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[[category:Discrete math]]
{{task|Prime Numbers}}
Write a function which says whether a number is perfect.
<br>
[[wp:Perfect_numbers|A perfect number]] is a positive integer that is the sum of its proper positive divisors excluding the number itself.
Equivalently, a perfect number is a number that is half the sum of all of its positive divisors (including itself).
Note: The faster [[Lucas-Lehmer test]] is used to find primes of the form <big> 2<sup>''n''</sup>-1</big>, all ''known'' perfect numbers can be derived from these primes
using the formula <big> (2<sup>''n''</sup> - 1) × 2<sup>''n'' - 1</sup></big>.
It is not known if there are any odd perfect numbers (any that exist are larger than <big>10<sup>2000</sup></big>).
The number of ''known'' perfect numbers is '''51''' (as of December, 2018), and the largest known perfect number contains '''49,724,095''' decimal digits.
;See also:
:* [[Rational Arithmetic]]
:* [[oeis:A000396|Perfect numbers on OEIS]]
:* [http://www.oddperfect.org/ Odd Perfect] showing the current status of bounds on odd perfect numbers.
<br><br>
=={{header|11l}}==
{{trans|Python}}
<syntaxhighlight lang="11l">F perf(n)
V sum = 0
L(i) 1 .< n
I n % i == 0
sum += i
R sum == n
L(i) 1..10000
I perf(i)
print(i, end' ‘ ’)</syntaxhighlight>
{{out}}
<pre>
6 28 496 8128
</pre>
=={{header|360 Assembly}}==
===Simple code===
{{trans|PL/I}}
For maximum compatibility, this program uses only the basic instruction set (S/360)
and two ASSIST macros (XDECO,XPRNT) to keep it as short as possible.
The only added optimization is the loop up to n/2 instead of n-1.
With 31 bit integers the limit is 2,147,483,647.
<syntaxhighlight lang="360asm">* Perfect numbers 15/05/2016
PERFECTN CSECT
USING PERFECTN,R13 prolog
SAVEAREA B STM-SAVEAREA(R15) "
DC 17F'0' "
STM STM R14,R12,12(R13) "
ST R13,4(R15) "
ST R15,8(R13) "
LR R13,R15 "
LA R6,2 i=2
LOOPI C R6,NN do i=2 to nn
BH ELOOPI
LR R1,R6 i
BAL R14,PERFECT
LTR R0,R0 if perfect(i)
BZ NOTPERF
XDECO R6,PG edit i
XPRNT PG,L'PG print i
NOTPERF LA R6,1(R6) i=i+1
B LOOPI
ELOOPI L R13,4(0,R13) epilog
LM R14,R12,12(R13) "
XR R15,R15 "
BR R14 exit
PERFECT SR R9,R9 function perfect(n); sum=0
LA R7,1 j
LR R8,R1 n
SRA R8,1 n/2
LOOPJ CR R7,R8 do j=1 to n/2
BH ELOOPJ
LR R4,R1 n
SRDA R4,32
DR R4,R7 n/j
LTR R4,R4 if mod(n,j)=0
BNZ NOTMOD
AR R9,R7 sum=sum+j
NOTMOD LA R7,1(R7) j=j+1
B LOOPJ
ELOOPJ SR R0,R0 r0=false
CR R9,R1 if sum=n
BNE NOTEQ
BCTR R0,0 r0=true
NOTEQ BR R14 return(r0); end perfect
NN DC F'10000'
PG DC CL12' ' buffer
YREGS
END PERFECTN</syntaxhighlight>
{{out}}
<pre>
6
28
496
8128
</pre>
===Some optimizations===
{{trans|REXX}}
Use of optimizations found in Rexx algorithms and use of packed decimal to have bigger numbers.
With 15 digit decimal integers the limit is 999,999,999,999,999.
<syntaxhighlight lang="360asm">* Perfect numbers 15/05/2016
PERFECPO CSECT
USING PERFECPO,R13 prolog
SAVEAREA B STM-SAVEAREA(R15) "
DC 17F'0' "
STM STM R14,R12,12(R13) "
ST R13,4(R15) "
ST R15,8(R13) "
LR R13,R15 "
ZAP I,I1 i=i1
LOOPI CP I,I2 do i=i1 to i2
BH ELOOPI
LA R1,I r1=@i
BAL R14,PERFECT perfect(i)
LTR R0,R0 if perfect(i)
BZ NOTPERF
UNPK PG(16),I unpack i
OI PG+15,X'F0'
XPRNT PG,16 print i
NOTPERF AP I,=P'1' i=i+1
B LOOPI
ELOOPI L R13,4(0,R13) epilog
LM R14,R12,12(R13) "
XR R15,R15 "
BR R14 exit
PERFECT EQU * function perfect(n);
ZAP N,0(8,R1) n=%r1
CP N,=P'6' if n=6
BNE NOT6
L R0,=F'-1' r0=true
B RETURN return(true)
NOT6 ZAP PW,N n
SP PW,=P'1' n-1
ZAP PW2,PW n-1
DP PW2,=PL8'9' (n-1)/9
ZAP R,PW2+8(8) if mod((n-1),9)<>0
BZ ZERO
SR R0,R0 r0=false
B RETURN return(false)
ZERO ZAP PW2,N n
DP PW2,=PL8'2' n/2
ZAP SUM,PW2(8) sum=n/2
AP SUM,=P'3' sum=n/2+3
ZAP J,=P'3' j=3
LOOPJ ZAP PW,J do loop on j
MP PW,J j*j
CP PW,N while j*j<=n
BH ELOOPJ
ZAP PW2,N n
DP PW2,J n/j
CP PW2+8(8),=P'0' if mod(n,j)<>0
BNE NEXTJ
AP SUM,J sum=sum+j
ZAP PW2,N n
DP PW2,J n/j
AP SUM,PW2(8) sum=sum+j+n/j
NEXTJ AP J,=P'1' j=j+1
B LOOPJ next j
ELOOPJ SR R0,R0 r0=false
CP SUM,N if sum=n
BNE RETURN
BCTR R0,0 r0=true
RETURN BR R14 return(r0); end perfect
I1 DC PL8'1'
I2 DC PL8'200000000000'
I DS PL8
PG DC CL16' ' buffer
N DS PL8
SUM DS PL8
J DS PL8
R DS PL8
C DS CL16
PW DS PL8
PW2 DS PL16
YREGS
END PERFECPO</syntaxhighlight>
{{out}}
<pre>
0000000000000006
0000000000000028
0000000000000496
0000000000008128
0000000033550337
0000008589869056
0000137438691328
</pre>
=={{header|AArch64 Assembly}}==
{{works with|as|Raspberry Pi 3B version Buster 64 bits}}
<syntaxhighlight lang="aarch64 assembly">
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program perfectNumber64.s */
/* use Euclide Formula : if M=(2puis p)-1 is prime M * (M+1)/2 is perfect see Wikipedia */
/*******************************************/
/* Constantes file */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeConstantesARM64.inc"
.equ MAXI, 63
/*********************************/
/* Initialized data */
/*********************************/
.data
sMessResult: .asciz "Perfect : @ \n"
szMessOverflow: .asciz "Overflow in function isPrime.\n"
szCarriageReturn: .asciz "\n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
sZoneConv: .skip 24
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: // entry of program
mov x4,2 // start 2
mov x3,1 // counter 2 power
1: // begin loop
lsl x4,x4,1 // 2 power
sub x0,x4,1 // - 1
bl isPrime // is prime ?
cbz x0,2f // no
sub x0,x4,1 // yes
mul x1,x0,x4 // multiply m by m-1
lsr x0,x1,1 // divide by 2
bl displayPerfect // and display
2:
add x3,x3,1 // next power of 2
cmp x3,MAXI
blt 1b
100: // standard end of the program
mov x0,0 // return code
mov x8,EXIT // request to exit program
svc 0 // perform the system call
qAdrszCarriageReturn: .quad szCarriageReturn
qAdrsMessResult: .quad sMessResult
/******************************************************************/
/* Display perfect number */
/******************************************************************/
/* x0 contains the number */
displayPerfect:
stp x1,lr,[sp,-16]! // save registers
ldr x1,qAdrsZoneConv
bl conversion10 // call décimal conversion
ldr x0,qAdrsMessResult
ldr x1,qAdrsZoneConv // insert conversion in message
bl strInsertAtCharInc
bl affichageMess // display message
100:
ldp x1,lr,[sp],16 // restaur 2 registers
ret // return to address lr x30
qAdrsZoneConv: .quad sZoneConv
/***************************************************/
/* is a number prime ? */
/***************************************************/
/* x0 contains the number */
/* x0 return 1 if prime else 0 */
//2147483647 OK
//4294967297 NOK
//131071 OK
//1000003 OK
//10001363 OK
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 // return zero
cmp x2,2 // for 1 and 2 return 1
ble 2f
mov x0,#2
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // no prime
cmp x2,3
beq 2f
mov x0,#3
bl moduloPuR64
blt 100f // error overflow
cmp x0,#1
bne 99f
cmp x2,5
beq 2f
mov x0,#5
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,7
beq 2f
mov x0,#7
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,11
beq 2f
mov x0,#11
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // Pas premier
cmp x2,13
beq 2f
mov x0,#13
bl moduloPuR64
bcs 100f // error overflow
cmp x0,#1
bne 99f // Pas premier
2:
cmn x0,0 // carry à zero no error
mov x0,1 // prime
b 100f
99:
cmn x0,0 // carry à zero no error
mov x0,#0 // prime
100:
ldp x2,x3,[sp],16 // restaur des 2 registres
ldp x1,lr,[sp],16 // restaur des 2 registres
ret
/**************************************************************/
/********************************************************/
/* Compute modulo de b power e modulo m */
/* Exemple 4 puissance 13 modulo 497 = 445 */
/********************************************************/
/* x0 number */
/* x1 exposant */
/* x2 modulo */
moduloPuR64:
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 // result
udiv x4,x8,x2
msub x9,x4,x2,x8 // remainder
1:
tst x7,1 // if bit = 1
beq 2f
mul x4,x9,x6
umulh x5,x9,x6
mov x6,x4
mov x0,x6
mov x1,x5
bl divisionReg128U // division 128 bits
cbnz x1,99f // overflow
mov x6,x3 // remainder
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 no error
b 100f
99:
ldr x0,qAdrszMessOverflow
bl affichageMess // display error message
cmp x0,0 // carry set error
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"
</syntaxhighlight>
<pre>
Perfect : 6
Perfect : 28
Perfect : 496
Perfect : 8128
Perfect : 33550336
Perfect : 8589869056
Perfect : 137438691328
Perfect : 2305843008139952128
Perfect : 8070450532247928832
</pre>
=={{header|Action!}}==
<syntaxhighlight lang="action!">PROC Main()
DEFINE MAXNUM="10000"
CARD ARRAY pds(MAXNUM+1)
CARD i,j
FOR i=2 TO MAXNUM
DO
pds(i)=1
OD
FOR i=2 TO MAXNUM
DO
FOR j=i+i TO MAXNUM STEP i
DO
pds(j)==+i
OD
OD
FOR i=2 TO MAXNUM
DO
IF pds(i)=i THEN
PrintCE(i)
FI
OD
RETURN</syntaxhighlight>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Perfect_numbers.png Screenshot from Atari 8-bit computer]
<pre>
6
28
496
8128
</pre>
=={{header|Ada}}==
<
Sum : Natural := 0;
begin
Line 20 ⟶ 514:
end loop;
return Sum = N;
end Is_Perfect;</
=={{header|ALGOL 60}}==
{{works with|A60}}
<syntaxhighlight lang="algol60">
begin
comment - return p mod q;
integer procedure mod(p, q);
value p, q; integer p, q;
begin
mod := p - q * entier(p / q);
end;
comment - return true if n is perfect, otherwise false;
boolean procedure isperfect(n);
value n; integer n;
begin
integer sum, f1, f2;
sum := 1;
f1 := 1;
for f1 := f1 + 1 while (f1 * f1) <= n do
begin
if mod(n, f1) = 0 then
begin
sum := sum + f1;
f2 := n / f1;
if f2 > f1 then sum := sum + f2;
end;
end;
isperfect := (sum = n);
end;
comment - exercise the procedure;
integer i, found;
outstring(1,"Searching up to 10000 for perfect numbers\n");
found := 0;
for i := 2 step 1 until 10000 do
if isperfect(i) then
begin
outinteger(1,i);
found := found + 1;
end;
outstring(1,"\n");
outinteger(1,found);
outstring(1,"perfect numbers were found");
end
</syntaxhighlight>
{{out}}
<pre>
Searching up to 10000 for perfect numbers
6 28 496 8128
4 perfect numbers were found
</pre>
=={{header|ALGOL 68}}==
{{works with|ALGOL 68|Revision 1 - no extensions to language used}}
{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny]}}
{{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d]}}
<syntaxhighlight lang="algol68">PROC is perfect = (INT candidate)BOOL: (
INT sum :=1;
FOR f1 FROM 2 TO ENTIER ( sqrt(candidate)*(1+2*small real) ) WHILE
Line 37 ⟶ 589:
sum=candidate
);
test:(
FOR i FROM 2 TO 33550336 DO
IF is perfect(i) THEN print((i, new line)) FI
OD
)</syntaxhighlight>
{{Out}}
<pre>
+
+28
+496
+8128
+33550336
</pre>
=={{header|ALGOL W}}==
Based on the Algol 68 version.
<syntaxhighlight lang="algolw">begin
% returns true if n is perfect, false otherwise %
% n must be > 0 %
logical procedure isPerfect ( integer value candidate ) ;
begin
integer sum;
sum := 1;
for f1 := 2 until round( sqrt( candidate ) ) do begin
if candidate rem f1 = 0 then begin
integer f2;
sum := sum + f1;
f2 := candidate div f1;
% avoid e.g. counting 2 twice as a factor of 4 %
if f2 > f1 then sum := sum + f2
end if_candidate_rem_f1_eq_0 ;
end for_f1 ;
sum = candidate
end isPerfect ;
% test isPerfect %
for n := 2 until 10000 do if isPerfect( n ) then write( n );
end.</syntaxhighlight>
{{out}}
<pre>
6
28
496
8128
</pre>
=={{header|AppleScript}}==
===Functional===
{{Trans|JavaScript}}
<syntaxhighlight lang="applescript">-- PERFECT NUMBERS -----------------------------------------------------------
-- perfect :: integer -> bool
on perfect(n)
-- isFactor :: integer -> bool
script isFactor
on |λ|(x)
n mod x = 0
end |λ|
end script
-- quotient :: number -> number
script quotient
on |λ|(x)
n / x
end |λ|
end script
-- sum :: number -> number -> number
script sum
on |λ|(a, b)
a + b
end |λ|
end script
-- Integer factors of n below the square root
set lows to filter(isFactor, enumFromTo(1, (n ^ (1 / 2)) as integer))
-- low and high factors (quotients of low factors) tested for perfection
(n > 1) and (foldl(sum, 0, (lows & map(quotient, lows))) / 2 = n)
end perfect
-- TEST ----------------------------------------------------------------------
on run
filter(perfect, enumFromTo(1, 10000))
--> {6, 28, 496, 8128}
end run
-- GENERIC FUNCTIONS ---------------------------------------------------------
-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m > n then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo
-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn</syntaxhighlight>
{{Out}}
<syntaxhighlight lang="applescript">{6, 28, 496, 8128}</syntaxhighlight>
----
===Idiomatic===
====Sum of proper divisors====
<syntaxhighlight lang="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
on isPerfect(n)
if (n > 1.37438691328E+11) then return missing value -- Too high for perfection to be determinable.
-- All the known perfect numbers listed in Wikipedia end with either 6 or 28.
-- These endings are either preceded by odd digits or are the numbers themselves.
tell (n mod 10) to ¬
return ((((it = 6) and ((n mod 20 = 16) or (n = 6))) or ¬
((it = 8) and ((n mod 200 = 128) or (n = 28)))) and ¬
(my aliquotSum(n) = n))
end isPerfect
local output, n
set output to {}
repeat with n from 1 to 10000
if (isPerfect(n)) then set end of output to n
end repeat
return output</syntaxhighlight>
{{output}}
<syntaxhighlight lang="applescript">{6, 28, 496, 8128}</syntaxhighlight>
====Euclid====
<syntaxhighlight lang="applescript">on isPerfect(n)
-- All the known perfect numbers listed in Wikipedia end with either 6 or 28.
-- These endings are either preceded by odd digits or are the numbers themselves.
tell (n mod 10) to ¬
if not (((it = 6) and ((n mod 20 = 16) or (n = 6))) or ((it = 8) and ((n mod 200 = 128) or (n = 28)))) then ¬
return false
-- Work through the only seven primes p where (2 ^ p - 1) is also prime
-- and (2 ^ p - 1) * (2 ^ (p - 1)) is a number that AppleScript can handle.
repeat with p in {2, 3, 5, 7, 13, 17, 19}
tell (2 ^ p - 1) * (2 ^ (p - 1))
if (it < n) then
else
return (it = n)
end if
end tell
end repeat
return missing value
end isPerfect
local output, n
set output to {}
repeat with n from 2 to 33551000 by 2
if (isPerfect(n)) then set end of output to n
end repeat
return output</syntaxhighlight>
{{output}}
<syntaxhighlight lang="applescript">{6, 28, 496, 8128, 33550336}</syntaxhighlight>
====Practical====
But since AppleScript can only physically manage seven of the known perfect numbers, they may as well be in a look-up list for maximum efficiency:
<syntaxhighlight lang="applescript">on isPerfect(n)
if (n > 1.37438691328E+11) then return missing value -- Too high for perfection to be determinable.
return (n is in {6, 28, 496, 8128, 33550336, 8.589869056E+9, 1.37438691328E+11})
end isPerfect</syntaxhighlight>
=={{header|ARM Assembly}}==
{{works with|as|Raspberry Pi}}
<syntaxhighlight lang="arm assembly">
/* ARM assembly Raspberry PI */
/* program perfectNumber.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 MAXI, 1<<31
/*********************************/
/* Initialized data */
/*********************************/
.data
sMessResultPerf: .asciz "Perfect : @ \n"
szCarriageReturn: .asciz "\n"
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
sZoneConv: .skip 24
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: @ entry of program
mov r2,#2 @ begin first number
1: @ begin loop
mov r5,#1 @ sum
mov r4,#2 @ first divisor 1
2:
udiv r0,r2,r4 @ compute divisor 2
mls r3,r0,r4,r2 @ remainder
cmp r3,#0
bne 3f @ remainder = 0 ?
add r5,r5,r0 @ add divisor 2
add r5,r5,r4 @ add divisor 1
3:
add r4,r4,#1 @ increment divisor
cmp r4,r0 @ divisor 1 < divisor 2
blt 2b @ yes -> loop
cmp r2,r5 @ compare number and divisors sum
bne 4f @ not equal
mov r0,r2 @ equal -> display
ldr r1,iAdrsZoneConv
bl conversion10 @ call décimal conversion
ldr r0,iAdrsMessResultPerf
ldr r1,iAdrsZoneConv @ insert conversion in message
bl strInsertAtCharInc
bl affichageMess @ display message
4:
add r2,#2 @ no perfect number odd < 10 puis 1500
cmp r2,#MAXI @ end ?
blo 1b @ no -> loop
100: @ standard end of the program
mov r0, #0 @ return code
mov r7, #EXIT @ request to exit program
svc #0 @ perform the system call
iAdrszCarriageReturn: .int szCarriageReturn
iAdrsMessResultPerf: .int sMessResultPerf
iAdrsZoneConv: .int sZoneConv
/***************************************************/
/* ROUTINES INCLUDE */
/***************************************************/
.include "../affichage.inc"
</syntaxhighlight>
<pre>
Perfect : 6
Perfect : 28
Perfect : 496
Perfect : 8128
Perfect : 33550336
</pre>
=={{header|Arturo}}==
<syntaxhighlight lang="rebol">divisors: $[n][ select 1..(n/2)+1 'i -> 0 = n % i ]
perfect?: $[n][ n = sum divisors n ]
loop 2..1000 'i [
if perfect? i -> print i
]</syntaxhighlight>
=={{header|AutoHotkey}}==
This will find the first 8 perfect numbers.
<
If isMersennePrime(A_Index + 1)
res .= "Perfect number: " perfectNum(A_Index + 1) "`n"
Line 70 ⟶ 943:
Return false
Return true
}</
=={{header|AWK}}==
<
BEGIN{for(i=1;i<10000;i++)if(perf(i))print i}'
6
28
496
8128</
=={{header|Axiom}}==
{{trans|Mathematica}}
Using the interpreter, define the function:
<syntaxhighlight lang="axiom">perfect?(n:Integer):Boolean == reduce(+,divisors n) = 2*n</syntaxhighlight>
Alternatively, using the Spad compiler:
<syntaxhighlight lang="axiom">)abbrev package TESTP TestPackage
TestPackage() : withma
perfect?: Integer -> Boolean
==
add
import IntegerNumberTheoryFunctions
perfect? n == reduce("+",divisors n) = 2*n</syntaxhighlight>
Examples (testing 496, testing 128, finding all perfect numbers in 1...10000):
<syntaxhighlight lang="axiom">perfect? 496
perfect? 128
[i for i in 1..10000 | perfect? i]</syntaxhighlight>
{{Out}}
<syntaxhighlight lang="axiom">true
false
[6,28,496,8128]</syntaxhighlight>
=={{header|BASIC}}==
{{works with|QuickBasic|4.5}}
<
sum = 0
for i = 1 to n - 1
Line 94 ⟶ 989:
perf = 0
END IF
END FUNCTION</
==={{header|BASIC256}}===
{{trans|FreeBASIC}}
<syntaxhighlight lang="basic256">
function isPerfect(n)
if (n < 2) or (n mod 2 = 1) then return False
#asumimos que los números impares no son perfectos
sum = 1
for i = 2 to sqr(n)
if n mod i = 0 then
sum += i
q = n \ i
if q > i then sum += q
end if
next
return n = sum
end function
print "Los primeros 5 números perfectos son:"
for i = 2 to 233550336
if isPerfect(i) then print i; " ";
next i
end
</syntaxhighlight>
==={{header|Craft Basic}}===
<syntaxhighlight lang="basic">for n = 1 to 10000
let s = 0
for i = 1 to n / 2
if n % i = 0 then
let s = s + i
endif
next i
if s = n then
print n, " ",
endif
wait
next n</syntaxhighlight>
{{out| Output}}<pre>6 28 496 8128 </pre>
==={{header|IS-BASIC}}===
<syntaxhighlight lang="is-basic">100 PROGRAM "PerfectN.bas"
110 FOR X=1 TO 10000
120 IF PERFECT(X) THEN PRINT X;
130 NEXT
140 DEF PERFECT(N)
150 IF N<2 OR MOD(N,2)<>0 THEN LET PERFECT=0:EXIT DEF
160 LET S=1
170 FOR I=2 TO SQR(N)
180 IF MOD(N,I)=0 THEN LET S=S+I+N/I
190 NEXT
200 LET PERFECT=N=S
210 END DEF</syntaxhighlight>
==={{header|Sinclair ZX81 BASIC}}===
Call this subroutine and it will (eventually) return <tt>PERFECT</tt> = 1 if <tt>N</tt> is perfect or <tt>PERFECT</tt> = 0 if it is not.
<syntaxhighlight lang="basic">2000 LET SUM=0
2010 FOR F=1 TO N-1
2020 IF N/F=INT (N/F) THEN LET SUM=SUM+F
2030 NEXT F
2040 LET PERFECT=SUM=N
2050 RETURN</syntaxhighlight>
==={{header|True BASIC}}===
<syntaxhighlight lang="basic">
FUNCTION perf(n)
IF n < 2 or ramainder(n,2) = 1 then LET perf = 0
LET sum = 0
FOR i = 1 to n-1
IF remainder(n,i) = 0 then LET sum = sum+i
NEXT i
IF sum = n then
LET perf = 1
ELSE
LET perf = 0
END IF
END FUNCTION
PRINT "Los primeros 5 números perfectos son:"
FOR i = 1 to 33550336
IF perf(i) = 1 then PRINT i; " ";
NEXT i
PRINT
PRINT "Presione cualquier tecla para salir"
END
</syntaxhighlight>
=={{header|BBC BASIC}}==
===BASIC version===
<syntaxhighlight lang="bbcbasic"> FOR n% = 2 TO 10000 STEP 2
IF FNperfect(n%) PRINT n%
NEXT
END
DEF FNperfect(N%)
LOCAL I%, S%
S% = 1
FOR I% = 2 TO SQR(N%)-1
IF N% MOD I% = 0 S% += I% + N% DIV I%
NEXT
IF I% = SQR(N%) S% += I%
= (N% = S%)</syntaxhighlight>
{{Out}}
<pre>
6
28
496
8128
</pre>
===Assembler version===
{{works with|BBC BASIC for Windows}}
<syntaxhighlight lang="bbcbasic"> DIM P% 100
[OPT 2 :.S% xor edi,edi
.perloop mov eax,ebx : cdq : div ecx : or edx,edx : loopnz perloop : inc ecx
add edi,ecx : add edi,eax : loop perloop : mov eax,edi : shr eax,1 : ret : ]
FOR B% = 2 TO 35000000 STEP 2
C% = SQRB%
IF B% = USRS% PRINT B%
NEXT
END</syntaxhighlight>
{{Out}}
<pre>
4
6
28
496
8128
33550336
</pre>
=={{header|Bracmat}}==
<syntaxhighlight lang="bracmat">( ( perf
= sum i
. 0:?sum
& 0:?i
& whl
' ( !i+1:<!arg:?i
& ( mod$(!arg.!i):0&!sum+!i:?sum
|
)
)
& !sum:!arg
)
& 0:?n
& whl
' ( !n+1:~>10000:?n
& (perf$!n&out$!n|)
)
);</syntaxhighlight>
{{Out}}
<pre>6
28
496
8128</pre>
=={{header|Burlesque}}==
<syntaxhighlight lang="burlesque">Jfc++\/2.*==</syntaxhighlight>
<syntaxhighlight lang="burlesque">blsq) 8200ro{Jfc++\/2.*==}f[
{6 28 496 8128}</syntaxhighlight>
=={{header|C}}==
{{trans|D}}
<syntaxhighlight lang="c">#include "stdio.h"
#include "math.h"
Line 125 ⟶ 1,195:
return 0;
}</
Using functions from [[Factors of an integer#Prime factoring]]:
<syntaxhighlight lang="c">int main()
{
int j;
ulong fac[10000], n, sum;
sieve();
for (n = 2; n < 33550337; n++) {
j = get_factors(n, fac) - 1;
for (sum = 0; j && sum <= n; sum += fac[--j]);
if (sum == n) printf("%lu\n", n);
}
return 0;
}</syntaxhighlight>
=={{header|C sharp|C#}}==
{{trans|C++}}
<syntaxhighlight lang="csharp">static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
int sum = 0;
for (int i = 1; i < num; i++)
{
if (num % i == 0)
sum += i;
}
return sum == num ;
}</syntaxhighlight>
===Version using Lambdas, will only work from version 3 of C# on===
<syntaxhighlight lang="csharp">static void Main(string[] args)
{
Console.WriteLine("Perfect numbers from 1 to 33550337:");
for (int x = 0; x < 33550337; x++)
{
if (IsPerfect(x))
Console.WriteLine(x + " is perfect.");
}
Console.ReadLine();
}
static bool IsPerfect(int num)
{
return Enumerable.Range(1, num - 1).Sum(n => num % n == 0 ? n : 0 ) == num;
}</syntaxhighlight>
=={{header|C++}}==
{{works with|gcc}}
<
using namespace std ;
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum;
}
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if (
cout << num << '\n' ;
}
return 0 ;
}
</syntaxhighlight>
=={{header|Clojure}}==
<
(if (< n 4)
(
(filter #(zero? (rem n %)))
(cons 1))))
(defn perfect? [n]
(
{{trans|Haskell}}
<syntaxhighlight lang="clojure">(defn perfect? [n]
(->> (for [i (range 1 n)] :when (zero? (rem n i))] i)
(reduce +)
(= n)))</syntaxhighlight>
===Functional version===
<syntaxhighlight lang="clojure">(defn perfect? [n]
(= (reduce + (filter #(zero? (rem n %)) (range 1 n))) n))</syntaxhighlight>
=={{header|COBOL}}==
{{trans|D}}
{{works with|Visual COBOL}}
main.cbl:
<syntaxhighlight lang="cobol"> $set REPOSITORY "UPDATE ON"
IDENTIFICATION DIVISION.
PROGRAM-ID. perfect-main.
ENVIRONMENT DIVISION.
CONFIGURATION SECTION.
REPOSITORY.
FUNCTION perfect
.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 i PIC 9(8).
PROCEDURE DIVISION.
PERFORM VARYING i FROM 2 BY 1 UNTIL 33550337 = i
IF FUNCTION perfect(i) = 0
DISPLAY i
END-IF
END-PERFORM
GOBACK
.
END PROGRAM perfect-main.</syntaxhighlight>
perfect.cbl:
<syntaxhighlight lang="cobol"> IDENTIFICATION DIVISION.
FUNCTION-ID. perfect.
DATA DIVISION.
LOCAL-STORAGE SECTION.
01 max-val PIC 9(8).
01 total PIC 9(8) VALUE 1.
01 i PIC 9(8).
01 q PIC 9(8).
LINKAGE SECTION.
01 n PIC 9(8).
01 is-perfect PIC 9.
PROCEDURE DIVISION USING VALUE n RETURNING is-perfect.
COMPUTE max-val = FUNCTION INTEGER(FUNCTION SQRT(n)) + 1
PERFORM VARYING i FROM 2 BY 1 UNTIL i = max-val
IF FUNCTION MOD(n, i) = 0
ADD i TO total
DIVIDE n BY i GIVING q
IF q > i
ADD q TO total
END-IF
END-IF
END-PERFORM
IF total = n
MOVE 0 TO is-perfect
ELSE
MOVE 1 TO is-perfect
END-IF
GOBACK
.
END FUNCTION perfect.</syntaxhighlight>
=={{header|CoffeeScript}}==
Optimized version, for fun.
<syntaxhighlight lang="coffeescript">is_perfect_number = (n) ->
do_factors_add_up_to n, 2*n
do_factors_add_up_to = (n, desired_sum) ->
# We mildly optimize here, by taking advantage of
# the fact that the sum_of_factors( (p^m) * x)
# is (1 + ... + p^m-1 + p^m) * sum_factors(x) when
# x is not itself a multiple of p.
p = smallest_prime_factor(n)
if p == n
return desired_sum == p + 1
# ok, now sum up all powers of p that
# divide n
sum_powers = 1
curr_power = 1
while n % p == 0
curr_power *= p
sum_powers += curr_power
n /= p
# if desired_sum does not divide sum_powers, we
# can short circuit quickly
return false unless desired_sum % sum_powers == 0
# otherwise, recurse
do_factors_add_up_to n, desired_sum / sum_powers
smallest_prime_factor = (n) ->
for i in [2..n]
return n if i*i > n
return i if n % i == 0
# tests
do ->
# This is pretty fast...
for n in [2..100000]
console.log n if is_perfect_number n
# For big numbers, let's just sanity check the known ones.
known_perfects = [
33550336
8589869056
137438691328
]
for n in known_perfects
throw Error("fail") unless is_perfect_number(n)
throw Error("fail") if is_perfect_number(n+1)</syntaxhighlight>
{{Out}}
<pre>
> coffee perfect_numbers.coffee
6
28
496
8128
</pre>
=={{header|Common Lisp}}==
{{trans|Haskell}}
<
(= n (loop for i from 1 below n when (= 0 (mod n i)) sum i)))</
=={{header|D}}==
===Functional Version===
<
bool isPerfectNumber1(in uint n) pure nothrow
in {
assert(n > 0);
} body {
return n == iota(1, n - 1).filter!(i => n % i == 0).sum;
}
void main() {
iota(1, 10_000).filter!isPerfectNumber1.writeln;
}</syntaxhighlight>
{{out}}
<pre>[6, 28, 496, 8128]</pre>
===Faster Imperative Version===
{{trans|Algol}}
<syntaxhighlight lang="d">import std.stdio, std.math, std.range, std.algorithm;
bool
if (n < 2)
return false;
foreach (immutable i; 2 .. cast(int)real(n).sqrt + 1)
if (n % i == 0) {
if (q > i)
}
return
}
void main() {
10_000.iota.filter!isPerfectNumber2.writeln;
}</syntaxhighlight>
{{out}}
<pre>[6, 28, 496, 8128]</pre>
With a <code>33_550_337.iota</code> it outputs:
<pre>[6, 28, 496, 8128, 33550336]</pre>
=={{header|
=== Explicit Iterative Version ===
<syntaxhighlight lang="d">/*
* Function to test if a number is a perfect number
* A number is a perfect number if it is equal to the sum of all its divisors
* Input: Positive integer n
* Output: true if n is a perfect number, false otherwise
*/
bool isPerfect(int n){
//Generate a list of integers in the range 1 to n-1 : [1, 2, ..., n-1]
List<int> range = new List<int>.generate(n-1, (int i) => i+1);
//Create a list that filters the divisors of n from range
List<int> divisors = new List.from(range.where((i) => n%i == 0));
//Sum the all the divisors
int sumOfDivisors = 0;
for (int i = 0; i < divisors.length; i++){
sumOfDivisors = sumOfDivisors + divisors[i];
}
// A number is a perfect number if it is equal to the sum of its divisors
// We return the test if n is equal to sumOfDivisors
return n == sumOfDivisors;
}</syntaxhighlight>
=== Compact Version ===
{{trans|Julia}}
<syntaxhighlight lang="d">isPerfect(n) =>
n == new List.generate(n-1, (i) => n%(i+1) == 0 ? i+1 : 0).fold(0, (p,n)=>p+n);</syntaxhighlight>
In either case, if we test to find all the perfect numbers up to 1000, we get:
<syntaxhighlight lang="d">main() =>
new List.generate(1000,(i)=>i+1).where(isPerfect).forEach(print);</syntaxhighlight>
{{out}}
<pre>6
28
496</pre>
=={{header|Delphi}}==
See [[#Pascal]].
=={{header|Dyalect}}==
<syntaxhighlight lang="dyalect">func isPerfect(num) {
var sum = 0
for i in 1..<num {
if !i {
break
}
if num % i == 0 {
sum += i
}
}
return sum == num
}
let max = 33550337
print("Perfect numbers from 0 to \(max):")
for x in 0..max {
if isPerfect(x) {
print("\(x) is perfect")
}
}</syntaxhighlight>
=={{header|E}}==
<syntaxhighlight lang="e">pragma.enable("accumulator")
def isPerfectNumber(x :int) {
var sum := 0
Line 203 ⟶ 1,553:
}
return sum <=> x
}</
=={{header|EasyLang}}==
<syntaxhighlight lang=easylang>
func perf n .
for i = 1 to n - 1
if n mod i = 0
sum += i
.
.
return if sum = n
.
for i = 2 to 10000
if perf i = 1
print i
.
.
</syntaxhighlight>
=={{header|Eiffel}}==
<syntaxhighlight lang="eiffel">
class
APPLICATION
create
make
feature
make
do
io.put_string (" 6 is perfect...%T")
io.put_boolean (is_perfect_number (6))
io.new_line
io.put_string (" 77 is perfect...%T")
io.put_boolean (is_perfect_number (77))
io.new_line
io.put_string ("128 is perfect...%T")
io.put_boolean (is_perfect_number (128))
io.new_line
io.put_string ("496 is perfect...%T")
io.put_boolean (is_perfect_number (496))
end
is_perfect_number (n: INTEGER): BOOLEAN
-- Is 'n' a perfect number?
require
n_positive: n > 0
local
sum: INTEGER
do
across
1 |..| (n - 1) as c
loop
if n \\ c.item = 0 then
sum := sum + c.item
end
end
Result := sum = n
end
end
</syntaxhighlight>
{{out}}
<pre>
6 is perfect... True
77 is perfect... False
128 is perfect... False
496 is perfect... True
</pre>
=={{header|Elena}}==
ELENA 6.x:
<syntaxhighlight lang="elena">import system'routines;
import system'math;
import extensions;
extension extension
{
isPerfect()
= new Range(1, self - 1).selectBy::(n => (self.mod(n) == 0).iif(n,0) ).summarize(new Integer()) == self;
}
public program()
{
for(int n := 1; n < 10000; n += 1)
{
if(n.isPerfect())
{ console.printLine(n," is perfect") }
};
console.readChar()
}</syntaxhighlight>
{{out}}
<pre>
6 is perfect
28 is perfect
496 is perfect
8128 is perfect
</pre>
=={{header|Elixir}}==
<syntaxhighlight lang="elixir">defmodule RC do
def is_perfect(1), do: false
def is_perfect(n) when n > 1 do
Enum.sum(factor(n, 2, [1])) == n
end
defp factor(n, i, factors) when n < i*i , do: factors
defp factor(n, i, factors) when n == i*i , do: [i | factors]
defp factor(n, i, factors) when rem(n,i)==0, do: factor(n, i+1, [i, div(n,i) | factors])
defp factor(n, i, factors) , do: factor(n, i+1, factors)
end
IO.inspect (for i <- 1..10000, RC.is_perfect(i), do: i)</syntaxhighlight>
{{out}}
<pre>
[6, 28, 496, 8128]
</pre>
=={{header|Erlang}}==
<
X == lists:sum([N || N <- lists:seq(1,X-1), X rem N == 0]).</
=={{header|ERRE}}==
<syntaxhighlight lang="erre">PROGRAM PERFECT
PROCEDURE PERFECT(N%->OK%)
LOCAL I%,S%
S%=1
FOR I%=2 TO SQR(N%)-1 DO
IF N% MOD I%=0 THEN S%+=I%+N% DIV I%
END FOR
IF I%=SQR(N%) THEN S%+=I%
OK%=(N%=S%)
END PROCEDURE
BEGIN
PRINT(CHR$(12);) ! CLS
FOR N%=2 TO 10000 STEP 2 DO
PERFECT(N%->OK%)
IF OK% THEN PRINT(N%)
END FOR
END PROGRAM</syntaxhighlight>
{{Out}}
<pre>
6
28
496
8128
</pre>
=={{header|F_Sharp|F#}}==
<syntaxhighlight lang="fsharp">let perf n = n = List.fold (+) 0 (List.filter (fun i -> n % i = 0) [1..(n-1)])
for i in 1..10000 do if (perf i) then printfn "%i is perfect" i</syntaxhighlight>
{{Out}}
<pre>6 is perfect
28 is perfect
496 is perfect
8128 is perfect</pre>
=={{header|Factor}}==
<syntaxhighlight lang="factor">USING: kernel math math.primes.factors sequences ;
IN: rosettacode.perfect-numbers
: perfect? ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;</syntaxhighlight>
=={{header|FALSE}}==
<syntaxhighlight lang="false">[0\1[\$@$@-][\$@$@$@$@\/*=[@\$@+@@]?1+]#%=]p:
45p;!." "28p;!. { 0 -1 }</syntaxhighlight>
=={{header|Forth}}==
<
1
over 2/ 1+ 2 ?do
over i mod 0= if i + then
loop
= ;</
=={{header|Fortran}}==
{{works with|Fortran|90 and later}}
<
LOGICAL :: isPerfect
INTEGER, INTENT(IN) :: n
Line 230 ⟶ 1,747:
END DO
IF (factorsum == n) isPerfect = .TRUE.
END FUNCTION isPerfect</
=={{header|FreeBASIC}}==
{{trans|C (with some modifications)}}
<syntaxhighlight lang="freebasic">' FB 1.05.0 Win64
Function isPerfect(n As Integer) As Boolean
If n < 2 Then Return False
If n Mod 2 = 1 Then Return False '' we can assume odd numbers are not perfect
Dim As Integer sum = 1, q
For i As Integer = 2 To Sqr(n)
If n Mod i = 0 Then
sum += i
q = n \ i
If q > i Then sum += q
End If
Next
Return n = sum
End Function
Print "The first 5 perfect numbers are : "
For i As Integer = 2 To 33550336
If isPerfect(i) Then Print i; " ";
Next
Print
Print "Press any key to quit"
Sleep</syntaxhighlight>
{{out}}
<pre>
The first 5 perfect numbers are :
6 28 496 8128 33550336
</pre>
=={{header|Frink}}==
<syntaxhighlight lang="frink">isPerfect = {|n| sum[allFactors[n, true, false]] == n}
println[select[1 to 1000, isPerfect]]</syntaxhighlight>
{{out}}
<pre>[1, 6, 28, 496]
</pre>
=={{header|FunL}}==
<syntaxhighlight lang="funl">def perfect( n ) = sum( d | d <- 1..n if d|n ) == 2n
println( (1..500).filter(perfect) )</syntaxhighlight>
{{out}}
<pre>
(6, 28, 496)
</pre>
=={{header|FutureBasic}}==
<syntaxhighlight lang="futurebasic">
_maxNum = 10000
local fn IsPerfectNumber( n as long ) as BOOL
—————————————————————————————————————————————
if ( n < 2 ) then exit fn = NO
if ( n mod 2 == 1 ) then exit fn = NO
long sum = 1, q, i
for i = 2 to sqr(n)
if ( n mod i == 0 )
sum += i
q = n / i
if ( q > i ) then sum += q
end if
next
end fn = ( n == sum )
printf @"Perfect numbers in range %ld..%ld",2,_maxNum
long i
for i = 2 To _maxNum
if ( fn IsPerfectNumber(i) ) then print i
next
HandleEvents
</syntaxhighlight>
{{out}}
<pre>
Perfect numbers in range 2..10000
6
28
496
8128
</pre>
=={{header|GAP}}==
<syntaxhighlight lang="gap">Filtered([1 .. 10000], n -> Sum(DivisorsInt(n)) = 2*n);
# [ 6, 28, 496, 8128 ]</syntaxhighlight>
=={{header|Go}}==
<syntaxhighlight lang="go">package main
import "fmt"
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
// following function satisfies the task, returning true for all
// perfect numbers representable in the argument type
func isPerfect(n int64) bool {
switch n {
case 6, 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128:
return true
}
return false
}
// validation
func main() {
for n := int64(1); ; n++ {
if isPerfect(n) != computePerfect(n) {
panic("bug")
}
if n%1e3 == 0 {
fmt.Println("tested", n)
}
}
}
</syntaxhighlight>
{{Out}}
<pre>
tested 1000
tested 2000
tested 3000
...
</pre>
=={{header|Groovy}}==
Solution:
<
n > 4 && (n == (2..Math.sqrt(n)).findAll { n % it == 0 }.inject(1) { factorSum, i -> factorSum += i + n/i })
}</
Test program:
<
{{Out}}
<pre>6
28
496
8128</pre>
=={{header|Haskell}}==
<syntaxhighlight lang="haskell">perfect n =
n == sum [i | i <- [1..n-1], n `mod` i == 0]</syntaxhighlight>
Create a list of known perfects:
<syntaxhighlight lang="haskell">perfect =
(\x -> (2 ^ x - 1) * (2 ^ (x - 1))) <$>
filter (\x -> isPrime x && isPrime (2 ^ x - 1)) maybe_prime
where
maybe_prime = scanl1 (+) (2 : 1 : cycle [2, 2, 4, 2, 4, 2, 4, 6])
isPrime n = all ((/= 0) . (n `mod`)) $ takeWhile (\x -> x * x <= n) maybe_prime
isPerfect n = f n perfect
where
f n (p:ps) =
case compare n p of
EQ -> True
LT -> False
GT -> f n ps
main :: IO ()
main = do
mapM_ print $ take 10 perfect
mapM_ (print . (\x -> (x, isPerfect x))) [6, 27, 28, 29, 496, 8128, 8129]</syntaxhighlight>
or, restricting the search space to improve performance:
<syntaxhighlight lang="haskell">isPerfect :: Int -> Bool
isPerfect n =
let lows = filter ((0 ==) . rem n) [1 .. floor (sqrt (fromIntegral n))]
in 1 < n &&
n ==
quot
(sum
(lows ++
[ y
| x <- lows
, let y = quot n x
, x /= y ]))
2
main :: IO ()
main = print $ filter isPerfect [1 .. 10000]</syntaxhighlight>
{{Out}}
<pre>[6,28,496,8128]</pre>
=={{header|HicEst}}==
<syntaxhighlight lang="hicest"> DO i = 1, 1E4
IF( perfect(i) ) WRITE() i
ENDDO
END ! end of "main"
FUNCTION perfect(n)
sum = 1
DO i = 2, n^0.5
sum = sum + (MOD(n, i) == 0) * (i + INT(n/i))
ENDDO
perfect = sum == n
END</syntaxhighlight>
=={{header|Icon}} and {{header|Unicon}}==
<syntaxhighlight lang="icon">procedure main(arglist)
limit := \arglist[1] | 100000
write("Perfect numbers from 1 to ",limit,":")
every write(isperfect(1 to limit))
write("Done.")
end
procedure isperfect(n) #: returns n if n is perfect
local sum,i
every (sum := 0) +:= (n ~= divisors(n))
if sum = n then return n
end
link factors</syntaxhighlight>
{{libheader|Icon Programming Library}} [http://www.cs.arizona.edu/icon/library/src/procs/factors.icn Uses divisors from factors]
{{Out}}
<pre>Perfect numbers from 1 to 100000:
6
28
496
8128
Done.</pre>
=={{header|J}}==
<
Examples of use, including extensions beyond those assumptions:
<
1
6 28 496 8128
Line 263 ⟶ 2,004:
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29
is_perfect zero_through_twentynine
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0
is_perfect 191561942608236107294793378084303638130997321548169216x
1</syntaxhighlight>
More efficient version based on [http://jsoftware.com/pipermail/programming/2014-June/037695.html comments] by Henry Rich and Roger Hui (comment train seeded by Jon Hough).
=={{header|Java}}==
<
int sum= 0;
for(int i= 1;i < n;i++){
Line 278 ⟶ 2,022:
}
return sum == n;
}</
Or for arbitrary precision:[[Category:Arbitrary precision]]
<
public static boolean perf(BigInteger n){
Line 286 ⟶ 2,030:
for(BigInteger i= BigInteger.ONE;
i.compareTo(n) < 0;i=i.add(BigInteger.ONE)){
if(n.mod(i).
sum= sum.add(i);
}
}
return sum.
}</
=={{header|JavaScript}}==
===Imperative===
{{trans|Java}}
<syntaxhighlight lang
{
for (i = sqrt-1; i>1; i--)
{
if (n % i == 0)
sum
}
}
if(n % sqrt == 0)
sum += sqrt + (sqrt*sqrt == n ? 0 : n/sqrt);
return sum === n;
}
var i;
for (i = 1; i < 10000; i++)
{
if (is_perfect(i))
print(i);
}</syntaxhighlight>
{{Out}}
<pre>6
28
Line 317 ⟶ 2,070:
8128</pre>
===Functional===
====ES5====
Naive version (brute force)
<syntaxhighlight lang="javascript">(function (nFrom, nTo) {
function perfect(n) {
return n === range(1, n - 1).reduce(
function (a, x) {
return n % x ? a : a + x;
}, 0
);
}
function range(m, n) {
return Array.apply(null, Array(n - m + 1)).map(function (x, i) {
return m + i;
});
}
return range(nFrom, nTo).filter(perfect);
})(1, 10000);</syntaxhighlight>
Output:
<syntaxhighlight lang="javascript">[6, 28, 496, 8128]</syntaxhighlight>
Much faster (more efficient factorisation)
<syntaxhighlight lang="javascript">(function (nFrom, nTo) {
function perfect(n) {
var lows = range(1, Math.floor(Math.sqrt(n))).filter(function (x) {
return (n % x) === 0;
});
return n > 1 && lows.concat(lows.map(function (x) {
return n / x;
})).reduce(function (a, x) {
return a + x;
}, 0) / 2 === n;
}
function range(m, n) {
return Array.apply(null, Array(n - m + 1)).map(function (x, i) {
return m + i;
});
}
return range(nFrom, nTo).filter(perfect)
})(1, 10000);</syntaxhighlight>
Output:
<syntaxhighlight lang="javascript">[6, 28, 496, 8128]</syntaxhighlight>
Note that the filter function, though convenient and well optimised, is not strictly necessary.
We can always replace it with a more general monadic bind (chain) function, which is essentially just concat map
(Monadic return/inject for lists is simply lambda x --> [x], inlined here, and fail is [].)
<syntaxhighlight lang="javascript">(function (nFrom, nTo) {
// MONADIC CHAIN (bind) IN LIEU OF FILTER
// ( monadic return for lists is just lambda x -> [x] )
return chain(
rng(nFrom, nTo),
function mPerfect(n) {
return (chain(
rng(1, Math.floor(Math.sqrt(n))),
function (y) {
return (n % y) === 0 && n > 1 ? [y, n / y] : [];
}
).reduce(function (a, x) {
return a + x;
}, 0) / 2 === n) ? [n] : [];
}
);
/******************************************************************/
// Monadic bind (chain) for lists
function chain(xs, f) {
return [].concat.apply([], xs.map(f));
}
function rng(m, n) {
return Array.apply(null, Array(n - m + 1)).map(function (x, i) {
return m + i;
});
}
})(1, 10000);</syntaxhighlight>
Output:
<syntaxhighlight lang="javascript">[6, 28, 496, 8128]</syntaxhighlight>
====ES6====
<syntaxhighlight lang="javascript">(() => {
const main = () =>
enumFromTo(1, 10000).filter(perfect);
// perfect :: Int -> Bool
const perfect = n => {
const
lows = enumFromTo(1, Math.floor(Math.sqrt(n)))
.filter(x => (n % x) === 0);
return n > 1 && lows.concat(lows.map(x => n / x))
.reduce((a, x) => (a + x), 0) / 2 === n;
};
// GENERIC --------------------------------------------
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: n - m + 1
}, (_, i) => i + m)
// MAIN ---
return main();
})();</syntaxhighlight>
{{Out}}
<syntaxhighlight lang="javascript">[6, 28, 496, 8128]</syntaxhighlight>
=={{header|jq}}==
<syntaxhighlight lang="jq">
def is_perfect:
. as $in
| $in == reduce range(1;$in) as $i
(0; if ($in % $i) == 0 then $i + . else . end);
# Example:
range(1;10001) | select( is_perfect )</syntaxhighlight>
{{Out}}
$ jq -n -f is_perfect.jq
6
28
496
8128
=={{header|Julia}}==
{{works with|Julia|0.6}}
<syntaxhighlight lang="julia">isperfect(n::Integer) = n == sum([n % i == 0 ? i : 0 for i = 1:(n - 1)])
perfects(n::Integer) = filter(isperfect, 1:n)
@show perfects(10000)</syntaxhighlight>
{{out}}
<pre>perfects(10000) = [6, 28, 496, 8128]</pre>
=={{header|K}}==
{{trans|J}}
<syntaxhighlight lang="k"> perfect:{(x>2)&x=+/-1_{d:&~x!'!1+_sqrt x;d,_ x%|d}x}
perfect 33550336
1
a@&perfect'a:!10000
6 28 496 8128
m:3 10#!30
(0 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29)
perfect'/: m
(0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0)</syntaxhighlight>
=={{header|Kotlin}}==
{{trans|C}}
<syntaxhighlight lang="scala">// version 1.0.6
fun isPerfect(n: Int): Boolean = when {
n < 2 -> false
n % 2 == 1 -> false // there are no known odd perfect numbers
else -> {
var tot = 1
var q: Int
for (i in 2 .. Math.sqrt(n.toDouble()).toInt()) {
if (n % i == 0) {
tot += i
q = n / i
if (q > i) tot += q
}
}
n == tot
}
}
fun main(args: Array<String>) {
// expect a run time of about 6 minutes on a typical laptop
println("The first five perfect numbers are:")
for (i in 2 .. 33550336) if (isPerfect(i)) print("$i ")
}</syntaxhighlight>
{{out}}
<pre>
The first five perfect numbers are:
6 28 496 8128 33550336
</pre>
=={{header|LabVIEW}}==
{{VI solution|LabVIEW_Perfect_numbers.png}}
=={{header|Lambdatalk}}==
===simple & slow===
<syntaxhighlight lang="scheme">
{def perf
{def perf.sum
{lambda {:n :sum :i}
{if {>= :i :n}
then {= :sum :n}
else {perf.sum :n
{if {= {% :n :i} 0}
then {+ :sum :i}
else :sum}
{+ :i 1}} }}}
{lambda {:n}
{perf.sum :n 0 2} }}
-> perf
{S.replace \s by space in
{S.map {lambda {:i} {if {perf :i} then :i else}}
{S.serie 2 1000 2}}}
-> 6 28 496 // 5200ms
</syntaxhighlight>
Too slow (and stackoverflow) to go further.
===improved===
<syntaxhighlight lang="scheme">
{def lt_perfect
{def lt_perfect.sum
{lambda {:n :sum :i}
{if {> :i 1}
then {lt_perfect.sum :n
{if {= {% :n :i} 0}
then {+ :sum :i {floor {/ :n :i}}}
else :sum}
{- :i 1}}
else :sum }}}
{lambda {:n}
{let { {:n :n}
{:sqrt {floor {sqrt :n}}}
{:sum {lt_perfect.sum :n 1 {- {floor {sqrt :n}} 0} }}
{:foo {if {= {* :sqrt :sqrt} :n}
then 0
else {floor {/ :n :sqrt}}}}
} {= :n {if {= {% :n :sqrt} 0}
then {+ :sum :sqrt :foo}
else :sum}} }}}
-> lt_perfect
-> {S.replace \s by space in
{S.map {lambda {:i} {if {lt_perfect :i} then :i else}}
{S.serie 6 10000 2}}}
-> 28 496 8128 // 7500ms
</syntaxhighlight>
===calling javascript===
Following the javascript entry.
<syntaxhighlight lang="scheme">
{S.replace \s by space in
{S.map {lambda {:i} {if {js_perfect :i} then :i else}}
{S.serie 2 10000}}}
-> 6 28 496 8128 // 80ms
{script
LAMBDATALK.DICT["js_perfect"] = function() {
function js_perfect(n) {
var sum = 1, i, sqrt=Math.floor(Math.sqrt(n));
for (i = sqrt-1; i>1; i--) {
if (n % i == 0)
sum += i + n/i;
}
if(n % sqrt == 0)
sum += sqrt + (sqrt*sqrt == n ? 0 : n/sqrt);
return sum === n;
}
var args = arguments[0].trim();
return (js_perfect( Number(args) )) ? "true" : "false"
};
}
</syntaxhighlight>
=={{header|Lasso}}==
<syntaxhighlight lang="lasso">#!/usr/bin/lasso9
define isPerfect(n::integer) => {
#n < 2 ? return false
return #n == (
with i in generateSeries(1, math_floor(math_sqrt(#n)) + 1)
where #n % #i == 0
let q = #n / #i
sum (#q > #i ? (#i == 1 ? 1 | #q + #i) | 0)
)
}
with x in generateSeries(1, 10000)
where isPerfect(#x)
select #x</syntaxhighlight>
{{Out}}
<syntaxhighlight lang="lasso">6, 28, 496, 8128</syntaxhighlight>
=={{header|Liberty BASIC}}==
<syntaxhighlight lang="lb">for n =1 to 10000
if perfect( n) =1 then print n; " is perfect."
next n
end
function perfect( n)
sum =0
for i =1 TO n /2
if n mod i =0 then
sum =sum +i
end if
next i
if sum =n then
perfect= 1
else
perfect =0
end if
end function</syntaxhighlight>
=={{header|Lingo}}==
<syntaxhighlight lang="lingo">on isPercect (n)
sum = 1
cnt = n/2
repeat with i = 2 to cnt
if n mod i = 0 then sum = sum + i
end repeat
return sum=n
end</syntaxhighlight>
=={{header|Logo}}==
<
output equal? :n apply "sum filter [equal? 0 modulo :n ?] iseq 1 :n/2
end</
=={{header|Lua}}==
<syntaxhighlight lang="lua">function isPerfect(x)
local sum = 0
for i = 1, x-1 do
sum = (x % i) == 0 and sum + i or sum
end
return sum == x
end</syntaxhighlight>
=={{header|M2000 Interpreter}}==
<syntaxhighlight lang="m2000 interpreter">
Module PerfectNumbers {
Function Is_Perfect(n as decimal) {
s=1 : sN=Sqrt(n)
last= n=sN*sN
t=n
If n mod 2=0 then s+=2+n div 2
i=3 : sN--
While i<sN {
if n mod i=0 then t=n div i :i=max.data(n div t, i): s+=t+ i
i++
}
=n=s
}
Inventory Known1=2@, 3@
IsPrime=lambda Known1 (x as decimal) -> {
=0=1
if exist(Known1, x) then =1=1 : exit
if x<=5 OR frac(x) then {if x == 2 OR x == 3 OR x == 5 then Append Known1, x : =1=1
Break}
if frac(x/2) else exit
if frac(x/3) else exit
x1=sqrt(x):d = 5@
{if frac(x/d ) else exit
d += 2: if d>x1 then Append Known1, x : =1=1 : exit
if frac(x/d) else exit
d += 4: if d<= x1 else Append Known1, x : =1=1: exit
loop}
}
\\ Check a perfect and a non perfect number
p=2 : n=3 : n1=2
Document Doc$
IsPerfect( 0, 28)
IsPerfect( 0, 1544)
While p<32 { ' max 32
if isprime(2^p-1@) then {
perf=(2^p-1@)*2@^(p-1@)
Rem Print perf
\\ decompose pretty fast the Perferct Numbers
\\ all have a series of 2 and last a prime equal to perf/2^(p-1)
inventory queue factors
For i=1 to p-1 {
Append factors, 2@
}
Append factors, perf/2^(p-1)
\\ end decompose
Rem Print factors
IsPerfect(factors, Perf)
}
p++
}
Clipboard Doc$
\\ exit here. No need for Exit statement
Sub IsPerfect(factors, n)
s=false
if n<10000 or type$(factors)<>"Inventory" then {
s=Is_Perfect(n)
} else {
local mm=each(factors, 1, -2), f =true
while mm {if eval(mm)<>2 then f=false
}
if f then if n/2@**(len(mm)-1)= factors(len(factors)-1!) then s=true
}
Local a$=format$("{0} is {1}perfect number", n, If$(s->"", "not "))
Doc$=a$+{
}
Print a$
End Sub
}
PerfectNumbers
</syntaxhighlight>
{{out}}
<pre style="height:30ex;overflow:scroll">
28 is perfect number
1544 is not perfect number
6 is perfect number
28 is perfect number
496 is perfect number
8128 is perfect number
33550336 is perfect number
8589869056 is perfect number
137438691328 is perfect number
2305843008139952128 is perfect number
</pre >
=={{header|M4}}==
<
`ifelse($#,0,``$0'',
`ifelse(eval($2<=$3),1,
Line 342 ⟶ 2,547:
for(`x',`2',`33550336',
`ifelse(isperfect(x),1,`x
')')</
=={{header|
<syntaxhighlight lang="mad"> NORMAL MODE IS INTEGER
R FUNCTION THAT CHECKS IF NUMBER IS PERFECT
INTERNAL FUNCTION(N)
ENTRY TO PERFCT.
DSUM = 0
THROUGH SUMMAT, FOR CAND=1, 1, CAND.GE.N
SUMMAT WHENEVER N/CAND*CAND.E.N, DSUM = DSUM+CAND
FUNCTION RETURN DSUM.E.N
END OF FUNCTION
R PRINT PERFECT NUMBERS UP TO 10,000
THROUGH SHOW, FOR I=1, 1, I.G.10000
SHOW WHENEVER PERFCT.(I), PRINT FORMAT FMT,I
VECTOR VALUES FMT = $I5*$
PRINT COMMENT $ $
END OF PROGRAM
</syntaxhighlight>
{{out}}
<pre> 6
28
496
8128
</pre>
=={{header|Maple}}==
<syntaxhighlight lang="maple">isperfect := proc(n) return evalb(NumberTheory:-SumOfDivisors(n) = 2*n); end proc:
isperfect(6);
true</syntaxhighlight>
=={{header|Mathematica}} / {{header|Wolfram Language}}==
Custom function:
<
Examples (testing 496, testing 128, finding all perfect numbers in 1...10000):
<
PerfectQ[128]
Flatten[PerfectQ/@Range[10000]//Position[#,True]&]</
gives back:
<syntaxhighlight lang="mathematica">True
False
{6,28,496,8128}</
=={{header|MATLAB}}==
Standard algorithm:
<syntaxhighlight lang="matlab">function perf = isPerfect(n)
total = 0;
for k = 1:n-1
if ~mod(n, k)
total = total+k;
end
end
perf = total == n;
end</syntaxhighlight>
Faster algorithm:
<syntaxhighlight lang="matlab">function perf = isPerfect(n)
if n < 2
perf = false;
else
total = 1;
k = 2;
quot = n;
while k < quot && total <= n
if ~mod(n, k)
total = total+k;
quot = n/k;
if quot ~= k
total = total+quot;
end
end
k = k+1;
end
perf = total == n;
end
end</syntaxhighlight>
=={{header|Maxima}}==
<syntaxhighlight lang="maxima">".."(a, b) := makelist(i, i, a, b)$
infix("..")$
perfectp(n) := is(divsum(n) = 2*n)$
sublist(1 .. 10000, perfectp);
/* [6, 28, 496, 8128] */</syntaxhighlight>
=={{header|MAXScript}}==
<
(
local sum = 0
Line 368 ⟶ 2,650:
)
sum == n
)</
=={{header|Microsoft Small Basic}}==
{{trans|BBC BASIC}}
<syntaxhighlight lang="microsoftsmallbasic">
For n = 2 To 10000 Step 2
VerifyIfPerfect()
If isPerfect = 1 Then
TextWindow.WriteLine(n)
EndIf
EndFor
Sub VerifyIfPerfect
s = 1
sqrN = Math.SquareRoot(n)
If Math.Remainder(n, 2) = 0 Then
s = s + 2 + Math.Floor(n / 2)
EndIf
i = 3
while i <= sqrN - 1
If Math.Remainder(n, i) = 0 Then
s = s + i + Math.Floor(n / i)
EndIf
i = i + 1
EndWhile
If i * i = n Then
s = s + i
EndIf
If n = s Then
isPerfect = 1
Else
isPerfect = 0
EndIf
EndSub
</syntaxhighlight>
=={{header|Modula-2}}==
{{trans|BBC BASIC}}
{{works with|ADW Modula-2|any (Compile with the linker option ''Console Application'').}}
<syntaxhighlight lang="modula2">
MODULE PerfectNumbers;
FROM SWholeIO IMPORT
WriteCard;
FROM STextIO IMPORT
WriteLn;
FROM RealMath IMPORT
sqrt;
VAR
N: CARDINAL;
PROCEDURE IsPerfect(N: CARDINAL): BOOLEAN;
VAR
S, I: CARDINAL;
SqrtN: REAL;
BEGIN
S := 1;
SqrtN := sqrt(FLOAT(N));
IF N REM 2 = 0 THEN
S := S + 2 + N / 2;
END;
I := 3;
WHILE FLOAT(I) <= SqrtN - 1.0 DO
IF N REM I = 0 THEN
S := S + I + N / I;
END;
I := I + 1;
END;
IF I * I = N THEN
S := S + I;
END;
RETURN (N = S);
END IsPerfect;
BEGIN
FOR N := 2 TO 10000 BY 2 DO
IF IsPerfect(N) THEN
WriteCard(N, 5);
WriteLn;
END;
END;
END PerfectNumbers.
</syntaxhighlight>
=={{header|Nanoquery}}==
{{trans|Python}}
<syntaxhighlight lang="nanoquery">def perf(n)
sum = 0
for i in range(1, n - 1)
if (n % i) = 0
sum += i
end
end
return sum = n
end</syntaxhighlight>
=={{header|Nim}}==
<syntaxhighlight lang="nim">import math
proc isPerfect(n: int): bool =
var sum: int = 1
for d in 2 .. int(n.toFloat.sqrt):
if n mod d == 0:
inc sum, d
let q = n div d
if q != d: inc sum, q
result = n == sum
for n in 2..10_000:
if n.isPerfect:
echo n</syntaxhighlight>
{{out}}
<pre>6
28
496
8128</pre>
=={{header|Objeck}}==
<syntaxhighlight lang="objeck">bundle Default {
class Test {
function : Main(args : String[]) ~ Nil {
"Perfect numbers from 1 to 33550337:"->PrintLine();
for(num := 1 ; num < 33550337; num += 1;) {
if(IsPerfect(num)) {
num->PrintLine();
};
};
}
function : native : IsPerfect(number : Int) ~ Bool {
sum := 0 ;
for(i := 1; i < number; i += 1;) {
if (number % i = 0) {
sum += i;
};
};
return sum = number;
}
}
}</syntaxhighlight>
=={{header|OCaml}}==
<
let sum = ref 0 in
for i = 1 to n-1 do
Line 377 ⟶ 2,801:
sum := !sum + i
done;
!sum = n</
Functional style:
<
let rec (--) a b =
if a > b then
Line 386 ⟶ 2,810:
a :: (a+1) -- b
let perf n = n = List.fold_left (+) 0 (List.filter (fun i -> n mod i = 0) (1 -- (n-1)))</
=={{header|Oforth}}==
<syntaxhighlight lang="oforth">: isPerfect(n) | i | 0 n 2 / loop: i [ n i mod ifZero: [ i + ] ] n == ; </syntaxhighlight>
{{out}}
<pre>
#isPerfect 10000 seq filter .
[6, 28, 496, 8128]
</pre>
=={{header|Odin}}==
<syntaxhighlight lang="Go">
package perfect_numbers
import "core:fmt"
main :: proc() {
fmt.println("\nPerfect numbers from 1 to 100,000:\n")
for num in 1 ..< 100001 {
if divisor_sum(num) == num {
fmt.print("num:", num, "\n")
}
if num % 10000 == 0 {
fmt.print("Count:", num, "\n")
}
}
}
divisor_sum :: proc(number: int) -> int {
sum := 0
for i in 1 ..< number {
if number % i == 0 {
sum += i}
}
return sum
}
</syntaxhighlight>
{{out}}
<pre>
Perfect numbers from 1 to 100,000:
num: 6
num: 28
num: 496
num: 8128
</pre>
=={{header|ooRexx}}==
<syntaxhighlight lang="oorexx">-- first perfect number over 10000 is 33550336...let's not be crazy
loop i = 1 to 10000
if perfectNumber(i) then say i "is a perfect number"
end
::routine perfectNumber
use strict arg n
sum = 0
-- the largest possible factor is n % 2, so no point in
-- going higher than that
loop i = 1 to n % 2
if n // i == 0 then sum += i
end
return sum = n</syntaxhighlight>
{{out}}
<pre>6 is a perfect number
28 is a perfect number
496 is a perfect number
8128 is a perfect number</pre>
=={{header|Oz}}==
<syntaxhighlight lang="oz">declare
fun {IsPerfect N}
fun {IsNFactor I} N mod I == 0 end
Factors = {Filter {List.number 1 N-1 1} IsNFactor}
in
{Sum Factors} == N
end
fun {Sum Xs} {FoldL Xs Number.'+' 0} end
in
{Show {Filter {List.number 1 10000 1} IsPerfect}}
{Show {IsPerfect 33550336}}</syntaxhighlight>
=={{header|PARI/GP}}==
===Using built-in methods===
<syntaxhighlight lang="parigp">
isPerfect(n)=sigma(n,-1)==2
</syntaxhighlight>
or
<syntaxhighlight lang="parigp">
isPerfect(n)=sigma(n)==2*n
</syntaxhighlight>
Show perfect numbers
<syntaxhighlight lang="parigp">
forprime(p=2, 2281,
if(isprime(2^p-1),
print(p"\t",(2^p-1)*2^(p-1))))
</syntaxhighlight>
faster alternative showing them still using built-in methods
<syntaxhighlight lang="parigp">
[n|n<-[1..10^4],sigma(n,-1)==2]
</syntaxhighlight>
{{Out}}
<pre>
[6, 28, 496, 8128]
</pre>
===Faster with Lucas-Lehmer test===
<syntaxhighlight lang="parigp">p=2;n=3;n1=2;
while(p<2281,
if(isprime(p),
s=Mod(4,n);
for(i=3,p,
s=s*s-2);
if(s==0 || p==2,
print("(2^"p"-1)2^("p"-1)=\t"n1*n"\n")));
p++; n1=n+1; n=2*n+1)</syntaxhighlight>
{{Out}}
<pre>(2^2-1)2^(2-1)= 6
(2^3-1)2^(3-1)= 28
(2^5-1)2^(5-1)= 496
(2^7-1)2^(7-1)= 8128
(2^13-1)2^(13-1)= 33550336
(2^17-1)2^(17-1)= 8589869056
(2^19-1)2^(19-1)= 137438691328
(2^31-1)2^(31-1)= 2305843008139952128
(2^61-1)2^(61-1)= 2658455991569831744654692615953842176
(2^89-1)2^(89-1)= 191561942608236107294793378084303638130997321548169216</pre>
=={{header|Pascal}}==
<syntaxhighlight lang="pascal">program PerfectNumbers;
function isPerfect(number: longint): boolean;
var
i, sum: longint;
begin
sum := 1;
for i := 2 to round(sqrt(real(number))) do
if (number mod i = 0) then
sum := sum + i + (number div i);
isPerfect := (sum = number);
end;
var
candidate: longint;
begin
writeln('Perfect numbers from 1 to 33550337:');
for candidate := 2 to 33550337 do
if isPerfect(candidate) then
writeln (candidate, ' is a perfect number.');
end.</syntaxhighlight>
{{Out}}
<pre>
Perfect numbers from 1 to 33550337:
6 is a perfect number.
28 is a perfect number.
496 is a perfect number.
8128 is a perfect number.
33550336 is a perfect number.
</pre>
=={{header|Perl}}==
=== Functions ===
<syntaxhighlight lang="perl">sub perf {
my $n = shift;
my $sum = 0;
Line 398 ⟶ 2,989:
}
return $sum == $n;
}</
Functional style:
<
sub perf {
my $n = shift;
$n == sum(0, grep {$n % $_ == 0} 1..$n-1);
}</
=== Modules ===
The functions above are terribly slow. As usual, this is easier and faster with modules. Both ntheory and Math::Pari have useful functions for this.
{{libheader|ntheory}}
A simple predicate:
<syntaxhighlight lang="perl">use ntheory qw/divisor_sum/;
sub is_perfect { my $n = shift; divisor_sum($n) == 2*$n; }</syntaxhighlight>
Use this naive method to show the first 5. Takes about 15 seconds:
<syntaxhighlight lang="perl">use ntheory qw/divisor_sum/;
for (1..33550336) {
print "$_\n" if divisor_sum($_) == 2*$_;
}</syntaxhighlight>
Or we can be clever and look for 2^(p-1) * (2^p-1) where 2^p -1 is prime. The first 20 takes about a second.
<syntaxhighlight lang="perl">use ntheory qw/forprimes is_prime/;
use bigint;
forprimes {
my $n = 2**$_ - 1;
print "$_\t", $n * 2**($_-1),"\n" if is_prime($n);
} 2, 4500;</syntaxhighlight>
{{out}}
<pre>
2 6
3 28
5 496
7 8128
13 33550336
17 8589869056
19 137438691328
31 2305843008139952128
61 2658455991569831744654692615953842176
89 191561942608236107294793378084303638130997321548169216
... 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423 ...
</pre>
We can speed this up even more using a faster program for printing the large results, as well as a faster primality solution. The first 38 in about 1 second with most of the time printing the large results. Caveat: this goes well past the current bound for odd perfect numbers and does not check for them.
<syntaxhighlight lang="perl">use ntheory qw/forprimes is_mersenne_prime/;
use Math::GMP qw/:constant/;
forprimes {
print "$_\t", (2**$_-1)*2**($_-1),"\n" if is_mersenne_prime($_);
} 7_000_000;</syntaxhighlight>
In addition to generating even perfect numbers, we can also have a fast function which returns true when a given even number is perfect:
<syntaxhighlight lang="perl">use ntheory qw(is_mersenne_prime valuation);
sub is_even_perfect {
my ($n) = @_;
my $v = valuation($n, 2) || return;
my $m = ($n >> $v);
($m & ($m + 1)) && return;
($m >> $v) == 1 || return;
is_mersenne_prime($v + 1);
}</syntaxhighlight>
=={{header|Phix}}==
<!--(phixonline)-->
=== naive/native ===
<syntaxhighlight lang="phix">
function is_perfect(integer n)
return sum(factors(n,-1))=n
end function
for i=2 to 100000 do
if is_perfect(i) then ?i end if
end for
</syntaxhighlight>
{{out}}
<pre>
6
28
496
8128
</pre>
=== gmp version ===
{{libheader|Phix/mpfr}}
<syntaxhighlight lang="phix">
with javascript_semantics
-- demo\rosetta\Perfect_numbers.exw (includes native and cheat versions)
include mpfr.e
atom t0 = time(), t1 = t0+1
integer maxprime = 4423, -- 19937 (rather slow)
lim = length(get_primes_le(maxprime))
mpz n = mpz_init(), m = mpz_init()
for i=1 to lim do
integer p = get_prime(i)
mpz_ui_pow_ui(n, 2, p)
mpz_sub_ui(n, n, 1)
if mpz_prime(n) then
mpz_ui_pow_ui(m, 2, p-1)
mpz_mul(n, n, m)
string ns = mpz_get_short_str(n,comma_fill:=true),
et = elapsed_short(time()-t0,5,"(%s)")
printf(1, "%d %s %s\n",{p,ns,et})
elsif time()>t1 then
progress("%d/%d (%.1f%%)\r",{p,maxprime,i/lim*100})
t1 = time()+1
end if
end for
?elapsed(time()-t0)
</syntaxhighlight>
{{out}}
<pre>
2 6
3 28
5 496
7 8,128
13 33,550,336
17 8,589,869,056
19 137,438,691,328
31 2,305,843,008,139,952,128
61 2,658,455,991,569,831,744,654,692,615,953,842,176
89 191,561,942,608,236,...,997,321,548,169,216 (54 digits)
107 13,164,036,458,569,6...,943,117,783,728,128 (65 digits)
127 14,474,011,154,664,5...,349,131,199,152,128 (77 digits)
521 23,562,723,457,267,3...,492,160,555,646,976 (314 digits)
607 141,053,783,706,712,...,570,759,537,328,128 (366 digits)
1279 54,162,526,284,365,8...,345,764,984,291,328 (770 digits)
2203 1,089,258,355,057,82...,580,834,453,782,528 (1,327 digits)
2281 99,497,054,337,086,4...,375,675,139,915,776 (1,373 digits)
3217 33,570,832,131,986,7...,888,332,628,525,056 (1,937 digits) (9s)
4253 18,201,749,040,140,4...,848,437,133,377,536 (2,561 digits) (24s)
4423 40,767,271,711,094,4...,020,642,912,534,528 (2,663 digits) (28s)
"28.4s"
</pre>
Beyond that it gets rather slow:
<pre>
9689 11,434,731,753,038,6...,982,558,429,577,216 (5,834 digits) (6:28)
9941 598,885,496,387,336,...,478,324,073,496,576 (5,985 digits) (7:31)
11213 3,959,613,212,817,94...,255,702,691,086,336 (6,751 digits) (11:32)
19937 931,144,559,095,633,...,434,790,271,942,656 (12,003 digits) (1:22:32)
</pre>
=== cheating ===
{{trans|Picat}}
<syntaxhighlight lang="phix">
include mpfr.e
atom t0 = time()
mpz n = mpz_init(), m = mpz_init()
sequence validp = {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607,
1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213,
19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091,
756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593,
13466917, 20996011, 24036583, 25964951, 30402457, 32582657,
37156667, 42643801, 43112609, 57885161,
74207281, 77232917, 82589933}
if platform()=JS then validp = validp[1..35] end if -- (keep it under 5s)
for p in validp do
mpz_ui_pow_ui(n, 2, p)
mpz_sub_ui(n, n, 1)
mpz_ui_pow_ui(m, 2, p-1)
mpz_mul(n, n, m)
string ns = mpz_get_short_str(n,comma_fill:=true),
et = elapsed_short(time()-t0,5,"(%s)")
printf(1, "%d %s %s\n",{p,ns,et})
end for
?elapsed(time()-t0)
</syntaxhighlight>
<pre>
2 6
3 28
5 496
7 8,128
13 33,550,336
17 8,589,869,056
19 137,438,691,328
31 2,305,843,008,139,952,128
61 2,658,455,991,569,831,744,654,692,615,953,842,176
89 191,561,942,608,236,...,997,321,548,169,216 (54 digits)
107 13,164,036,458,569,6...,943,117,783,728,128 (65 digits)
127 14,474,011,154,664,5...,349,131,199,152,128 (77 digits)
521 23,562,723,457,267,3...,492,160,555,646,976 (314 digits)
607 141,053,783,706,712,...,570,759,537,328,128 (366 digits)
1279 54,162,526,284,365,8...,345,764,984,291,328 (770 digits)
2203 1,089,258,355,057,82...,580,834,453,782,528 (1,327 digits)
2281 99,497,054,337,086,4...,375,675,139,915,776 (1,373 digits)
3217 33,570,832,131,986,7...,888,332,628,525,056 (1,937 digits)
4253 18,201,749,040,140,4...,848,437,133,377,536 (2,561 digits)
4423 40,767,271,711,094,4...,020,642,912,534,528 (2,663 digits)
9689 11,434,731,753,038,6...,982,558,429,577,216 (5,834 digits)
9941 598,885,496,387,336,...,478,324,073,496,576 (5,985 digits)
11213 3,959,613,212,817,94...,255,702,691,086,336 (6,751 digits)
19937 931,144,559,095,633,...,434,790,271,942,656 (12,003 digits)
21701 1,006,564,970,546,40...,865,255,141,605,376 (13,066 digits)
23209 81,153,776,582,351,0...,048,603,941,666,816 (13,973 digits)
44497 365,093,519,915,713,...,965,353,031,827,456 (26,790 digits)
86243 144,145,836,177,303,...,480,957,360,406,528 (51,924 digits)
110503 13,620,458,213,388,4...,255,233,603,862,528 (66,530 digits)
132049 13,145,129,545,436,9...,438,491,774,550,016 (79,502 digits)
216091 27,832,745,922,032,7...,263,416,840,880,128 (130,100 digits)
756839 15,161,657,022,027,0...,971,600,565,731,328 (455,663 digits)
859433 83,848,822,675,015,7...,651,540,416,167,936 (517,430 digits)
1257787 849,732,889,343,651,...,394,028,118,704,128 (757,263 digits)
1398269 331,882,354,881,177,...,668,017,723,375,616 (841,842 digits)
2976221 194,276,425,328,791,...,106,724,174,462,976 (1,791,864 digits)
3021377 811,686,848,628,049,...,147,573,022,457,856 (1,819,050 digits)
6972593 9,551,760,305,212,09...,914,475,123,572,736 (4,197,919 digits)
13466917 42,776,415,902,185,6...,230,460,863,021,056 (8,107,892 digits)
20996011 7,935,089,093,651,70...,903,578,206,896,128 (12,640,858 digits)
24036583 44,823,302,617,990,8...,680,460,572,950,528 (14,471,465 digits) (5s)
25964951 7,462,098,419,004,44...,245,874,791,088,128 (15,632,458 digits) (8s)
30402457 49,743,776,545,907,0...,934,536,164,704,256 (18,304,103 digits) (10s)
32582657 77,594,685,533,649,8...,428,476,577,120,256 (19,616,714 digits) (13s)
37156667 20,453,422,553,410,5...,147,975,074,480,128 (22,370,543 digits) (16s)
42643801 1,442,850,579,600,99...,314,837,377,253,376 (25,674,127 digits) (20s)
43112609 50,076,715,684,982,3...,909,221,145,378,816 (25,956,377 digits) (24s)
57885161 169,296,395,301,618,...,179,626,270,130,176 (34,850,340 digits) (29s)
74207281 45,112,996,270,669,0...,008,557,930,315,776 (44,677,235 digits) (36s)
77232917 10,920,015,213,433,6...,001,402,016,301,056 (46,498,850 digits) (43s)
82589933 1,108,477,798,641,48...,798,007,191,207,936 (49,724,095 digits) (50s)
"50.6s"
</pre>
=={{header|PHP}}==
{{trans|C++}}
<syntaxhighlight lang="php">function is_perfect($number)
{
$sum = 0;
for($i = 1; $i < $number; $i++)
{
if($number % $i == 0)
$sum += $i;
}
return $sum == $number;
}
echo "Perfect numbers from 1 to 33550337:" . PHP_EOL;
for($num = 1; $num < 33550337; $num++)
{
if(is_perfect($num))
echo $num . PHP_EOL;
}</syntaxhighlight>
=={{header|Picat}}==
===Simple divisors/1 function===
First is the slow <code>perfect1/1</code> that use the simple divisors/1 function:
<syntaxhighlight lang="picat">go =>
println(perfect1=[I : I in 1..10_000, perfect1(I)]),
nl.
perfect1(N) => sum(divisors(N)) == N.
divisors(N) = [J: J in 1..1+N div 2, N mod J == 0].</syntaxhighlight>
{{out}}
<pre>perfect1 = [1,6,28,496,8128]</pre>
===Using formula for perfect number candidates===
The formula for perfect number candidates is: 2^(p-1)*(2^p-1) for prime p. This is used to find some more perfect numbers in reasonable time. <code>perfect2/1</code> is a faster version of checking if a number is perfect.
<syntaxhighlight lang="picat">go2 =>
println("Using the formula: 2^(p-1)*(2^p-1) for prime p"),
foreach(P in primes(32))
PF=perfectf(P),
% Check that it is really a perfect number
if perfect2(PF) then
printf("%w (prime %w)\n",PF,P)
end
end,
nl.
% Formula for perfect number candidates:
% 2^(p-1)*(2^p-1) where p is a prime
%
perfectf(P) = (2**(P-1))*((2**P)-1).
% Faster check of a perfect number
perfect2(N) => sum_divisors(N) == N.
% Sum of divisors
table
sum_divisors(N) = Sum =>
sum_divisors(2,N,1,Sum).
sum_divisors(I,N,Sum0,Sum), I > floor(sqrt(N)) =>
Sum = Sum0.
% 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).
% I is not a divisor of N.
sum_divisors(I,N,Sum0,Sum) =>
sum_divisors(I+1,N,Sum0,Sum).</syntaxhighlight>
{{out}}
<pre>6 (prime 2)
28 (prime 3)
496 (prime 5)
8128 (prime 7)
33550336 (prime 13)
8589869056 (prime 17)
137438691328 (prime 19)
2305843008139952128 (prime 31)
CPU time 118.039 seconds. Backtracks: 0</pre>
===Using list of the primes generating the perfect numbers===
Now let's cheat a little. At https://en.wikipedia.org/wiki/Perfect_number there is a list of the first 48 primes that generates perfect numbers according to the formula 2^(p-1)*(2^p-1) for prime p.
The perfect numbers are printed only if they has < 80 digits, otherwise the number of digits are shown. The program stops when reaching a number with more than 100 000 digits. (Note: The major time running this program is getting the number of digits.)
<syntaxhighlight lang="picat">go3 =>
ValidP = [2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607,
1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213,
19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091,
756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593,
13466917, 20996011, 24036583, 25964951, 30402457, 32582657,
37156667, 42643801, 43112609, 57885161],
foreach(P in ValidP)
printf("prime %w: ", P),
PF = perfectf(P),
Len = PF.to_string.len,
if Len < 80 then
println(PF)
else
println(len=Len)
end,
if Len >= 100_000 then
fail
end
end,
nl.</syntaxhighlight>
{{out}}
<pre>prime 2: 6
prime 3: 28
prime 5: 496
prime 7: 8128
prime 13: 33550336
prime 17: 8589869056
prime 19: 137438691328
prime 31: 2305843008139952128
prime 61: 2658455991569831744654692615953842176
prime 89: 191561942608236107294793378084303638130997321548169216
prime 107: 13164036458569648337239753460458722910223472318386943117783728128
prime 127: 14474011154664524427946373126085988481573677491474835889066354349131199152128
prime 521: len = 314
prime 607: len = 366
prime 1279: len = 770
prime 2203: len = 1327
prime 2281: len = 1373
prime 3217: len = 1937
prime 4253: len = 2561
prime 4423: len = 2663
prime 9689: len = 5834
prime 9941: len = 5985
prime 11213: len = 6751
prime 19937: len = 12003
prime 21701: len = 13066
prime 23209: len = 13973
prime 44497: len = 26790
prime 86243: len = 51924
prime 110503: len = 66530
prime 132049: len = 79502
prime 216091: len = 130100</pre>
=={{header|PicoLisp}}==
<syntaxhighlight lang="picolisp">(de perfect (N)
(let C 0
(for I (/ N 2)
(and (=0 (% N I)) (inc 'C I)) )
(= C N) ) )</syntaxhighlight>
<syntaxhighlight lang="picolisp">(de faster (N)
(let (C 1 Stop (sqrt N))
(for (I 2 (<= I Stop) (inc I))
(and
(=0 (% N I))
(inc 'C (+ (/ N I) I)) ) )
(= C N) ) )</syntaxhighlight>
=={{header|PL/I}}==
<syntaxhighlight lang="pl/i">perfect: procedure (n) returns (bit(1));
declare n fixed;
declare sum fixed;
declare i fixed binary;
sum = 0;
do i = 1 to n-1;
if mod(n, i) = 0 then sum = sum + i;
end;
return (sum=n);
end perfect;</syntaxhighlight>
==={{header|PL/I-80}}===
<syntaxhighlight lang="pl/i">perfect_search: procedure options (main);
%replace
search_limit by 10000,
true by '1'b,
false by '0'b;
dcl (k, found) fixed bin;
put skip list ('Searching for perfect numbers up to ');
put edit (search_limit) (f(5));
found = 0;
do k = 2 to search_limit;
if isperfect(k) then
do;
put skip list(k);
found = found + 1;
end;
end;
put skip list (found, ' perfect numbers were found');
/* return true if n is perfect, otherwise false */
isperfect: procedure(n) returns (bit(1));
dcl (n, sum, f1, f2) fixed bin;
sum = 1; /* 1 is a proper divisor of every number */
f1 = 2;
do while ((f1 * f1) <= n);
if mod(n, f1) = 0 then
do;
sum = sum + f1;
f2 = n / f1;
/* don't double count identical co-factors! */
if f2 > f1 then sum = sum + f2;
end;
f1 = f1 + 1;
end;
return (sum = n);
end isperfect;
end perfect_search;</syntaxhighlight>
{{out}}
<pre>
Searching for perfect numbers up to 10000
6
28
496
8128
4 perfect numbers were found
</pre>
=={{header|PL/M}}==
{{works with|8080 PL/M Compiler}} ... under CP/M (or an emulator)
<syntaxhighlight lang="pli">100H: /* FIND SOME PERFECT NUMBERS: NUMBERS EQUAL TO THE SUM OF THEIR PROPER */
/* DIVISORS */
/* CP/M SYSTEM CALL AND I/O ROUTINES */
BDOS: PROCEDURE( FN, ARG ); /* CP/M BDOS SYSTEM CALL */
DECLARE FN BYTE, ARG ADDRESS;
GOTO 5;
END BDOS;
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
PR$NUMBER: PROCEDURE( N );
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR( 6 ) BYTE, W BYTE;
V = N;
W = LAST( N$STR );
N$STR( W ) = '$';
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
/* TASK */
/* RETURNS TRUE IF N IS PERFECT, 0 OTHERWISE */
IS$PERFECT: PROCEDURE( N )BYTE;
DECLARE N ADDRESS;
DECLARE ( F1, F2, SUM ) ADDRESS;
SUM = 1;
F1 = 2;
F2 = N;
DO WHILE( F1 * F1 <= N );
IF N MOD F1 = 0 THEN DO;
SUM = SUM + F1;
F2 = N / F1;
/* AVOID COUNTING E.G., 2 TWICE AS A FACTOR OF 4 */
IF F2 > F1 THEN SUM = SUM + F2;
END;
F1 = F1 + 1;
END;
RETURN SUM = N;
END IS$PERFECT ;
/* TEST IS$PERFECT */
DECLARE N ADDRESS;
DO N = 2 TO 10$000;
IF IS$PERFECT( N ) THEN DO;
CALL PR$CHAR( ' ' );
CALL PR$NUMBER( N );
END;
END;
EOF</syntaxhighlight>
{{out}}
<pre>
6 28 496 8128
</pre>
Alternative, much faster version.
{{Trans|Action!}}
{{works with|8080 PL/M Compiler}} ... under CP/M (or an emulator)
<syntaxhighlight lang="pli">100H: /* FIND SOME PERFECT NUMBERS: NUMBERS EQUAL TO THE SUM OF THEIR PROPER */
/* DIVISORS */
/* CP/M SYSTEM CALL AND I/O ROUTINES */
BDOS: PROCEDURE( FN, ARG ); /* CP/M BDOS SYSTEM CALL */
DECLARE FN BYTE, ARG ADDRESS;
GOTO 5;
END BDOS;
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
PR$NUMBER: PROCEDURE( N );
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR( 6 ) BYTE, W BYTE;
V = N;
W = LAST( N$STR );
N$STR( W ) = '$';
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
/* TASK - TRANSLATION OF ACTION! */
DECLARE MAX$NUM LITERALLY '10$000';
DECLARE PDS( 10$001 ) ADDRESS;
DECLARE ( I, J ) ADDRESS;
DO I = 2 TO MAX$NUM;
PDS( I ) = 1;
END;
DO I = 2 TO MAX$NUM;
DO J = I + I TO MAX$NUM BY I;
PDS( J ) = PDS( J ) + I;
END;
END;
DO I = 2 TO MAX$NUM;
IF PDS( I ) = I THEN DO;
CALL PR$NUMBER( I );
CALL PR$NL;
END;
END;
EOF</syntaxhighlight>
{{out}}
<pre>
6
28
496
8128
</pre>
=={{header|PowerShell}}==
<syntaxhighlight lang="powershell">Function IsPerfect($n)
{
$sum=0
for($i=1;$i-lt$n;$i++)
{
if($n%$i -eq 0)
{
$sum += $i
}
}
return $sum -eq $n
}
Returns "True" if the given number is perfect and "False" if it's not.</syntaxhighlight>
=={{header|Prolog}}==
===Classic approach===
Works with SWI-Prolog
<syntaxhighlight lang="prolog">tt_divisors(X, N, TT) :-
Q is X / N,
( 0 is X mod N -> (Q = N -> TT1 is N + TT;
TT1 is N + Q + TT);
TT = TT1),
( sqrt(X) > N + 1 -> N1 is N+1, tt_divisors(X, N1, TT1);
TT1 = X).
perfect(X) :-
tt_divisors(X, 2, 1).
perfect_numbers(N, L) :-
numlist(2, N, LN),
include(perfect, LN, L).</syntaxhighlight>
===Faster method===
Since a perfect number is of the form 2^(n-1) * (2^n - 1), we can eliminate a lot of candidates by merely factoring out the 2s and seeing if the odd portion is (2^(n+1)) - 1.
<syntaxhighlight lang="prolog">
perfect(N) :-
factor_2s(N, Chk, Exp),
Chk =:= (1 << (Exp+1)) - 1,
prime(Chk).
factor_2s(N, S, D) :- factor_2s(N, 0, S, D).
factor_2s(D, S, D, S) :- getbit(D, 0) =:= 1, !.
factor_2s(N, E, D, S) :-
E2 is E + 1, N2 is N >> 1, factor_2s(N2, E2, D, S).
% check if a number is prime
%
wheel235(L) :-
W = [4, 2, 4, 2, 4, 6, 2, 6 | W],
L = [1, 2, 2 | W].
prime(N) :-
N >= 2,
wheel235(W),
prime(N, 2, W).
prime(N, D, _) :- D*D > N, !.
prime(N, D, [A|As]) :-
N mod D =\= 0,
D2 is D + A, prime(N, D2, As).
</syntaxhighlight>
{{out}}
<pre>
?- between(1, 10_000, N), perfect(N).
N = 6 ;
N = 28 ;
N = 496 ;
N = 8128 ;
false.
</pre>
===Functional approach===
Works with SWI-Prolog and module lambda, written by <b>Ulrich Neumerkel</b> found there http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/lambda.pl
<syntaxhighlight lang="prolog">:- use_module(library(lambda)).
is_divisor(V, N) :-
0 =:= V mod N.
is_perfect(N) :-
N1 is floor(N/2),
numlist(1, N1, L),
f_compose_1(foldl((\X^Y^Z^(Z is X+Y)), 0), filter(is_divisor(N)), F),
call(F, L, N).
f_perfect_numbers(N, L) :-
numlist(2, N, LN),
filter(is_perfect, LN, L).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% functionnal predicates
%% foldl(Pred, Init, List, R).
%
foldl(_Pred, Val, [], Val).
foldl(Pred, Val, [H | T], Res) :-
call(Pred, Val, H, Val1),
foldl(Pred, Val1, T, Res).
%% filter(Pred, LstIn, LstOut)
%
filter(_Pre, [], []).
filter(Pred, [H|T], L) :-
filter(Pred, T, L1),
( call(Pred,H) -> L = [H|L1]; L = L1).
%% f_compose_1(Pred1, Pred2, Pred1(Pred2)).
%
f_compose_1(F,G, \X^Z^(call(G,X,Y), call(F,Y,Z))).</syntaxhighlight>
=={{header|PureBasic}}==
<syntaxhighlight lang="purebasic">Procedure is_Perfect_number(n)
Protected summa, i=1, result=#False
Repeat
If Not n%i
summa+i
EndIf
i+1
Until i>=n
If summa=n
result=#True
EndIf
ProcedureReturn result
EndProcedure</syntaxhighlight>
=={{header|Python}}==
;Relative timings:
Relative timings for sifting the integers from 1 to 50_000 inclusive for perfect numbers.
{| class="wikitable"
! style="font-weight:bold;" | Function
! style="font-weight:bold;" | Time
! style="font-weight:bold;" | Type
|-
| perf4
| 1
| Optimised procedural
|-
| perfect
| 1.6
| Optimised functional
|-
| perf1
| 259
| Procedural
|-
| perf2
| 273
| Functional
|}
===Python: Procedural===
<syntaxhighlight lang="python">def perf1(n):
sum = 0
for i in
if n % i == 0:
sum += i
return sum == n</
===Python: Optimised Procedural===
<syntaxhighlight lang="python">from itertools import chain, cycle, accumulate
def factor2(n):
def prime_powers(n):
# c goes through 2, 3, 5, then the infinite (6n+1, 6n+5) series
for c in accumulate(chain([2, 1, 2], cycle([2,4]))):
if c*c > n: break
if n%c: continue
d,p = (), c
while not n%c:
n,p,d = n//c, p*c, d + (p,)
yield(d)
if n > 1: yield((n,))
r = [1]
for e in prime_powers(n):
r += [a*b for a in r for b in e]
return r
def perf4(n):
"Using most efficient prime factoring routine from: http://rosettacode.org/wiki/Factors_of_an_integer#Python"
return 2 * n == sum(factor2(n))</syntaxhighlight>
===Python: Functional===
<syntaxhighlight lang="python">def perf2(n):
return n == sum(i for i in range(1, n) if n % i == 0)
print (
list(filter(perf2, range(1, 10001)))
)</syntaxhighlight>
<syntaxhighlight lang="python">'''Perfect numbers'''
from math import sqrt
# perfect :: Int - > Bool
def perfect(n):
'''Is n the sum of its proper divisors other than 1 ?'''
root = sqrt(n)
lows = [x for x in enumFromTo(2)(int(root)) if 0 == (n % x)]
return 1 < n and (
n == 1 + sum(lows + [n / x for x in lows if root != x])
)
# main :: IO ()
def main():
'''Test'''
print([
x for x in enumFromTo(1)(10000) if perfect(x)
])
# GENERIC -------------------------------------------------
# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))
if __name__ == '__main__':
main()</syntaxhighlight>
{{Out}}
<pre>[6, 28, 496, 8128]</pre>
=={{header|Quackery}}==
<code>factors</code> is defined at [http://rosettacode.org/wiki/Factors_of_an_integer#Quackery Factors of an integer].
<syntaxhighlight lang="quackery"> [ 0 swap witheach + ] is sum ( [ --> n )
[ factors -1 pluck dip sum = ] is perfect ( n --> n )
say "Perfect numbers less than 10000:" cr
10000 times
[ i^ 1+ perfect if [ i^ 1+ echo cr ] ]
</syntaxhighlight>
{{out}}
<pre>Perfect numbers less than 10000:
6
28
496
8128
</pre>
=={{header|R}}==
<
if (n==0|n==1) return(FALSE)
s <- seq (1,n-1)
Line 428 ⟶ 3,817:
# Usage - Warning High Memory Usage
is.perf(28)
sapply(c(6,28,496,8128,33550336),is.perf)</
=={{header|Racket}}==
<syntaxhighlight lang="racket">#lang racket
(require math)
(define (perfect? n)
(=
(* n 2)
(sum (divisors n))))
; filtering to only even numbers for better performance
(filter perfect? (filter even? (range 1e5)))
;-> '(0 6 28 496 8128)</syntaxhighlight>
=={{header|Raku}}==
(formerly Perl 6)
Naive (very slow) version
<syntaxhighlight lang="raku" line>sub is-perf($n) { $n == [+] grep $n %% *, 1 .. $n div 2 }
# used as
put ((1..Inf).hyper.grep: {.&is-perf})[^4];</syntaxhighlight>
{{out}}
<pre>6 28 496 8128</pre>
Much, much faster version:
<syntaxhighlight lang="raku" line>my @primes = lazy (2,3,*+2 … Inf).grep: { .is-prime };
my @perfects = lazy gather for @primes {
my $n = 2**$_ - 1;
take $n * 2**($_ - 1) if $n.is-prime;
}
.put for @perfects[^12];</syntaxhighlight>
{{out}}
<pre>6
28
496
8128
33550336
8589869056
137438691328
2305843008139952128
2658455991569831744654692615953842176
191561942608236107294793378084303638130997321548169216
13164036458569648337239753460458722910223472318386943117783728128
14474011154664524427946373126085988481573677491474835889066354349131199152128</pre>
=={{header|REBOL}}==
<syntaxhighlight lang="rebol">perfect?: func [n [integer!] /local sum] [
sum: 0
repeat i (n - 1) [
if zero? remainder n i [
sum: sum + i
]
]
sum = n
]</syntaxhighlight>
=={{header|REXX}}==
===Classic REXX version of ooRexx===
This version is a '''Classic Rexx''' version of the '''ooRexx''' program as of 14-Sep-2013.
<syntaxhighlight lang="rexx">/*REXX version of the ooRexx program (the code was modified to run with Classic REXX).*/
do i=1 to 10000 /*statement changed: LOOP ──► DO*/
if perfectNumber(i) then say i "is a perfect number"
end
exit
perfectNumber: procedure; parse arg n /*statements changed: ROUTINE,USE*/
sum=0
do i=1 to n%2 /*statement changed: LOOP ──► DO*/
if n//i==0 then sum=sum+i /*statement changed: sum += i */
end
return sum=n</syntaxhighlight>
'''output''' when using the default of 10000:
<pre>
6 is a perfect number
28 is a perfect number
496 is a perfect number
8128 is a perfect number
</pre>
===Classic REXX version of PL/I===
This version is a '''Classic REXX''' version of the '''PL/I''' program as of 14-Sep-2013, a REXX '''say''' statement
<br>was added to display the perfect numbers. Also, an epilog was written for the re-worked function.
<syntaxhighlight lang="rexx">/*REXX version of the PL/I program (code was modified to run with Classic REXX). */
parse arg low high . /*obtain the specified number(s).*/
if high=='' & low=='' then high=34000000 /*if no arguments, use a range. */
if low=='' then low=1 /*if no LOW, then assume unity.*/
if high=='' then high=low /*if no HIGH, then assume LOW. */
do i=low to high /*process the single # or range. */
if perfect(i) then say i 'is a perfect number.'
end /*i*/
exit
perfect: procedure; parse arg n /*get the number to be tested. */
sum=0 /*the sum of the factors so far. */
do i=1 for n-1 /*starting at 1, find all factors*/
if n//i==0 then sum=sum+i /*I is a factor of N, so add it.*/
end /*i*/
return sum=n /*if the sum matches N, perfect! */</syntaxhighlight>
'''output''' when using the input defaults of: <tt> 1 10000 </tt>
The output is the same as for the ooRexx version (above).
===traditional method===
Programming note: this traditional method takes advantage of a few shortcuts:
:::* testing only goes up to the (integer) square root of '''X'''
:::* testing bypasses the test of the first and last factors
:::* the ''corresponding factor'' is also used when a factor is found
<syntaxhighlight lang="rexx">/*REXX program tests if a number (or a range of numbers) is/are perfect. */
parse arg low high . /*obtain optional arguments from the CL*/
if high=='' & low=="" then high=34000000 /*if no arguments, then use a range. */
if low=='' then low=1 /*if no LOW, then assume unity. */
if high=='' then high=low /*if no HIGH, then assume LOW. */
w=length(high) /*use W for formatting the output. */
numeric digits max(9,w+2) /*ensure enough digits to handle number*/
do i=low to high /*process the single number or a range.*/
if isPerfect(i) then say right(i,w) 'is a perfect number.'
end /*i*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isPerfect: procedure; parse arg x /*obtain the number to be tested. */
if x<6 then return 0 /*perfect numbers can't be < six. */
s=1 /*the first factor of X. ___*/
do j=2 while j*j<=x /*starting at 2, find the factors ≤√ X */
if x//j\==0 then iterate /*J isn't a factor of X, so skip it.*/
s = s + j + x%j /* ··· add it and the other factor. */
end /*j*/ /*(above) is marginally faster. */
return s==x /*if the sum matches X, it's perfect! */</syntaxhighlight>
'''output''' when using the default inputs:
<pre>
6 is a perfect number.
28 is a perfect number.
496 is a perfect number.
8128 is a perfect number.
33550336 is a perfect number.
</pre>
For 10,000 numbers tested, this version is '''19.6''' times faster than the ooRexx program logic.<br>
For 10,000 numbers tested, this version is '''25.6''' times faster than the PL/I program logic.
<br><br>Note: For the above timings, only 10,000 numbers were tested.
===optimized using digital root===
This REXX version makes use of the fact that all ''known'' perfect numbers > 6 have a ''digital root'' of '''1'''.
<syntaxhighlight lang="rexx">/*REXX program tests if a number (or a range of numbers) is/are perfect. */
parse arg low high . /*obtain the specified number(s). */
if high=='' & low=="" then high=34000000 /*if no arguments, then use a range. */
if low=='' then low=1 /*if no LOW, then assume unity. */
if high=='' then high=low /*if no HIGH, then assume LOW. */
w=length(high) /*use W for formatting the output. */
numeric digits max(9,w+2) /*ensure enough digits to handle number*/
do i=low to high /*process the single number or a range.*/
if isPerfect(i) then say right(i,w) 'is a perfect number.'
end /*i*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isPerfect: procedure; parse arg x 1 y /*obtain the number to be tested. */
if x==6 then return 1 /*handle the special case of six. */
/*[↓] perfect number's digitalRoot = 1*/
do until y<10 /*find the digital root of Y. */
parse var y r 2; do k=2 for length(y)-1; r=r+substr(y,k,1); end /*k*/
y=r /*find digital root of the digit root. */
end /*until*/ /*wash, rinse, repeat ··· */
if r\==1 then return 0 /*Digital root ¬ 1? Then ¬ perfect. */
s=1 /*the first factor of X. ___*/
do j=2 while j*j<=x /*starting at 2, find the factors ≤√ X */
if x//j\==0 then iterate /*J isn't a factor of X, so skip it. */
s = s + j + x%j /*··· add it and the other factor. */
end /*j*/ /*(above) is marginally faster. */
return s==x /*if the sum matches X, it's perfect! */</syntaxhighlight>
'''output''' is the same as the traditional version and is about '''5.3''' times faster (testing '''34,000,000''' numbers).
===optimized using only even numbers===
This REXX version uses the fact that all ''known'' perfect numbers are ''even''.
<syntaxhighlight lang="rexx">/*REXX program tests if a number (or a range of numbers) is/are perfect. */
parse arg low high . /*obtain optional arguments from the CL*/
if high=='' & low=="" then high=34000000 /*if no arguments, then use a range. */
if low=='' then low=1 /*if no LOW, then assume unity. */
low=low+low//2 /*if LOW is odd, bump it by one. */
if high=='' then high=low /*if no HIGH, then assume LOW. */
w=length(high) /*use W for formatting the output. */
numeric digits max(9,w+2) /*ensure enough digits to handle number*/
do i=low to high by 2 /*process the single number or a range.*/
if isPerfect(i) then say right(i,w) 'is a perfect number.'
end /*i*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isPerfect: procedure; parse arg x 1 y /*obtain the number to be tested. */
if x==6 then return 1 /*handle the special case of six. */
do until y<10 /*find the digital root of Y. */
parse var y 1 r 2; do k=2 for length(y)-1; r=r+substr(y,k,1); end /*k*/
y=r /*find digital root of the digital root*/
end /*until*/ /*wash, rinse, repeat ··· */
if r\==1 then return 0 /*Digital root ¬ 1 ? Then ¬ perfect.*/
s=3 + x%2 /*the first 3 factors of X. ___*/
do j=3 while j*j<=x /*starting at 3, find the factors ≤√ X */
if x//j\==0 then iterate /*J isn't a factor o f X, so skip it.*/
s = s + j + x%j /* ··· add it and the other factor. */
end /*j*/ /*(above) is marginally faster. */
return s==x /*if sum matches X, then it's perfect!*/</syntaxhighlight>
'''output''' is the same as the traditional version and is about '''11.5''' times faster (testing '''34,000,000''' numbers).
===Lucas-Lehmer method===
This version uses memoization to implement a fast version of the Lucas-Lehmer test.
<syntaxhighlight lang="rexx">/*REXX program tests if a number (or a range of numbers) is/are perfect. */
parse arg low high . /*obtain the optional arguments from CL*/
if high=='' & low=="" then high=34000000 /*if no arguments, then use a range. */
if low=='' then low=1 /*if no LOW, then assume unity. */
low=low+low//2 /*if LOW is odd, bump it by one. */
if high=='' then high=low /*if no HIGH, then assume LOW. */
w=length(high) /*use W for formatting the output. */
numeric digits max(9,w+2) /*ensure enough digits to handle number*/
@.=0; @.1=2 /*highest magic number and its index. */
do i=low to high by 2 /*process the single number or a range.*/
if isPerfect(i) then say right(i,w) 'is a perfect number.'
end /*i*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isPerfect: procedure expose @.; parse arg x /*obtain the number to be tested. */
/*Lucas-Lehmer know that perfect */
/* numbers can be expressed as: */
/* [2**n - 1] * [2** (n-1) ] */
if @.0<x then do @.1=@.1 while @._<=x; _=(2**@.1-1)*2**(@.1-1); @.0=_; @._=_
end /*@.1*/ /*uses memoization for the formula. */
if @.x==0 then return 0 /*Didn't pass Lucas-Lehmer test? */
s = 3 + x%2 /*we know the following factors: */
/* 1 ('cause Mama said so.) */
/* 2 ('cause it's even.) */
/* x÷2 ( " " " ) ___*/
do j=3 while j*j<=x /*starting at 3, find the factors ≤√ X */
if x//j\==0 then iterate /*J divides X evenly, so ··· */
s=s + j + x%j /*··· add it and the other factor. */
end /*j*/ /*(above) is marginally faster. */
return s==x /*if the sum matches X, it's perfect!*/</syntaxhighlight>
'''output''' is the same as the traditional version and is about '''75''' times faster (testing '''34,000,000''' numbers).
===Lucas-Lehmer + other optimizations===
This version uses the Lucas-Lehmer method, digital roots, and restricts itself to ''even'' numbers, and
<br>also utilizes a check for the last-two-digits as per François Édouard Anatole Lucas (in 1891).
Also, in the first '''do''' loop, the index <big>'''i'''</big> is ''fast advanced'' according to the last number tested.
An integer square root function was added to limit the factorization of a number.
<syntaxhighlight lang="rexx">/*REXX program tests if a number (or a range of numbers) is/are perfect. */
parse arg low high . /*obtain optional arguments from the CL*/
if high=='' & low=="" then high=34000000 /*No arguments? Then use a range. */
if low=='' then low=1 /*if no LOW, then assume unity. */
low=low+low//2 /*if LOW is odd, bump it by one. */
if high=='' then high=low /*if no HIGH, then assume LOW. */
w=length(high) /*use W for formatting the output. */
numeric digits max(9,w+2) /*ensure enough decimal digits for nums*/
@. =0; @.1=2; !.=2; _=' 6' /*highest magic number and its index.*/
!._=22; !.16=12; !.28=8; !.36=20; !.56=20; !.76=20; !.96=20
/* [↑] "Lucas' numbers, in 1891. */
do i=low to high by 0 /*process the single number or a range.*/
if isPerfect(i) then say right(i,w) 'is a perfect number.'
i=i+!.? /*use a fast advance for the DO index. */
end /*i*/ /* [↑] note: the DO index is modified.*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isPerfect: procedure expose @. !. ? /*expose (make global) some variables. */
parse arg x 1 y '' -2 ? /*# (and copy), and the last 2 digits.*/
if x==6 then return 1 /*handle the special case of six. */
if !.?==2 then return 0 /*test last two digits: François Lucas.*/
/*╔═════════════════════════════════════════════╗
║ Lucas─Lehmer know that perfect numbers can ║
║ be expressed as: [2^n -1] * {2^(n-1) } ║
╚═════════════════════════════════════════════╝*/
if @.0<x then do @.1=@.1 while @._<=x; _=(2**@.1-1)*2**(@.1-1); @.0=_; @._=_
end /*@.1*/ /* [↑] uses memoization for formula. */
if @.x==0 then return 0 /*Didn't pass Lucas-Lehmer? Not perfect*/
/*[↓] perfect numbers digital root = 1*/
do until y<10 /*find the digital root of Y. */
parse var y d 2; do k=2 for length(y)-1; d=d+substr(y,k,1); end /*k*/
y=d /*find digital root of the digital root*/
end /*until*/ /*wash, rinse, repeat ··· */
if d\==1 then return 0 /*Is digital root ¬ 1? Then ¬ perfect.*/
s=3 + x%2 /*we know the following factors: unity,*/
z=x /*2, and x÷2 (x is even). */
q=1; do while q<=z; q=q*4 ; end /*while q≤z*/ /* _____*/
r=0 /* [↓] R will be the integer √ X */
do while q>1; q=q%4; _=z-r-q; r=r%2; if _>=0 then do; z=_; r=r+q; end
end /*while q>1*/ /* [↑] compute the integer SQRT of X.*/
/* _____*/
do j=3 to r /*starting at 3, find factors ≤ √ X */
if x//j==0 then s=s+j+x%j /*J divisible by X? Then add J and X÷J*/
end /*j*/
return s==x /*if the sum matches X, then perfect! */</syntaxhighlight>
'''output''' is the same as the traditional version and is about '''500''' times faster (testing '''34,000,000''' numbers). <br><br>
=={{header|Ring}}==
<syntaxhighlight lang="ring">
for i = 1 to 10000
if perfect(i) see i + nl ok
next
func perfect n
sum = 0
for i = 1 to n - 1
if n % i = 0 sum = sum + i ok
next
if sum = n return 1 else return 0 ok
return sum
</syntaxhighlight>
=={{header|RPL}}==
≪ 0 SWAP 1
'''WHILE''' DUP2 > '''REPEAT'''
'''IF''' DUP2 MOD NOT '''THEN''' ROT OVER + ROT ROT '''END'''
1 + '''END'''
DROP ==
≫ ''''PFCT?'''' STO
≪
{ } 1 1000 '''FOR''' n
'''IF''' n '''PFCT?''' '''THEN''' n + '''END''' '''NEXT'''
≫ ''''TASK'''' STO
{{out}}
<pre>
1: { 6 28 496 }
</pre>
A vintage HP-28S needs 157 seconds to collect all perfect numbers under 100...
=={{header|Ruby}}==
<
sum += i if n % i == 0
end
sum == n
end</syntaxhighlight>
Functional style:
<syntaxhighlight lang="ruby">def perf(n)
n == (1...n).select {|i| n % i == 0}.inject(:+)
end</syntaxhighlight>
Faster version:
<syntaxhighlight lang="ruby">def perf(n)
divisors = []
for i in 1..Integer.sqrt(n)
divisors << i << n/i if n % i == 0
end
divisors.uniq.inject(:+) == 2*n
end</syntaxhighlight>
Test:
<syntaxhighlight lang="ruby">for n in 1..10000
puts n if perf(n)
end</syntaxhighlight>
{{out}}
<pre>
6
28
496
8128
</pre>
===Fast (Lucas-Lehmer)===
Generate and memoize perfect numbers as needed.
<syntaxhighlight lang="ruby">require "prime"
def mersenne_prime_pow?(p)
# Lucas-Lehmer test; expects prime as argument
return true if p == 2
m_p = ( 1 << p ) - 1
s = 4
(p-2).times{ s = (s**2 - 2) % m_p }
s == 0
end
@perfect_numerator = Prime.each.lazy.select{|p| mersenne_prime_pow?(p)}.map{|p| 2**(p-1)*(2**p-1)}
@perfects = @perfect_numerator.take(1).to_a
def perfect?(num)
@perfects << @perfect_numerator.next until @perfects.last >= num
@perfects.include? num
end
# demo
p (1..10000).select{|num| perfect?(num)}
t1 = Time.now
p perfect?(13164036458569648337239753460458722910223472318386943117783728128)
p Time.now - t1
</syntaxhighlight>
{{out}}
<pre>
[6, 28, 496, 8128]
true
0.001053954
</pre>
As the task states, it is not known if there are any odd perfect numbers (any that exist are larger than 10**2000). This program tests 10**2001 in about 30 seconds - but only for even perfects.
=={{header|Run BASIC}}==
<syntaxhighlight lang="runbasic">for i = 1 to 10000
if perf(i) then print i;" ";
next i
FUNCTION perf(n)
for i = 1 TO n - 1
IF n MOD i = 0 THEN sum = sum + i
next i
IF sum = n THEN perf = 1
END FUNCTION</syntaxhighlight>
{{Out}}
<pre>6 28 496 8128</pre>
=={{header|Rust}}==
<syntaxhighlight lang="rust">
fn main ( ) {
fn factor_sum(n: i32) -> i32 {
let mut v = Vec::new(); //create new empty array
for x in 1..n-1 { //test vaules 1 to n-1
if n%x == 0 { //if current x is a factor of n
v.push(x); //add x to the array
}
}
let mut sum = v.iter().sum(); //iterate over array and sum it up
return sum;
}
fn perfect_nums(n: i32) {
for x in 2..n { //test numbers from 1-n
if factor_sum(x) == x {//call factor_sum on each value of x, if return value is = x
println!("{} is a perfect number.", x); //print value of x
}
}
}
perfect_nums(10000);
}
</syntaxhighlight>
=={{header|SASL}}==
Copied from the SASL manual, page 22:
<syntaxhighlight lang="sasl">
|| The function which takes a number and returns a list of its factors (including one but excluding itself)
|| can be written
factors n = { a <- 1.. n/2; n rem a = 0 }
|| If we define a perfect number as one which is equal to the sum of its factors (for example 6 = 3 + 2 + 1 is perfect)
|| we can write the list of all perfect numbers as
perfects = { n <- 1... ; n = sum(factors n) }
</syntaxhighlight>
=={{header|S-BASIC}}==
<syntaxhighlight lang="basic">
$lines
rem - return p mod q
function mod(p, q = integer) = integer
end = p - q * (p/q)
rem - return true if n is perfect, otherwise false
function isperfect(n = integer) = integer
var sum, f1, f2 = integer
sum = 1
f1 = 2
while (f1 * f1) <= n do
begin
if mod(n, f1) = 0 then
begin
sum = sum + f1
f2 = n / f1
if f2 > f1 then sum = sum + f2
end
f1 = f1 + 1
end
end =
rem - exercise the function
var k, found = integer
print "Searching up to"; search_limit; " for perfect numbers ..."
found = 0
for k = 2 to search_limit
if isperfect(k) then
begin
print k
found = found + 1
end
next k
print found; " were found"
end
</syntaxhighlight>
{{out}}
<pre>
Searching up to 10000 for perfect numbers ...
6
28
496
8128
4 were found
</pre>
=={{header|Scala}}==
<syntaxhighlight lang="scala">def perfectInt(input: Int) = ((2 to sqrt(input).toInt).collect {case x if input % x == 0 => x + input / x}).sum == input - 1</syntaxhighlight>
'''or'''
<syntaxhighlight lang="scala">def perfect(n: Int) =
(for (x <- 2 to n/2 if n % x == 0) yield x).sum + 1 == n
</syntaxhighlight>
=={{header|Scheme}}==
<
(let loop ((i 1)
(sum 0))
Line 453 ⟶ 4,337:
(loop (+ i 1) (+ sum i)))
(else
(loop (+ i 1) sum)))))</
=={{header|Seed7}}==
<syntaxhighlight lang="seed7">$ include "seed7_05.s7i";
const func boolean: isPerfect (in integer: n) is func
result
var boolean: isPerfect is FALSE;
local
var integer: i is 0;
var integer: sum is 1;
var integer: q is 0;
begin
for i range 2 to sqrt(n) do
if n rem i = 0 then
sum +:= i;
q := n div i;
if q > i then
sum +:= q;
end if;
end if;
end for;
isPerfect := sum = n;
end func;
const proc: main is func
local
var integer: n is 0;
begin
for n range 2 to 33550336 do
if isPerfect(n) then
writeln(n);
end if;
end for;
end func;</syntaxhighlight>
{{Out}}
<pre>
6
28
496
8128
33550336
</pre>
=={{header|Sidef}}==
<syntaxhighlight lang="ruby">func is_perfect(n) {
n.sigma == 2*n
}
for n in (1..10000) {
say n if is_perfect(n)
}</syntaxhighlight>
Alternatively, a more efficient check for even perfect numbers:
<syntaxhighlight lang="ruby">func is_even_perfect(n) {
var square = (8*n + 1)
square.is_square || return false
var t = ((square.isqrt + 1) / 2)
t.is_smooth(2) || return false
t-1 -> is_prime
}
for n in (1..10000) {
say n if is_even_perfect(n)
}</syntaxhighlight>
{{out}}
<pre>
6
28
496
8128
</pre>
=={{header|Simula}}==
<syntaxhighlight lang="simula">BOOLEAN PROCEDURE PERF(N); INTEGER N;
BEGIN
INTEGER SUM;
FOR I := 1 STEP 1 UNTIL N-1 DO
IF MOD(N, I) = 0 THEN
SUM := SUM + I;
PERF := SUM = N;
END PERF;</syntaxhighlight>
=={{header|Slate}}==
<
[
(((2 to: n // 2 + 1) select: [| :m | (n rem: m) isZero])
inject: 1 into: #+ `er) = n
].</
=={{header|Smalltalk}}==
<syntaxhighlight lang="smalltalk">Integer extend [
"Translation of the C version; this is faster..."
Line 484 ⟶ 4,452:
inject: 1 into: [ :a :b | a + b ] ) = self
]
].</
<
=={{header|SparForte}}==
As a structured script.
<syntaxhighlight lang="ada">#!/usr/local/bin/spar
pragma annotate( summary, "perfect" );
pragma annotate( description, "In mathematics, a perfect number is a positive integer" );
pragma annotate( description, "that is the sum of its proper positive divisors, that is," );
pragma annotate( description, "the sum of the positive divisors excluding the number" );
pragma annotate( description, "itself." );
pragma annotate( see_also, "http://en.wikipedia.org/wiki/Perfect_number" );
pragma annotate( author, "Ken O. Burtch" );
pragma license( unrestricted );
pragma restriction( no_external_commands );
procedure perfect is
function is_perfect( n : positive ) return boolean is
total : natural := 0;
begin
for i in 1..n-1 loop
if n mod i = 0 then
total := @+i;
end if;
end loop;
return total = natural( n );
end is_perfect;
number : positive;
result : boolean;
begin
number := 6;
result := is_perfect( number );
put( number ) @ ( " : " ) @ ( result );
new_line;
number := 18;
result := is_perfect( number );
put( number ) @ ( " : " ) @ ( result );
new_line;
number := 28;
result := is_perfect( number );
put( number ) @ ( " : " ) @ ( result );
new_line;
end perfect;</syntaxhighlight>
=={{header|Swift}}==
{{trans|Java}}
<syntaxhighlight lang="swift">func perfect(n:Int) -> Bool {
var sum = 0
for i in 1..<n {
if n % i == 0 {
sum += i
}
}
return sum == n
}
for i in 1..<10000 {
if perfect(i) {
println(i)
}
}</syntaxhighlight>
{{out}}
<pre>
6
28
496
8128
</pre>
=={{header|Tcl}}==
<
set sum 0
for {set i 1} {$i <= $n} {incr i} {
Line 495 ⟶ 4,535:
}
expr {$sum == 2*$n}
}</
=={{header|Ursala}}==
<
#import nat
is_perfect = ~&itB&& ^(~&,~&t+ iota); ^E/~&l sum:-0+ ~| not remainder</
This test program applies the function to a list of the first five hundred natural
numbers and deletes the imperfect ones.
<
examples = is_perfect*~ iota 500</
{{Out}}
<pre><6,28,496></pre>
=={{header|VBA}}==
{{trans|Phix}}
Using [[Factors_of_an_integer#VBA]], slightly adapted.
<syntaxhighlight lang="vb">Private Function Factors(x As Long) As String
Application.Volatile
Dim i As Long
Dim cooresponding_factors As String
Factors = 1
corresponding_factors = x
For i = 2 To Sqr(x)
If x Mod i = 0 Then
Factors = Factors & ", " & i
If i <> x / i Then corresponding_factors = x / i & ", " & corresponding_factors
End If
Next i
If x <> 1 Then Factors = Factors & ", " & corresponding_factors
End Function
Private Function is_perfect(n As Long)
fs = Split(Factors(n), ", ")
Dim f() As Long
ReDim f(UBound(fs))
For i = 0 To UBound(fs)
f(i) = Val(fs(i))
Next i
is_perfect = WorksheetFunction.Sum(f) - n = n
End Function
Public Sub main()
Dim i As Long
For i = 2 To 100000
If is_perfect(i) Then Debug.Print i
Next i
End Sub</syntaxhighlight>{{out}}
<pre> 6
28
496
8128 </pre>
=={{header|VBScript}}==
<syntaxhighlight lang="vb">Function IsPerfect(n)
IsPerfect = False
i = n - 1
sum = 0
Do While i > 0
If n Mod i = 0 Then
sum = sum + i
End If
i = i - 1
Loop
If sum = n Then
IsPerfect = True
End If
End Function
WScript.StdOut.Write IsPerfect(CInt(WScript.Arguments(0)))
WScript.StdOut.WriteLine</syntaxhighlight>
{{out}}
<pre>
C:\>cscript /nologo perfnum.vbs 6
True
C:\>cscript /nologo perfnum.vbs 29
False
C:\>
</pre>
=={{header|V (Vlang)}}==
{{trans|go}}
<syntaxhighlight lang="v (vlang)">fn compute_perfect(n i64) bool {
mut sum := i64(0)
for i := i64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == n
}
// following fntion satisfies the task, returning true for all
// perfect numbers representable in the argument type
fn is_perfect(n i64) bool {
return n in [i64(6), 28, 496, 8128, 33550336, 8589869056,
137438691328, 2305843008139952128]
}
// validation
fn main() {
for n := i64(1); ; n++ {
if is_perfect(n) != compute_perfect(n) {
panic("bug")
}
if n%i64(1e3) == 0 {
println("tested $n")
}
}
}</syntaxhighlight>
{{Out}}
<pre>
tested 1000
tested 2000
tested 3000
...
</pre>
=={{header|Wren}}==
===Version 1===
{{trans|D}}
Restricted to the first four perfect numbers as the fifth one is very slow to emerge.
<syntaxhighlight lang="wren">var isPerfect = Fn.new { |n|
if (n <= 2) return false
var tot = 1
for (i in 2..n.sqrt.floor) {
if (n%i == 0) {
tot = tot + i
var q = (n/i).floor
if (q > i) tot = tot + q
}
}
return n == tot
}
System.print("The first four perfect numbers are:")
var count = 0
var i = 2
while (count < 4) {
if (isPerfect.call(i)) {
System.write("%(i) ")
count = count + 1
}
i = i + 2 // there are no known odd perfect numbers
}
System.print()</syntaxhighlight>
{{out}}
<pre>
6 28 496 8128
</pre>
===Version 2===
{{libheader|Wren-math}}
This makes use of the fact that all known perfect numbers are of the form <big> (2<sup>''n''</sup> - 1) × 2<sup>''n'' - 1</sup></big> where <big> (2<sup>''n''</sup> - 1)</big> is prime and finds the first seven perfect numbers instantly. The numbers are too big after that to be represented accurately by Wren.
<syntaxhighlight lang="wren">import "./math" for Int
var isPerfect = Fn.new { |n|
if (n <= 2) return false
var tot = 1
for (i in 2..n.sqrt.floor) {
if (n%i == 0) {
tot = tot + i
var q = (n/i).floor
if (q > i) tot = tot + q
}
}
return n == tot
}
System.print("The first seven perfect numbers are:")
var count = 0
var p = 2
while (count < 7) {
var n = 2.pow(p) - 1
if (Int.isPrime(n)) {
n = n * 2.pow(p-1)
if (isPerfect.call(n)) {
System.write("%(n) ")
count = count + 1
}
}
p = p + 1
}
System.print()</syntaxhighlight>
{{out}}
<pre>
6 28 496 8128 33550336 8589869056 137438691328
</pre>
=={{header|XPL0}}==
<syntaxhighlight lang="xpl0">include c:\cxpl\codes; \intrinsic 'code' declarations
func Perfect(N); \Return 'true' if N is a perfect number
int N, S, I, Q;
[S:= 1;
for I:= 2 to sqrt(N) do
[Q:= N/I;
if rem(0)=0 then S:= S+I+Q;
];
return S=N & N#1;
];
int A, N;
[for A:= 1 to 16 do
[N:= (1<<A - 1) * 1<<(A-1);
if Perfect(N) then [IntOut(0, N); CrLf(0)];
];
]</syntaxhighlight>
{{out}}
<pre>
6
28
496
8128
33550336
</pre>
=={{header|Yabasic}}==
{{trans|True BASIC}}
<syntaxhighlight lang="basic">
sub isPerfect(n)
if (n < 2) or mod(n, 2) = 1 then return false : endif
// asumimos que los números impares no son perfectos
sum = 0
for i = 1 to n-1
if mod(n,i) = 0 then sum = sum + i : endif
next i
if sum = n then return true else return false : endif
end sub
print "Los primeros 5 numeros perfectos son:"
for i = 1 to 33550336
if isPerfect(i) then print i, " ", : endif
next i
print
end
</syntaxhighlight>
=={{header|Zig}}==
<syntaxhighlight lang="zig">
const std = @import("std");
const expect = std.testing.expect;
const stdout = std.io.getStdOut().outStream();
pub fn main() !void {
var i: u32 = 2;
try stdout.print("The first few perfect numbers are: ", .{});
while (i <= 10_000) : (i += 2) if (propersum(i) == i)
try stdout.print("{} ", .{i});
try stdout.print("\n", .{});
}
fn propersum(n: u32) u32 {
var sum: u32 = 1;
var d: u32 = 2;
while (d * d <= n) : (d += 1) if (n % d == 0) {
sum += d;
const q = n / d;
if (q > d)
sum += q;
};
return sum;
}
test "Proper divisors" {
expect(propersum(28) == 28);
expect(propersum(71) == 1);
expect(propersum(30) == 42);
}
</syntaxhighlight>
{{Out}}
<pre>
The first few perfect numbers are: 6 28 496 8128
</pre>
=={{header|zkl}}==
{{trans|D}}
<syntaxhighlight lang="zkl">fcn isPerfectNumber1(n)
{ n == [1..n-1].filter('wrap(i){ n % i == 0 }).sum(); }</syntaxhighlight>
{{out}}
<pre>
[1..0d10_000].filter(isPerfectNumber1).println();
L(6,28,496,8128)
</pre>
| |||