1: I is a divisor of N sum_divisors(I,N,Sum0,Sum), N mod I == 0 => Sum1 = Sum0 + I, (I != N div I -> Sum2 = Sum1 + N div I ; Sum2 = Sum1 ), sum_divisors(I+1,N,Sum2,Sum). % Part 2: I is not a divisor of N. sum_divisors(I,N,Sum0,Sum) => sum_divisors(I+1,N,Sum0,Sum). Output: (map)[perfect = 4,deficient = 15043,abundant = 4953] ## PicoLisp (de accud (Var Key) (if (assoc Key (val Var)) (con @ (inc (cdr @))) (push Var (cons Key 1)) ) Key ) (de **sum (L) (let S 1 (for I (cdr L) (inc 'S (** (car L) I)) ) S ) ) (de factor-sum (N) (if (=1 N) 0 (let (R NIL D 2 L (1 2 2 . (4 2 4 2 4 6 2 6 .)) M (sqrt N) N1 N S 1 ) (while (>= M D) (if (=0 (% N1 D)) (setq M (sqrt (setq N1 (/ N1 (accud 'R D)))) ) (inc 'D (pop 'L)) ) ) (accud 'R N1) (for I R (setq S (* S (**sum I))) ) (- S N) ) ) ) (bench (let (A 0 D 0 P 0 ) (for I 20000 (setq @@ (factor-sum I)) (cond ((< @@ I) (inc 'D)) ((= @@ I) (inc 'P)) ((> @@ I) (inc 'A)) ) ) (println D P A) ) ) (bye) Output: 15043 4 4953 0.110 sec  ## PL/I *process source xref; apd: Proc Options(main); p9a=time(); Dcl (p9a,p9b) Pic'(9)9'; Dcl cnt(3) Bin Fixed(31) Init((3)0); Dcl x Bin Fixed(31); Dcl pd(300) Bin Fixed(31); Dcl sumpd Bin Fixed(31); Dcl npd Bin Fixed(31); Do x=1 To 20000; Call proper_divisors(x,pd,npd); sumpd=sum(pd,npd); Select; When(x<sumpd) cnt(1)+=1; /* abundant */ When(x=sumpd) cnt(2)+=1; /* perfect */ Otherwise cnt(3)+=1; /* deficient */ End; End; Put Edit('In the range 1 - 20000')(Skip,a); Put Edit(cnt(1),' numbers are abundant ')(Skip,f(5),a); Put Edit(cnt(2),' numbers are perfect ')(Skip,f(5),a); Put Edit(cnt(3),' numbers are deficient')(Skip,f(5),a); p9b=time(); Put Edit((p9b-p9a)/1000,' seconds elapsed')(Skip,f(6,3),a); Return; proper_divisors: Proc(n,pd,npd); Dcl (n,pd(300),npd) Bin Fixed(31); Dcl (d,delta) Bin Fixed(31); npd=0; If n>1 Then Do; If mod(n,2)=1 Then /* odd number */ delta=2; Else /* even number */ delta=1; Do d=1 To n/2 By delta; If mod(n,d)=0 Then Do; npd+=1; pd(npd)=d; End; End; End; End; sum: Proc(pd,npd) Returns(Bin Fixed(31)); Dcl (pd(300),npd) Bin Fixed(31); Dcl sum Bin Fixed(31) Init(0); Dcl i Bin Fixed(31); Do i=1 To npd; sum+=pd(i); End; Return(sum); End; End; Output: In the range 1 - 20000 4953 numbers are abundant 4 numbers are perfect 15043 numbers are deficient 0.560 seconds elapsed  ## PL/M 100H: BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS; EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT; PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT; PRINT$NUMBER: PROCEDURE (N);
DECLARE S (6) BYTE INITIAL ('.....$'); DECLARE (N, P) ADDRESS, C BASED P BYTE; P = .S(5); DIGIT: P = P - 1; C = N MOD 10 + '0'; N = N / 10; IF N > 0 THEN GO TO DIGIT; CALL PRINT(P); END PRINT$NUMBER;

DECLARE LIMIT LITERALLY '20$000'; DECLARE (PBASE, P BASED PBASE) ADDRESS; DECLARE (I, J) ADDRESS; PBASE = .MEMORY; DO I=0 TO LIMIT; P(I)=0; END; DO I=1 TO LIMIT/2; DO J=I+I TO LIMIT BY I; P(J) = P(J)+I; END; END; DECLARE (DEF, PER, AB) ADDRESS INITIAL (0, 0, 0); DO I=1 TO LIMIT; IF P(I)<I THEN DEF = DEF+1; ELSE IF P(I)=I THEN PER = PER+1; ELSE IF P(I)>I THEN AB = AB+1; END; CALL PRINT$NUMBER(DEF);
CALL PRINT(.(' DEFICIENT',13,10,'$')); CALL PRINT$NUMBER(PER);
CALL PRINT(.(' PERFECT',13,10,'$')); CALL PRINT$NUMBER(AB);
CALL PRINT(.(' ABUNDANT',13,10,'$')); CALL EXIT; EOF Output: 15043 DEFICIENT 4 PERFECT 4953 ABUNDANT ## PowerShell Works with: PowerShell version 2 function Get-ProperDivisorSum ( [int]$N )
{
If ( $N -lt 2 ) { return 0 }$Sum = 1
If ( $N -gt 3 ) {$SqrtN = [math]::Sqrt( $N ) ForEach ($Divisor in 2..$SqrtN ) { If ($N % $Divisor -eq 0 ) {$Sum += $Divisor +$N / $Divisor } } If ($N % $SqrtN -eq 0 ) {$Sum -= $SqrtN } } return$Sum
}

$Deficient =$Perfect = $Abundant = 0 ForEach ($N in 1..20000 )
{
Switch ( [math]::Sign( ( Get-ProperDivisorSum $N ) -$N ) )
{
-1 { $Deficient++ } 0 {$Perfect++   }
1 { $Abundant++ } } } "Deficient:$Deficient"
"Perfect  : $Perfect" "Abundant :$Abundant"

Output:
Deficient: 15043
Perfect  : 4
Abundant : 4953


### As a single function

Using the Get-ProperDivisorSum as a helper function in an advanced function:

function Get-NumberClassification
{
[CmdletBinding()]
[OutputType([PSCustomObject])]
Param
(
[Parameter(Mandatory=$true, ValueFromPipeline=$true,
ValueFromPipelineByPropertyName=$true, Position=0)] [int]$Number
)

Begin
{
function Get-ProperDivisorSum ([int]$Number) { if ($Number -lt 2) {return 0}

$sum = 1 if ($Number -gt 3)
{
$sqrtNumber = [Math]::Sqrt($Number)

foreach ($divisor in 2..$sqrtNumber)
{
if ($Number %$divisor -eq 0) {$sum +=$divisor + $Number /$divisor}
}

if ($Number %$sqrtNumber -eq 0) {$sum -=$sqrtNumber}
}

$sum } [System.Collections.ArrayList]$numbers = @()
}
Process
{
switch ([Math]::Sign((Get-ProperDivisorSum $Number) -$Number))
{
-1 { [void]$numbers.Add([PSCustomObject]@{Class="Deficient"; Number=$Number}) }
0 { [void]$numbers.Add([PSCustomObject]@{Class="Perfect" ; Number=$Number}) }
1 { [void]$numbers.Add([PSCustomObject]@{Class="Abundant" ; Number=$Number}) }
}
}
End
{
$numbers | Group-Object -Property Class | Select-Object -Property Count, @{Name='Class' ; Expression={$_.Name}},
@{Name='Number'; Expression={$_.Group.Number}} } }  1..20000 | Get-NumberClassification  Output: Count Class Number ----- ----- ------ 15043 Deficient {1, 2, 3, 4...} 4 Perfect {6, 28, 496, 8128} 4953 Abundant {12, 18, 20, 24...}  ## Processing void setup() { int deficient = 0, perfect = 0, abundant = 0; for (int i = 1; i <= 20000; i++) { int sum_divisors = propDivSum(i); if (sum_divisors < i) { deficient++; } else if (sum_divisors == i) { perfect++; } else { abundant++; } } println("Deficient numbers less than 20000: " + deficient); println("Perfect numbers less than 20000: " + perfect); println("Abundant numbers less than 20000: " + abundant); } int propDivSum(int n) { int sum = 0; for (int i = 1; i < n; i++) { if (n % i == 0) { sum += i; } } return sum; } Output: Deficient numbers less than 20000: 15043 Perfect numbers less than 20000: 4 Abundant numbers less than 20000: 4953 ## Prolog proper_divisors(1, []) :- !. proper_divisors(N, [1|L]) :- FSQRTN is floor(sqrt(N)), proper_divisors(2, FSQRTN, N, L). proper_divisors(M, FSQRTN, _, []) :- M > FSQRTN, !. proper_divisors(M, FSQRTN, N, L) :- N mod M =:= 0, !, MO is N//M, % must be integer L = [M,MO|L1], % both proper divisors M1 is M+1, proper_divisors(M1, FSQRTN, N, L1). proper_divisors(M, FSQRTN, N, L) :- M1 is M+1, proper_divisors(M1, FSQRTN, N, L). dpa(1, [1], [], []) :- !. dpa(N, D, P, A) :- N > 1, proper_divisors(N, PN), sum_list(PN, SPN), compare(VGL, SPN, N), dpa(VGL, N, D, P, A). dpa(<, N, [N|D], P, A) :- N1 is N-1, dpa(N1, D, P, A). dpa(=, N, D, [N|P], A) :- N1 is N-1, dpa(N1, D, P, A). dpa(>, N, D, P, [N|A]) :- N1 is N-1, dpa(N1, D, P, A). dpa(N) :- T0 is cputime, dpa(N, D, P, A), Dur is cputime-T0, length(D, LD), length(P, LP), length(A, LA), format("deficient: ~d~n abundant: ~d~n perfect: ~d~n", [LD, LA, LP]), format("took ~f seconds~n", [Dur]).  Output: ?- dpa(20000). deficient: 15036 abundant: 4960 perfect: 4 took 0.802559 seconds  ## PureBasic EnableExplicit Procedure.i SumProperDivisors(Number) If Number < 2 : ProcedureReturn 0 : EndIf Protected i, sum = 0 For i = 1 To Number / 2 If Number % i = 0 sum + i EndIf Next ProcedureReturn sum EndProcedure Define n, sum, deficient, perfect, abundant If OpenConsole() For n = 1 To 20000 sum = SumProperDivisors(n) If sum < n deficient + 1 ElseIf sum = n perfect + 1 Else abundant + 1 EndIf Next PrintN("The breakdown for the numbers 1 to 20,000 is as follows : ") PrintN("") PrintN("Deficient = " + deficient) PrintN("Pefect = " + perfect) PrintN("Abundant = " + abundant) PrintN("") PrintN("Press any key to close the console") Repeat: Delay(10) : Until Inkey() <> "" CloseConsole() EndIf Output: The breakdown for the numbers 1 to 20,000 is as follows : Deficient = 15043 Pefect = 4 Abundant = 4953  ## Python ### Python: Counter Importing Proper divisors from prime factors: >>> from proper_divisors import proper_divs >>> from collections import Counter >>> >>> rangemax = 20000 >>> >>> def pdsum(n): ... return sum(proper_divs(n)) ... >>> def classify(n, p): ... return 'perfect' if n == p else 'abundant' if p > n else 'deficient' ... >>> classes = Counter(classify(n, pdsum(n)) for n in range(1, 1 + rangemax)) >>> classes.most_common() [('deficient', 15043), ('abundant', 4953), ('perfect', 4)] >>>  Output: Between 1 and 20000: 4953 abundant numbers 15043 deficient numbers 4 perfect numbers  ### Python: Reduce Works with: Python version 3.7 In terms of a single fold: '''Abundant, deficient and perfect number classifications''' from itertools import accumulate, chain, groupby, product from functools import reduce from math import floor, sqrt from operator import mul # deficientPerfectAbundantCountsUpTo :: Int -> (Int, Int, Int) def deficientPerfectAbundantCountsUpTo(n): '''Counts of deficient, perfect, and abundant integers in the range [1..n]. ''' def go(dpa, x): deficient, perfect, abundant = dpa divisorSum = sum(properDivisors(x)) return ( succ(deficient), perfect, abundant ) if x > divisorSum else ( deficient, perfect, succ(abundant) ) if x < divisorSum else ( deficient, succ(perfect), abundant ) return reduce(go, range(1, 1 + n), (0, 0, 0)) # --------------------------TEST-------------------------- # main :: IO () def main(): '''Size of each sub-class of integers drawn from [1..20000]:''' print(main.__doc__) print( '\n'.join(map( lambda a, b: a.rjust(10) + ' -> ' + str(b), ['Deficient', 'Perfect', 'Abundant'], deficientPerfectAbundantCountsUpTo(20000) )) ) # ------------------------GENERIC------------------------- # primeFactors :: Int -> [Int] def primeFactors(n): '''A list of the prime factors of n. ''' def f(qr): r = qr[1] return step(r), 1 + r def step(x): return 1 + (x << 2) - ((x >> 1) << 1) def go(x): root = floor(sqrt(x)) def p(qr): q = qr[0] return root < q or 0 == (x % q) q = until(p)(f)( (2 if 0 == x % 2 else 3, 1) )[0] return [x] if q > root else [q] + go(x // q) return go(n) # properDivisors :: Int -> [Int] def properDivisors(n): '''The ordered divisors of n, excluding n itself. ''' def go(a, x): return [a * b for a, b in product( a, accumulate(chain([1], x), mul) )] return sorted( reduce(go, [ list(g) for _, g in groupby(primeFactors(n)) ], [1]) )[:-1] if 1 < n else [] # succ :: Int -> Int def succ(x): '''The successor of a value. For numeric types, (1 +). ''' return 1 + x # 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()  and the main function could be rewritten in terms of an nthArrow abstraction: # nthArrow :: (a -> b) -> Tuple -> Int -> Tuple def nthArrow(f): '''A simple function lifted to one which applies to a tuple, transforming only its nth value. ''' def go(v, n): m = n - 1 return v if n > len(v) else [ x if m != i else f(x) for i, x in enumerate(v) ] return lambda tpl: lambda n: tuple(go(tpl, n))  as something like: # deficientPerfectAbundantCountsUpTo :: Int -> (Int, Int, Int) def deficientPerfectAbundantCountsUpTo(n): '''Counts of deficient, perfect, and abundant integers in the range [1..n]. ''' def go(dpa, x): divisorSum = sum(properDivisors(x)) return nthArrow(succ)(dpa)( 1 if x > divisorSum else ( 3 if x < divisorSum else 2 ) ) return reduce(go, range(1, 1 + n), (0, 0, 0))  Output: Size of each sub-class of integers drawn from [1..20000]: Deficient -> 15043 Perfect -> 4 Abundant -> 4953 ### The Simple Way pn = 0 an = 0 dn = 0 tt = [] num = 20000 for n in range(1, num+1): for x in range(1,1+n//2): if n%x == 0: tt.append(x) if sum(tt) == n: pn += 1 elif sum(tt) > n: an += 1 elif sum(tt) < n: dn += 1 tt = [] print(str(pn) + " Perfect Numbers") print(str(an) + " Abundant Numbers") print(str(dn) + " Deficient Numbers")  Output: 4 Perfect Numbers 4953 Abundant Numbers 15043 Deficient Numbers ### Simple vs Optimized A few changes: Instead of obtaining the remainder of n divided by every number halfway up to n, stop just short of the square root of n and add both factors to the running sum. And then in the case that n is a perfect square, add the square root of n to the sum. Don't compute the square root of each n, increment the square root as each n becomes a perfect square. Switch the summed list of factors to a single variable. Initialize the sum to 1 and start checking factors from 2 and up, which cuts one iteration from each factor checking loop, (a 19,999 iteration savings). Resulting optimized code is thirty five times faster than the simplified code, and not nearly as complicated as the Counter or Reduce methods (as this optimized method requires no imports, other than time for the performance comparison to the simple way). from time import time st = time() pn, an, dn = 0, 0, 0 tt = [] num = 20000 for n in range(1, num + 1): for x in range(1, 1 + n // 2): if n % x == 0: tt.append(x) if sum(tt) == n: pn += 1 elif sum(tt) > n: an += 1 elif sum(tt) < n: dn += 1 tt = [] et1 = time() - st print(str(pn) + " Perfect Numbers") print(str(an) + " Abundant Numbers") print(str(dn) + " Deficient Numbers") print(et1, "sec\n") st = time() pn, an, dn = 0, 0, 1 sum = 1 r = 1 num = 20000 for n in range(2, num + 1): d = r * r - n if d < 0: r += 1 for x in range(2, r): if n % x == 0: sum += x + n // x if d == 0: sum += r if sum == n: pn += 1 elif sum > n: an += 1 elif sum < n: dn += 1 sum = 1 et2 = time() - st print(str(pn) + " Perfect Numbers") print(str(an) + " Abundant Numbers") print(str(dn) + " Deficient Numbers") print(et2 * 1000, "ms\n") print (et1 / et2,"times faster")  Output @ Tio.run using Python 3 (PyPy): 4 Perfect Numbers 4953 Abundant Numbers 15043 Deficient Numbers 1.312887191772461 sec 4 Perfect Numbers 4953 Abundant Numbers 15043 Deficient Numbers 37.12296485900879 ms 35.365903471307924 times faster ## Quackery factors is defined at Factors of an integer. dpa returns 0 if n is deficient, 1 if n is perfect and 2 if n is abundant.  [ 0 swap witheach + ] is sum ( [ --> n ) [ factors -1 pluck dip sum 2dup = iff [ 2drop 1 ] done < iff 0 else 2 ] is dpa ( n --> n ) 0 0 0 20000 times [ i 1+ dpa [ table [ 1+ ] [ dip 1+ ] [ rot 1+ unrot ] ] do ] say "Deficient = " echo cr say " Perfect = " echo cr say " Abundant = " echo cr Output: Deficient = 15043 Perfect = 4 Abundant = 4953 ## R Works with: R version 3.3.2 and above # Abundant, deficient and perfect number classifications. 12/10/16 aev require(numbers); propdivcls <- function(n) { V <- sapply(1:n, Sigma, proper = TRUE); c1 <- c2 <- c3 <- 0; for(i in 1:n){ if(V[i]<i){c1 = c1 +1} else if(V[i]==i){c2 = c2 +1} else{c3 = c3 +1} } cat(" *** Between 1 and ", n, ":\n"); cat(" * ", c1, "deficient numbers\n"); cat(" * ", c2, "perfect numbers\n"); cat(" * ", c3, "abundant numbers\n"); } propdivcls(20000);  Output: > require(numbers) Loading required package: numbers > propdivcls(20000); *** Between 1 and 20000 : * 15043 deficient numbers * 4 perfect numbers * 4953 abundant numbers >  ## Racket #lang racket (require math) (define (proper-divisors n) (drop-right (divisors n) 1)) (define classes '(deficient perfect abundant)) (define (classify n) (list-ref classes (add1 (sgn (- (apply + (proper-divisors n)) n))))) (let ([N 20000]) (define t (make-hasheq)) (for ([i (in-range 1 (add1 N))]) (define c (classify i)) (hash-set! t c (add1 (hash-ref t c 0)))) (printf "The range between 1 and ~a has:\n" N) (for ([c classes]) (printf " ~a ~a numbers\n" (hash-ref t c 0) c)))  Output: The range between 1 and 20000 has: 15043 deficient numbers 4 perfect numbers 4953 abundant numbers  ## Raku (formerly Perl 6) Works with: rakudo version 2018.12 sub propdivsum (\x) { my @l = 1 if x > 1; (2 .. x.sqrt.floor).map: -> \d { unless x % d { @l.push: d; my \y = x div d; @l.push: y if y != d } } sum @l } say bag (1..20000).map: { propdivsum($_) <=> $_ }  Output: Bag(Less(15043), More(4953), Same(4)) ## REXX ### version 1 /*REXX program counts the number of abundant/deficient/perfect numbers within a range.*/ parse arg low high . /*obtain optional arguments from the CL*/ high=word(high low 20000,1); low= word(low 1,1) /*obtain the LOW and HIGH values.*/ say center('integers from ' low " to " high, 45, "═") /*display a header.*/ !.= 0 /*define all types of sums to zero. */ do j=low to high;$= sigma(j)   /*get sigma for an integer in a range. */
if $<j then !.d= !.d + 1 /*Less? It's a deficient number.*/ else if$>j  then  !.a= !.a + 1    /*Greater?     "  "  abundant      "   */
else  !.p= !.p + 1    /*Equal?       "  "  perfect       "   */
end  /*j*/                                 /* [↑]  IFs are coded as per likelihood*/

say '   the number of perfect   numbers: '       right(!.p, length(high) )
say '   the number of abundant  numbers: '       right(!.a, length(high) )
say '   the number of deficient numbers: '       right(!.d, length(high) )
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
sigma: procedure; parse arg x; if x<2  then return 0;  odd=x // 2    /* // ◄──remainder.*/
s= 1                                      /* [↓]  only use  EVEN or ODD integers.*/
do k=2+odd  by 1+odd  while k*k<x   /*divide by all integers up to  √x.    */
if x//k==0  then  s= s + k +  x % k /*add the two divisors to (sigma) sum. */
end   /*k*/                         /* [↑]  %  is the REXX integer division*/
if k*k==x  then  return s + k             /*Was  X  a square?   If so, add  √ x  */
return s                 /*return (sigma) sum of the divisors.  */

output   when using the default input:
═════════integers from  1  to  20000═════════
the number of perfect   numbers:      4
the number of abundant  numbers:   4953
the number of deficient numbers:  15043


### version 1.5

This version is pretty much identical to the 1st version but uses an   integer square root   calculation to find the
limit of the   do   loop in the   sigma   function.

 For    20k  integers,  it's approximately  15%  faster.
"    100k     "         "        "        20%    "
"      1m     "         "        "        30%    "

/*REXX program counts the number of  abundant/deficient/perfect  numbers within a range.*/
parse arg low high .                             /*obtain optional arguments from the CL*/
high=word(high low 20000,1);  low=word(low 1, 1) /*obtain the   LOW  and  HIGH   values.*/
say center('integers from '   low    " to "    high,  45,  "═")      /*display a header.*/
!.= 0                                            /*define all types of  sums  to zero.  */
do j=low  to high;           $= sigma(j) /*get sigma for an integer in a range. */ if$<j  then               !.d= !.d + 1    /*Less?      It's a  deficient  number.*/
else if $>j then !.a= !.a + 1 /*Greater? " " abundant " */ else !.p= !.p + 1 /*Equal? " " perfect " */ end /*j*/ /* [↑] IFs are coded as per likelihood*/ say ' the number of perfect numbers: ' right(!.p, length(high) ) say ' the number of abundant numbers: ' right(!.a, length(high) ) say ' the number of deficient numbers: ' right(!.d, length(high) ) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ sigma: procedure; parse arg x 1 z; if x<5 then return max(0, x-1) /*sets X&Z to arg1.*/ q=1; do while q<=z; q= q * 4; end /* ◄──↓ compute integer sqrt of Z (=R)*/ r=0; do while q>1; q=q%4; _=z-r-q; r=r%2; if _>=0 then do; z=_; r=r+q; end; end odd= x//2 /* [↓] only use EVEN | ODD ints. ___*/ s= 1; do k=2+odd by 1+odd to r /*divide by all integers up to √ x */ if x//k==0 then s=s + k + x%k /*add the two divisors to (sigma) sum. */ end /*k*/ /* [↑] % is the REXX integer division*/ if r*r==x then return s - k /*Was X a square? If so, subtract √ x */ return s /*return (sigma) sum of the divisors. */  output is identical to the 1st REXX version. It is about 2,800% faster than the REXX version 2. ### version 2 /* REXX */ Call time 'R' cnt.=0 Do x=1 To 20000 pd=proper_divisors(x) sumpd=sum(pd) Select When x<sumpd Then cnt.abundant =cnt.abundant +1 When x=sumpd Then cnt.perfect =cnt.perfect +1 Otherwise cnt.deficient=cnt.deficient+1 End Select When npd>hi Then Do list.npd=x hi=npd End When npd=hi Then list.hi=list.hi x Otherwise Nop End End Say 'In the range 1 - 20000' Say format(cnt.abundant ,5) 'numbers are abundant ' Say format(cnt.perfect ,5) 'numbers are perfect ' Say format(cnt.deficient,5) 'numbers are deficient ' Say time('E') 'seconds elapsed' Exit proper_divisors: Procedure Parse Arg n Pd='' If n=1 Then Return '' If n//2=1 Then /* odd number */ delta=2 Else /* even number */ delta=1 Do d=1 To n%2 By delta If n//d=0 Then pd=pd d End Return space(pd) sum: Procedure Parse Arg list sum=0 Do i=1 To words(list) sum=sum+word(list,i) End Return sum  Output: In the range 1 - 20000 4953 numbers are abundant 4 numbers are perfect 15043 numbers are deficient 28.392000 seconds elapsed ## Ring The following classifies the first few numbers of each type. n = 30 perfect(n) func perfect n for i = 1 to n sum = 0 for j = 1 to i - 1 if i % j = 0 sum = sum + j ok next see i if sum = i see " is a perfect number" + nl but sum < i see " is a deficient number" + nl else see " is a abundant number" + nl ok next ### Task using modulo/division Translation of: Lua – Summing the factors using modulo/division a = 0 d = 0 p = 0 for n = 1 to 20000 Pn = sumDivs(n) if Pn > n a = a + 1 ok if Pn < n d = d + 1 ok if Pn = n p = p + 1 ok next see "Abundant : " + a + nl see "Deficient: " + d + nl see "Perfect : " + p + nl func sumDivs (n) if n < 2 return 0 else sum = 1 sr = sqrt(n) for d = 2 to sr if n % d = 0 sum = sum + d if d != sr sum = sum + n / d ok ok next return sum ok Output: Abundant : 4953 Deficient: 15043 Perfect : 4  ### Task using a table Translation of: Lus – Summiing the factors using a table maxNumber = 20000 abundantCount = 0 deficientCount = 0 perfectCount = 0 pds = list( maxNumber ) pds[ 1 ] = 0 for i = 2 to maxNumber pds[ i ] = 1 next for i = 2 to maxNumber for j = i + i to maxNumber step i pds[ j ] = pds[ j ] + i next next for n = 1 to maxNumber pdSum = pds[ n ] if pdSum < n deficientCount = deficientCount + 1 but pdSum = n perfectCount = perfectCount + 1 else # pdSum > n abundantCount = abundantCount + 1 ok next see "Abundant : " + abundantCount + nl see "Deficient: " + deficientCount + nl see "Perfect : " + perfectCount + nl Output: Abundant : 4953 Deficient: 15043 Perfect : 4  ## RPL Works with: HP version 49 ≪ [1 0 0] 2 20000 FOR n n DIVIS REVLIST TAIL @ get the list of divisors of n excluding n 0. + @ avoid ∑LIST and SIGN errors when n is prime ∑LIST n - SIGN 2 + @ turn P(n)-n into 1, 2 or 3 DUP2 GET 1 + PUT @ increment appropriate array element NEXT ≫ 'TASK' STO  Output: 1: [15042 4 4953]  ## Ruby Works with: ruby version 2.7 With proper_divisors#Ruby in place: res = (1 .. 20_000).map{|n| n.proper_divisors.sum <=> n }.tally puts "Deficient: #{res[-1]} Perfect: #{res[0]} Abundant: #{res[1]}"  Output:  Deficient: 15043 Perfect: 4 Abundant: 4953  ## Rust With proper_divisors#Rust in place: fn main() { // deficient starts at 1 because 1 is deficient but proper_divisors returns // and empty Vec let (mut abundant, mut deficient, mut perfect) = (0u32, 1u32, 0u32); for i in 1..20_001 { if let Some(divisors) = i.proper_divisors() { let sum: u64 = divisors.iter().sum(); if sum < i { deficient += 1 } else if sum > i { abundant += 1 } else { perfect += 1 } } } println!("deficient:\t{:5}\nperfect:\t{:5}\nabundant:\t{:5}", deficient, perfect, abundant); }  Output: deficient: 15043 perfect: 4 abundant: 4953  ## Scala def properDivisors(n: Int) = (1 to n/2).filter(i => n % i == 0) def classifier(i: Int) = properDivisors(i).sum compare i val groups = (1 to 20000).groupBy( classifier ) println("Deficient: " + groups(-1).length) println("Abundant: " + groups(1).length) println("Perfect: " + groups(0).length + " (" + groups(0).mkString(",") + ")")  Output: Deficient: 15043 Abundant: 4953 Perfect: 4 (6,28,496,8128) ## Scheme (define (classify n) (define (sum_of_factors x) (cond ((= x 1) 1) ((= (remainder n x) 0) (+ x (sum_of_factors (- x 1)))) (else (sum_of_factors (- x 1))))) (cond ((or (= n 1) (< (sum_of_factors (floor (/ n 2))) n)) -1) ((= (sum_of_factors (floor (/ n 2))) n) 0) (else 1))) (define n_perfect 0) (define n_abundant 0) (define n_deficient 0) (define (count n) (cond ((= n 1) (begin (display "perfect ") (display n_perfect) (newline) (display "abundant") (display n_abundant) (newline) (display "deficinet") (display n_perfect) (newline))) ((equal? (classify n) 0) (begin (set! n_perfect (+ 1 n_perfect)) (display n) (newline) (count (- n 1)))) ((equal? (classify n) 1) (begin (set! n_abundant (+ 1 n_abundant)) (count (- n 1)))) ((equal? (classify n) -1) (begin (set! n_deficient (+ 1 n_deficient)) (count (- n 1))))))  ## Seed7 $ include "seed7_05.s7i";

const func integer: sumProperDivisors (in integer: number) is func
result
var integer: sum is 0;
local
var integer: num is 0;
begin
if number >= 2 then
for num range 1 to number div 2 do
if number rem num = 0 then
sum +:= num;
end if;
end for;
end if;
end func;

const proc: main is func
local
var integer: sum is 0;
var integer: deficient is 0;
var integer: perfect is 0;
var integer: abundant is 0;
var integer: number is 0;
begin
for number range 1 to 20000 do
sum := sumProperDivisors(number);
if sum < number then
incr(deficient);
elsif sum = number then
incr(perfect);
else
incr(abundant);
end if;
end for;
writeln("Deficient: " <& deficient);
writeln("Perfect:   " <& perfect);
writeln("Abundant:  " <& abundant);
end func;
Output:
Deficient: 15043
Perfect:   4
Abundant:  4953


## SETL

program classifications;
P := properdivisorsums(20000);

print("Deficient:", #[n : n in [1..#P] | P(n) < n]);
print("  Perfect:", #[n : n in [1..#P] | P(n) = n]);
print(" Abundant:", #[n : n in [1..#P] | P(n) > n]);

proc properdivisorsums(n);
p := [0];
loop for i in [1..n] do
loop for j in [i*2, i*3..n] do
p(j) +:= i;
end loop;
end loop;
return p;
end proc;
end program;
Output:
Deficient: 15043
Perfect: 4
Abundant: 4953

## Sidef

func propdivsum(n) { n.sigma - n }

var h = Hash()
{|i| ++(h{propdivsum(i) <=> i} := 0) } << 1..20000
say "Perfect: #{h{0}}    Deficient: #{h{-1}}    Abundant: #{h{1}}"

Output:
Perfect: 4    Deficient: 15043    Abundant: 4953


## Swift

Translation of: C
var deficients = 0 // sumPd < n
var perfects = 0 // sumPd = n
var abundants = 0 // sumPd > n

// 1 is deficient (no proper divisor)
deficients++

for i in 2...20000 {

var sumPd = 1 // 1 is a proper divisor of all integer above 1

var maxPdToTest = i/2 // the max divisor to test

for var j = 2; j < maxPdToTest; j++ {

if (i%j) == 0 {
// j is a proper divisor
sumPd += j

// New maximum for divisibility check
maxPdToTest = i / j

if maxPdToTest != j {
sumPd += maxPdToTest
}
}
}

// Select type according to sum of Proper divisors
if sumPd < i {
deficients++
} else if sumPd > i {
abundants++
} else {
perfects++
}
}

println("There are \(deficients) deficient, \(perfects) perfect and \(abundants) abundant integers from 1 to 20000.")

Output:
There are 15043 deficient, 4 perfect and 4953 abundant integers from 1 to 20000.

## Tcl

proc ProperDivisors {n} {
if {$n == 1} {return 0} set divs 1 set sum 1 for {set i 2} {$i*$i <=$n} {incr i} {
if {! ($n %$i)} {
lappend divs $i incr sum$i
if {$i*$i<$n} { lappend divs [set d [expr {$n / $i}]] incr sum$d
}
}
}
list $sum$divs
}

proc cmp {i j} {    ;# analogous to [string compare], but for numbers
if {$i ==$j} {return 0}
if {$i >$j} {return 1}
return -1
}

proc classify {k} {
lassign [ProperDivisors $k] p ;# we only care about the first part of the result dict get { 1 abundant 0 perfect -1 deficient } [cmp$k $p] } puts "Classifying the integers in $1, 20_000$:" set classes {} ;# this will be a dict for {set i 1} {$i <= 20000} {incr i} {
set class [classify $i] dict incr classes$class
}

# using [lsort] to order the dictionary by value:
foreach {kind count} [lsort -stride 2 -index 1 -integer $classes] { puts "$kind: $count" }  Output: Classifying the integers in [1, 20_000]: perfect: 4 deficient: 4953 abundant: 15043 ## TypeScript function integer_classification(){ var sum:number=0, i:number,j:number; var try:number=0; var number_list:number[]={1,0,0}; for(i=2;i<=20000;i++){ try=i/2; sum=1; for(j=2;j<try;j++){ if (i%j) continue; try=i/j; sum+=j; if (j!=try) sum+=try; } if (sum<i){ number_list[d]++; continue; } else if (sum>i){ number_list[a]++; continue; } number_list[p]++; } console.log('There are '+number_list[d]+ ' deficient , ' + 'number_list[p] + ' perfect and '+ number_list[a]+ ' abundant numbers between 1 and 20000'); }  ## uBasic/4tH This is about the limit of what is feasible with uBasic/4tH performance wise, since a full run takes over 5 minutes. P = 0 : D = 0 : A = 0 For n= 1 to 20000 s = FUNC(_SumDivisors(n))-n If s = n Then P = P + 1 If s < n Then D = D + 1 If s > n Then A = A + 1 Next Print "Perfect: ";P;" Deficient: ";D;" Abundant: ";A End ' Return the least power of a@ that does not divide b@ _LeastPower Param(2) Local(1) c@ = a@ Do While (b@ % c@) = 0 c@ = c@ * a@ Loop Return (c@) ' Return the sum of the proper divisors of a@ _SumDivisors Param(1) Local(4) b@ = a@ c@ = 1 ' Handle two specially d@ = FUNC(_LeastPower (2,b@)) c@ = c@ * (d@ - 1) b@ = b@ / (d@ / 2) ' Handle odd factors For e@ = 3 Step 2 While (e@*e@) < (b@+1) d@ = FUNC(_LeastPower (e@,b@)) c@ = c@ * ((d@ - 1) / (e@ - 1)) b@ = b@ / (d@ / e@) Loop ' At this point, t must be one or prime If (b@ > 1) c@ = c@ * (b@+1) Return (c@)  Output: Perfect: 4 Deficient: 15043 Abundant: 4953 0 OK, 0:210 ## Vala Translation of: C enum Classification { DEFICIENT, PERFECT, ABUNDANT } void main() { var i = 0; var j = 0; var sum = 0; var try_max = 0; int[] count_list = {1, 0, 0}; for (i = 2; i <= 20000; i++) { try_max = i / 2; sum = 1; for (j = 2; j < try_max; j++) { if (i % j != 0) continue; try_max = i / j; sum += j; if (j != try_max) sum += try_max; } if (sum < i) { count_list[Classification.DEFICIENT]++; continue; } if (sum > i) { count_list[Classification.ABUNDANT]++; continue; } count_list[Classification.PERFECT]++; } print(@"There are$(count_list[Classification.DEFICIENT]) deficient,");
print(@" $(count_list[Classification.PERFECT]) perfect,"); print(@"$(count_list[Classification.ABUNDANT]) abundant numbers between 1 and 20000.\n");
}

Output:
There are 15043 deficient, 4 perfect, 4953 abundant numbers between 1 and 20000.


## VBA

Option Explicit

Public Sub Nb_Classifications()
Dim A As New Collection, D As New Collection, P As New Collection
Dim n As Long, l As Long, s As String, t As Single

t = Timer
'Start
For n = 1 To 20000
l = SumPropers(n): s = CStr(n)
Select Case n
Case Is > l: D.Add s, s
Case Is < l: A.Add s, s
End Select
Next

'End. Return :
Debug.Print "Execution Time : " & Timer - t & " seconds."
Debug.Print "-------------------------------------------"
Debug.Print "Deficient := " & D.Count
Debug.Print "Perfect := " & P.Count
Debug.Print "Abundant := " & A.Count
End Sub

Private Function SumPropers(n As Long) As Long
'returns the sum of the proper divisors of n
Dim j As Long
For j = 1 To n \ 2
If n Mod j = 0 Then SumPropers = j + SumPropers
Next
End Function
Output:
Execution Time : 2,6875 seconds.
-------------------------------------------
Deficient := 15043
Perfect := 4
Abundant := 4953

## VBScript

Deficient = 0
Perfect = 0
Abundant = 0
For i = 1 To 20000
sum = 0
For n = 1 To 20000
If n < i Then
If i Mod n = 0 Then
sum = sum + n
End If
End If
Next
If sum < i Then
Deficient = Deficient + 1
ElseIf sum = i Then
Perfect = Perfect + 1
ElseIf sum > i Then
Abundant = Abundant + 1
End If
Next
WScript.Echo "Deficient = " & Deficient & vbCrLf &_
"Perfect = " & Perfect & vbCrLf &_
"Abundant = " & Abundant

Output:
Deficient = 15043
Perfect = 4
Abundant = 4953

## Visual Basic .NET

Translation of: FreeBASIC
Module Module1

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

Sub Main()
Dim sum, deficient, perfect, abundant As Integer

For n As Integer = 1 To 20000
sum = SumProperDivisors(n)
If sum < n Then
deficient += 1
ElseIf sum = n Then
perfect += 1
Else
abundant += 1
End If
Next

Console.WriteLine("The classification of the numbers from 1 to 20,000 is as follows : ")
Console.WriteLine()
Console.WriteLine("Deficient = {0}", deficient)
Console.WriteLine("Perfect   = {0}", perfect)
Console.WriteLine("Abundant  = {0}", abundant)
End Sub

End Module

Output:
The classification of the numbers from 1 to 20,000 is as follows :

Deficient = 15043
Perfect   = 4
Abundant  = 4953

## V (Vlang)

Translation of: go
fn p_fac_sum(i int) int {
mut sum := 0
for p := 1; p <= i/2; p++ {
if i%p == 0 {
sum += p
}
}
return sum
}

fn main() {
mut d := 0
mut a := 0
mut p := 0
for i := 1; i <= 20000; i++ {
j := p_fac_sum(i)
if j < i {
d++
} else if j == i {
p++
} else {
a++
}
}
println("There are $d deficient numbers between 1 and 20000") println("There are$a abundant numbers  between 1 and 20000")
println("There are \$p perfect numbers between 1 and 20000")
}
Output:
There are 15043 deficient numbers between 1 and 20000
There are 4953 abundant numbers  between 1 and 20000
There are 4 perfect numbers between 1 and 20000


## VTL-2

10 M=20000
20 I=1
30 :I)=0
40 I=I+1
50 #=M>I*30
60 I=1
70 J=I*2
80 :J)=:J)+I
90 J=J+I
100 #=M>J*80
110 I=I+1
120 #=M/2>I*70
130 D=0
140 P=0
150 A=0
160 I=1
170 #=:I)<I*230
180 #=:I)=I*210
190 A=A+1
200 #=240
210 P=P+1
220 #=240
230 D=D+1
240 I=I+1
250 #=M>I*170
260 ?=D
270 ?=" deficient"
280 ?=P
290 ?=" perfect"
300 ?=A
310 ?=" abundant"
Output:
15043 deficient
4 perfect
4953 abundant

## Wren

### Using modulo/division

Library: Wren-math
import "./math" for Int, Nums

var d = 0
var a = 0
var p = 0
for (i in 1..20000) {
var j = Nums.sum(Int.properDivisors(i))
if (j < i) {
d = d + 1
} else if (j == i) {
p = p + 1
} else {
a = a + 1
}
}
System.print("There are %(d) deficient numbers between 1 and 20000")
System.print("There are %(a) abundant numbers  between 1 and 20000")
System.print("There are %(p) perfect numbers between 1 and 20000")

Output:
There are 15043 deficient numbers between 1 and 20000
There are 4953 abundant numbers  between 1 and 20000
There are 4 perfect numbers between 1 and 20000


### Using a table

Alternative version, computing a table of divisor sums.

Translation of: Lua – Summing the factors using a table
var maxNumber      = 20000
var abundantCount  = 0
var deficientCount = 0
var perfectCount   = 0

var pds = []
for (i in 2..maxNumber) {
}
for (i in 2..maxNumber) {
var j = i + i
while (j <= maxNumber) {
pds[j] = pds[j] + i
j = j + i
}
}
for (n in 1..maxNumber) {
var pdSum = pds[n]
if (pdSum < n) {
deficientCount = deficientCount + 1
} else if (pdSum == n) {
perfectCount = perfectCount + 1
} else { // pdSum >  n
abundantCount = abundantCount + 1
}
}

System.print("Abundant : %(abundantCount)")
System.print("Deficient: %(deficientCount)")
System.print("Perfect  : %(perfectCount)")

Output:
Abundant : 4953
Deficient: 15043
Perfect  : 4


## XPL0

int CntD, CntP, CntA, Num, Div, Sum;
[CntD:= 0;  CntP:= 0;  CntA:= 0;
for Num:= 1 to 20000 do
[Sum:= if Num = 1 then 0 else 1;
for Div:= 2 to Num-1 do
if rem(Num/Div) = 0 then
Sum:= Sum + Div;
case of
Sum < Num: CntD:= CntD+1;
Sum > Num: CntA:= CntA+1
other CntP:= CntP+1;
];
Text(0, "Deficient: ");  IntOut(0, CntD);  CrLf(0);
Text(0, "Perfect:   ");  IntOut(0, CntP);  CrLf(0);
Text(0, "Abundant:  ");  IntOut(0, CntA);  CrLf(0);
]
Output:
Deficient: 15043
Perfect:   4
Abundant:  4953


## Yabasic

Translation of: AWK
clear screen

Deficient = 0
Perfect = 0
Abundant = 0
For j=1 to 20000
sump = sumprop(j)
If sump < j Then
Deficient = Deficient + 1
ElseIf sump = j Then
Perfect = Perfect + 1
ElseIf sump > j Then
Abundant = Abundant + 1
End If
Next j

PRINT "Number deficient: ",Deficient
PRINT "Number perfect:   ",Perfect
PRINT "Number abundant:  ",Abundant

sub sumprop(num)
local i, sum, root

if num>1 then
sum=1
root=sqrt(num)
for i=2 to root
if mod(num,i) = 0 then
sum=sum+i
if (i*i)<>num sum=sum+num/i
end if
next i
end if
return sum
end sub

## zkl

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

fcn classify(n){
p:=properDivs(n).sum();
return(if(p<n) -1 else if(p==n) 0 else 1);
}

const rangeMax=20_000;
classified:=[1..rangeMax].apply(classify);
perfect   :=classified.filter('==(0)).len();
abundant  :=classified.filter('==(1)).len();
println("Deficient=%d, perfect=%d, abundant=%d".fmt(
classified.len()-perfect-abundant, perfect, abundant));
Output:
Deficient=15043, perfect=4, abundant=4953

## ZX Spectrum Basic

Solution 1:

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

Solution 2 (more efficient):

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