Perfect numbers: Difference between revisions
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Or, a little faster (by restricting the search space): |
Or, a little faster (by restricting the search space): |
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<lang python>from |
<lang python>from operator import (add) |
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from operator import (add) |
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from math import (sqrt) |
from math import (sqrt) |
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# perfect:: |
# perfect::Int - > Bool |
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def perfect(n): |
def perfect(n): |
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lows = filter( |
lows = filter( |
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xrange(1, 1 + int(sqrt(n))) |
xrange(1, 1 + int(sqrt(n))) |
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) |
) |
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return (1 < n) and |
return (1 < n) and( |
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n == sum( |
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lows + concatMap( |
lows + concatMap( |
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lambda x: ( |
lambda x: ( |
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lambda y=n / x: [y |
lambda y=n / x: [y] |
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if x != y |
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else [] |
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)() |
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)(lows) |
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) / 2 |
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) |
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# GENERIC |
# GENERIC-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- - |
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# concatMap |
# concatMap::(a - > [b]) - > [a] - > [b] |
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def concatMap(f): |
def concatMap(f): |
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return lambda xs: ( |
return lambda xs: ( |
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Revision as of 15:43, 8 September 2018
You are encouraged to solve this task according to the task description, using any language you may know.
Write a function which says whether a number is perfect.
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 2n-1, all known perfect numbers can be derived from these primes
using the formula (2n - 1) × 2n - 1.
It is not known if there are any odd perfect numbers (any that exist are larger than 102000).
The number of known perfect numbers is 50 (as of September, 2018), and contains over 46 million decimal digits.
- See also
-
- Rational Arithmetic
- Perfect numbers on OEIS
- Odd Perfect showing the current status of bounds on odd perfect numbers.
360 Assembly
Simple code
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. <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</lang>
- Output:
6
28
496
8128
Some optimizations
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. <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</lang>
- Output:
0000000000000006 0000000000000028 0000000000000496 0000000000008128 0000000033550337 0000008589869056 0000137438691328
Ada
<lang ada>function Is_Perfect(N : Positive) return Boolean is
Sum : Natural := 0;
begin
for I in 1..N - 1 loop
if N mod I = 0 then
Sum := Sum + I;
end if;
end loop;
return Sum = N;
end Is_Perfect;</lang>
ALGOL 68
<lang algol68>PROC is perfect = (INT candidate)BOOL: (
INT sum :=1;
FOR f1 FROM 2 TO ENTIER ( sqrt(candidate)*(1+2*small real) ) WHILE
IF candidate MOD f1 = 0 THEN
sum +:= f1;
INT f2 = candidate OVER f1;
IF f2 > f1 THEN
sum +:= f2
FI
FI;
- WHILE # sum <= candidate DO
SKIP OD; sum=candidate
);
test:(
FOR i FROM 2 TO 33550336 DO IF is perfect(i) THEN print((i, new line)) FI OD
)</lang>
- Output:
+6
+28
+496
+8128
+33550336
AppleScript
<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</lang>
- Output:
<lang AppleScript>{6, 28, 496, 8128}</lang>
AutoHotkey
This will find the first 8 perfect numbers. <lang autohotkey>Loop, 30 {
If isMersennePrime(A_Index + 1) res .= "Perfect number: " perfectNum(A_Index + 1) "`n"
}
MsgBox % res
perfectNum(N) {
Return 2**(N - 1) * (2**N - 1)
}
isMersennePrime(N) {
If (isPrime(N)) && (isPrime(2**N - 1)) Return true
}
isPrime(N) {
Loop, % Floor(Sqrt(N))
If (A_Index > 1 && !Mod(N, A_Index))
Return false
Return true
}</lang>
AWK
<lang awk>$ awk 'func perf(n){s=0;for(i=1;i<n;i++)if(n%i==0)s+=i;return(s==n)} BEGIN{for(i=1;i<10000;i++)if(perf(i))print i}' 6 28 496 8128</lang>
Axiom
Using the interpreter, define the function: <lang Axiom>perfect?(n:Integer):Boolean == reduce(+,divisors n) = 2*n</lang> Alternatively, using the Spad compiler: <lang Axiom>)abbrev package TESTP TestPackage TestPackage() : with
perfect?: Integer -> Boolean
==
add
import IntegerNumberTheoryFunctions
perfect? n == reduce("+",divisors n) = 2*n</lang>
Examples (testing 496, testing 128, finding all perfect numbers in 1...10000): <lang Axiom>perfect? 496 perfect? 128 [i for i in 1..10000 | perfect? i]</lang>
- Output:
<lang Axiom>true false [6,28,496,8128]</lang>
BASIC
<lang qbasic>FUNCTION perf(n) sum = 0 for i = 1 to n - 1 IF n MOD i = 0 THEN sum = sum + i END IF NEXT i IF sum = n THEN perf = 1 ELSE perf = 0 END IF END FUNCTION</lang>
Sinclair ZX81 BASIC
Call this subroutine and it will (eventually) return PERFECT = 1 if N is perfect or PERFECT = 0 if it is not. <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</lang>
BBC BASIC
BASIC version
<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%)</lang>
- Output:
6
28
496
8128
Assembler version
<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</lang>
- Output:
4
6
28
496
8128
33550336
Bracmat
<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|) )
);</lang>
- Output:
6 28 496 8128
C
<lang c>#include "stdio.h"
- include "math.h"
int perfect(int n) {
int max = (int)sqrt((double)n) + 1; int tot = 1; int i;
for (i = 2; i < max; i++)
if ( (n % i) == 0 ) {
tot += i;
int q = n / i;
if (q > i)
tot += q;
}
return tot == n;
}
int main() {
int n;
for (n = 2; n < 33550337; n++)
if (perfect(n))
printf("%d\n", n);
return 0;
}</lang> Using functions from Factors of an integer#Prime factoring: <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; }</lang>
C#
<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 ; }</lang>
Version using Lambdas, will only work from version 3 of C# on
<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; }</lang>
C++
<lang cpp>#include <iostream> using namespace std ;
int divisor_sum( int number ) {
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 (divisor_sum(num) == num)
cout << num << '\n' ;
}
return 0 ;
} </lang>
Clojure
<lang clojure>(defn proper-divisors [n]
(if (< n 4)
[1]
(->> (range 2 (inc (quot n 2)))
(filter #(zero? (rem n %)))
(cons 1))))
(defn perfect? [n]
(= (reduce + (proper-divisors n)) n))</lang>
<lang clojure>(defn perfect? [n]
(->> (for [i (range 1 n)] :when (zero? (rem n i))] i)
(reduce +)
(= n)))</lang>
Functional version
<lang clojure>(defn perfect? [n] (= (reduce + (filter #(zero? (rem n %)) (range 1 n))) n))</lang>
CoffeeScript
Optimized version, for fun. <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)</lang>
- Output:
> coffee perfect_numbers.coffee 6 28 496 8128
COBOL
main.cbl: <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.</lang>
perfect.cbl: <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.</lang>
Common Lisp
<lang lisp>(defun perfectp (n)
(= n (loop for i from 1 below n when (= 0 (mod n i)) sum i)))</lang>
D
Functional Version
<lang d>import std.stdio, std.algorithm, std.range;
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;
}</lang>
- Output:
[6, 28, 496, 8128]
Faster Imperative Version
<lang d>import std.stdio, std.math, std.range, std.algorithm;
bool isPerfectNumber2(in int n) pure nothrow {
if (n < 2)
return false;
int total = 1;
foreach (immutable i; 2 .. cast(int)real(n).sqrt + 1)
if (n % i == 0) {
immutable int q = n / i;
total += i;
if (q > i)
total += q;
}
return total == n;
}
void main() {
10_000.iota.filter!isPerfectNumber2.writeln;
}</lang>
- Output:
[6, 28, 496, 8128]
With a 33_550_337.iota it outputs:
[6, 28, 496, 8128, 33550336]
Dart
Explicit Iterative Version
<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;
}</lang>
Compact Version
<lang d>isPerfect(n) =>
n == new List.generate(n-1, (i) => n%(i+1) == 0 ? i+1 : 0).fold(0, (p,n)=>p+n);</lang>
In either case, if we test to find all the perfect numbers up to 1000, we get: <lang d>main() =>
new List.generate(1000,(i)=>i+1).where(isPerfect).forEach(print);</lang>
- Output:
6 28 496
E
<lang e>pragma.enable("accumulator") def isPerfectNumber(x :int) {
var sum := 0
for d ? (x % d <=> 0) in 1..!x {
sum += d
if (sum > x) { return false }
}
return sum <=> x
}</lang>
Eiffel
<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 </lang>
- Output:
6 is perfect... True 77 is perfect... False 128 is perfect... False 496 is perfect... True
Elena
ELENA 3.4: <lang elena>import system'routines. import system'math. import extensions.
extension extension {
isPerfect
= 1 till:self repeat(:n)( (self mod:n == 0) iif(n,0) ); summarize(Integer new) == self.
}
public program [
1 till:10000 do(:n)
[
if(n isPerfect)
[ console printLine(n," is perfect") ]
].
console readChar
]</lang>
- Output:
6 is perfect 28 is perfect 496 is perfect 8128 is perfect
Elixir
<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)</lang>
- Output:
[6, 28, 496, 8128]
Erlang
<lang erlang>is_perfect(X) ->
X == lists:sum([N || N <- lists:seq(1,X-1), X rem N == 0]).</lang>
ERRE
<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</lang>
- Output:
6
28
496
8128
F#
<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</lang>
- Output:
6 is perfect 28 is perfect 496 is perfect 8128 is perfect
FALSE
<lang false>[0\1[\$@$@-][\$@$@$@$@\/*=[@\$@+@@]?1+]#%=]p: 45p;!." "28p;!. { 0 -1 }</lang>
Factor
<lang factor>USING: kernel math math.primes.factors sequences ; IN: rosettacode.perfect-numbers
- perfect? ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;</lang>
Forth
<lang forth>: perfect? ( n -- ? )
1 over 2/ 1+ 2 ?do over i mod 0= if i + then loop = ;</lang>
Fortran
<lang fortran>FUNCTION isPerfect(n)
LOGICAL :: isPerfect INTEGER, INTENT(IN) :: n INTEGER :: i, factorsum
isPerfect = .FALSE.
factorsum = 1
DO i = 2, INT(SQRT(REAL(n)))
IF(MOD(n, i) == 0) factorsum = factorsum + i + (n / i)
END DO
IF (factorsum == n) isPerfect = .TRUE.
END FUNCTION isPerfect</lang>
FreeBASIC
<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</lang>
- Output:
The first 5 perfect numbers are : 6 28 496 8128 33550336
FunL
<lang funl>def perfect( n ) = sum( d | d <- 1..n if d|n ) == 2n
println( (1..500).filter(perfect) )</lang>
- Output:
(6, 28, 496)
GAP
<lang gap>Filtered([1 .. 10000], n -> Sum(DivisorsInt(n)) = 2*n);
- [ 6, 28, 496, 8128 ]</lang>
Go
<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)
}
}
}
</lang>
- Output:
tested 1000 tested 2000 tested 3000 ...
Groovy
Solution: <lang groovy>def isPerfect = { n ->
n > 4 && (n == (2..Math.sqrt(n)).findAll { n % it == 0 }.inject(1) { factorSum, i -> factorSum += i + n/i })
}</lang> Test program: <lang groovy>(0..10000).findAll { isPerfect(it) }.each { println it }</lang>
- Output:
6 28 496 8128
Haskell
<lang haskell>perfect n =
n == sum [i | i <- [1..n-1], n `mod` i == 0]</lang>
Create a list of known perfects: <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]</lang>
HicEst
<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</lang>
Icon and Unicon
<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</lang>
- Output:
Perfect numbers from 1 to 100000: 6 28 496 8128 Done.
J
<lang j>is_perfect=: +: = >:@#.~/.~&.q:@(6>.<.)</lang>
Examples of use, including extensions beyond those assumptions: <lang j> is_perfect 33550336 1
I. is_perfect i. 100000
6 28 496 8128
] zero_through_twentynine =. i. 3 10 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
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</lang>
More efficient version based on comments by Henry Rich and Roger Hui (comment train seeded by Jon Hough).
Java
<lang java>public static boolean perf(int n){ int sum= 0; for(int i= 1;i < n;i++){ if(n % i == 0){ sum+= i; } } return sum == n; }</lang> Or for arbitrary precision: <lang java>import java.math.BigInteger;
public static boolean perf(BigInteger n){ BigInteger sum= BigInteger.ZERO; for(BigInteger i= BigInteger.ONE; i.compareTo(n) < 0;i=i.add(BigInteger.ONE)){ if(n.mod(i).equals(BigInteger.ZERO)){ sum= sum.add(i); } } return sum.equals(n); }</lang>
JavaScript
Imperative
<lang javascript>function is_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 i;
for (i = 1; i < 10000; i++)
{
if (is_perfect(i)) print(i);
}</lang>
- Output:
6 28 496 8128
Functional
ES5
Naive version (brute force)
<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);</lang>
Output:
<lang JavaScript>[6, 28, 496, 8128]</lang>
Much faster (more efficient factorisation)
<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);</lang>
Output:
<lang JavaScript>[6, 28, 496, 8128]</lang>
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 [].)
<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);</lang>
Output: <lang JavaScript>[6, 28, 496, 8128]</lang>
ES6
<lang JavaScript>(() => {
const main = () =>
enumFromTo(1, 10000).filter(perfect);
// perfect :: Int -> Bool
const perfect = n => {
let 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)
return main();
})();</lang>
- Output:
<lang JavaScript>[6, 28, 496, 8128]</lang>
jq
<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 )</lang>
- Output:
$ jq -n -f is_perfect.jq 6 28 496 8128
Julia
<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)</lang>
- Output:
perfects(10000) = [6, 28, 496, 8128]
K
<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)</lang>
Kotlin
<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("$i ")
}</lang>
- Output:
The first five perfect numbers are: 6 28 496 8128 33550336
LabVIEW
This image is a VI Snippet, an executable image of LabVIEW code. The LabVIEW version is shown on the top-right hand corner. You can download it, then drag-and-drop it onto the LabVIEW block diagram from a file browser, and it will appear as runnable, editable code.
Lasso
<lang lasso>#!/usr/bin/lasso9
define isPerfect(n::integer) => {
#n < 2 ? return false
return #n == (
with i in generateSeries(1, math_floor(math_sqrt(#n)) + 1)
where #n % #i == 0
let q = #n / #i
sum (#q > #i ? (#i == 1 ? 1 | #q + #i) | 0)
)
}
with x in generateSeries(1, 10000)
where isPerfect(#x)
select #x</lang>
- Output:
<lang lasso>6, 28, 496, 8128</lang>
Liberty BASIC
<lang lb>for n =1 to 10000
if perfect( n) =1 then print n; " is perfect."
next n
end
function perfect( n)
sum =0
for i =1 TO n /2
if n mod i =0 then
sum =sum +i
end if
next i
if sum =n then
perfect= 1
else
perfect =0
end if
end function</lang>
Lingo
<lang lingo>on isPercect (n)
sum = 1 cnt = n/2 repeat with i = 2 to cnt if n mod i = 0 then sum = sum + i end repeat return sum=n
end</lang>
Logo
<lang logo>to perfect? :n
output equal? :n apply "sum filter [equal? 0 modulo :n ?] iseq 1 :n/2
end</lang>
Lua
<lang Lua>function isPerfect(x)
local sum = 0 for i = 1, x-1 do
sum = (x % i) == 0 and sum + i or sum
end return sum == x
end</lang>
M4
<lang M4>define(`for',
`ifelse($#,0,``$0, `ifelse(eval($2<=$3),1, `pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',incr($2),$3,`$4')')')')dnl
define(`ispart',
`ifelse(eval($2*$2<=$1),1,
`ifelse(eval($1%$2==0),1,
`ifelse(eval($2*$2==$1),1,
`ispart($1,incr($2),eval($3+$2))',
`ispart($1,incr($2),eval($3+$2+$1/$2))')',
`ispart($1,incr($2),$3)')',
$3)')
define(`isperfect',
`eval(ispart($1,2,1)==$1)')
for(`x',`2',`33550336',
`ifelse(isperfect(x),1,`x
')')</lang>
Mathematica / Wolfram Language
Custom function: <lang Mathematica>PerfectQ[i_Integer] := Total[Divisors[i]] == 2 i</lang> Examples (testing 496, testing 128, finding all perfect numbers in 1...10000): <lang Mathematica>PerfectQ[496] PerfectQ[128] Flatten[PerfectQ/@Range[10000]//Position[#,True]&]</lang> gives back: <lang Mathematica>True False {6,28,496,8128}</lang>
MATLAB
Standard algorithm: <lang MATLAB>function perf = isPerfect(n)
total = 0;
for k = 1:n-1
if ~mod(n, k)
total = total+k;
end
end
perf = total == n;
end</lang> Faster algorithm: <lang MATLAB>function perf = isPerfect(n)
if n < 2
perf = false;
else
total = 1;
k = 2;
quot = n;
while k < quot && total <= n
if ~mod(n, k)
total = total+k;
quot = n/k;
if quot ~= k
total = total+quot;
end
end
k = k+1;
end
perf = total == n;
end
end</lang>
Maxima
<lang maxima>".."(a, b) := makelist(i, i, a, b)$ infix("..")$
perfectp(n) := is(divsum(n) = 2*n)$
sublist(1 .. 10000, perfectp); /* [6, 28, 496, 8128] */</lang>
MAXScript
<lang maxscript>fn isPerfect n = (
local sum = 0
for i in 1 to (n-1) do
(
if mod n i == 0 then
(
sum += i
)
)
sum == n
)</lang>
Microsoft Small Basic
<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 </lang>
Modula-2
<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. </lang>
Nim
<lang nim>import math
proc isPerfect(n: int): bool =
var sum: int = 1
for i in 2 .. <(n.toFloat.sqrt+1).toInt:
if n mod i == 0:
sum += (i + n div i)
return (n == sum)
for i in 2..10_000:
if isPerfect(i):
echo(i)</lang>
Objeck
<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;
}
}
}</lang>
OCaml
<lang ocaml>let perf n =
let sum = ref 0 in
for i = 1 to n-1 do
if n mod i = 0 then
sum := !sum + i
done;
!sum = n</lang>
Functional style: <lang ocaml>(* range operator *) let rec (--) a b =
if a > b then [] else a :: (a+1) -- b
let perf n = n = List.fold_left (+) 0 (List.filter (fun i -> n mod i = 0) (1 -- (n-1)))</lang>
Oforth
<lang Oforth>: isPerfect(n) | i | 0 n 2 / loop: i [ n i mod ifZero: [ i + ] ] n == ; </lang>
- Output:
#isPerfect 10000 seq filter . [6, 28, 496, 8128]
ooRexx
<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</lang>
- Output:
6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number
Oz
<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}}</lang>
PARI/GP
Uses built-in method. Faster tests would use the LL test for evens and myriad results on OPNs otherwise. <lang parigp>isPerfect(n)=sigma(n,-1)==2</lang> Show perfect numbers <lang parigp>forprime(p=2, 2281, if(isprime(2^p-1), print(p"\t",(2^p-1)*2^(p-1))))</lang> Faster with Lucas-Lehmer test <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)</lang>
- Output:
(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
Pascal
<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.</lang>
- Output:
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.
Perl
Functions
<lang perl>sub perf {
my $n = shift;
my $sum = 0;
foreach my $i (1..$n-1) {
if ($n % $i == 0) {
$sum += $i;
}
}
return $sum == $n;
}</lang> Functional style: <lang perl>use List::Util qw(sum);
sub perf {
my $n = shift;
$n == sum(0, grep {$n % $_ == 0} 1..$n-1);
}</lang>
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.
A simple predicate: <lang perl>use ntheory qw/divisor_sum/; sub is_perfect { my $n = shift; divisor_sum($n) == 2*$n; }</lang> Use this naive method to show the first 5. Takes about 15 seconds: <lang perl>use ntheory qw/divisor_sum/; for (1..33550336) {
print "$_\n" if divisor_sum($_) == 2*$_;
}</lang> 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. <lang perl>use ntheory qw/forprimes is_prime/; use bigint; forprimes {
my $n = 2**$_ - 1; print "$_\t", $n * 2**($_-1),"\n" if is_prime($n);
} 2, 4500;</lang>
- Output:
2 6 3 28 5 496 7 8128 13 33550336 17 8589869056 19 137438691328 31 2305843008139952128 61 2658455991569831744654692615953842176 89 191561942608236107294793378084303638130997321548169216 ... 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423 ...
We can speed this up even more using a faster program for printing the large results, as well as a faster primality solution. The first 38 in about 1 second with most of the time printing the large results. Caveat: this goes well past the current bound for odd perfect numbers and does not check for them. <lang perl>use ntheory qw/forprimes is_mersenne_prime/; use Math::GMP qw/:constant/; forprimes {
print "$_\t", (2**$_-1)*2**($_-1),"\n" if is_mersenne_prime($_);
} 7_000_000;</lang>
In addition to generating even perfect numbers, we can also have a fast function which returns true when a given even number is perfect: <lang perl>use ntheory qw(is_mersenne_prime valuation);
sub is_even_perfect {
my ($n) = @_; my $v = valuation($n, 2) || return; my $m = ($n >> $v); ($m & ($m + 1)) && return; ($m >> $v) == 1 || return; is_mersenne_prime($v + 1);
}</lang>
Perl 6
<lang perl6>sub is-perf($n) { $n == [+] grep $n %% *, 1 .. $n div 2 }
- used as
put (grep {.&is-perf}, 1..Inf)[^4];</lang>
- Output:
6 28 496 8128
Phix
<lang Phix>function is_perfect(integer n)
return sum(factors(n,-1))=n
end function
for i=2 to 100000 do
if is_perfect(i) then ?i end if
end for</lang>
- Output:
6 28 496 8128
PHP
<lang php>function is_perfect($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 $num . PHP_EOL;
}</lang>
PicoLisp
<lang PicoLisp>(de perfect (N)
(let C 0
(for I (/ N 2)
(and (=0 (% N I)) (inc 'C I)) )
(= C N) ) )</lang>
<lang PicoLisp>(de faster (N)
(let (C 1 Stop (sqrt N))
(for (I 2 (<= I Stop) (inc I))
(and
(=0 (% N I))
(inc 'C (+ (/ N I) I)) ) )
(= C N) ) )</lang>
PL/I
<lang PL/I>perfect: procedure (n) returns (bit(1));
declare n fixed; declare sum fixed; declare i fixed binary;
sum = 0;
do i = 1 to n-1;
if mod(n, i) = 0 then sum = sum + i;
end;
return (sum=n);
end perfect;</lang>
PowerShell
<lang powershell>Function IsPerfect($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.</lang>
Prolog
Classic approach
Works with SWI-Prolog <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).</lang>
Functional approach
Works with SWI-Prolog and module lambda, written by Ulrich Neumerkel found there http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/lambda.pl <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))).</lang>
PureBasic
<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</lang>
Python
<lang python>def perf(n):
sum = 0
for i in xrange(1, n):
if n % i == 0:
sum += i
return sum == n</lang>
Functional style: <lang python>perf = lambda n: n == sum(i for i in xrange(1, n) if n % i == 0)</lang>
Or, a little faster (by restricting the search space):
<lang python>from operator import (add) from math import (sqrt)
def main():
print(filter(perfect, xrange(1, 10000)))
- perfect::Int - > Bool
def perfect(n):
lows = filter(
lambda x: 0 == (n % x),
xrange(1, 1 + int(sqrt(n)))
)
return (1 < n) and(
n == sum(
lows + concatMap(
lambda x: (
lambda y=n / x: [y]
if x != y
else []
)()
)(lows)
) / 2
)
- GENERIC-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -
- concatMap::(a - > [b]) - > [a] - > [b]
def concatMap(f):
return lambda xs: (
reduce(add, map(f, xs), [])
)
main()</lang>
- Output:
[6, 28, 496, 8128]
R
<lang R>is.perf <- function(n){ if (n==0|n==1) return(FALSE) s <- seq (1,n-1) x <- n %% s m <- data.frame(s,x) out <- with(m, s[x==0]) return(sum(out)==n) }
- Usage - Warning High Memory Usage
is.perf(28) sapply(c(6,28,496,8128,33550336),is.perf)</lang>
Racket
<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)</lang>
REBOL
<lang rebol>perfect?: func [n [integer!] /local sum] [
sum: 0
repeat i (n - 1) [
if zero? remainder n i [
sum: sum + i
]
]
sum = n
]</lang>
REXX
Classic REXX version of ooRexx
This version is a Classic Rexx version of the ooRexx program as of 14-Sep-2013. <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</lang> output when using the default of 10000:
6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number
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
was added to display the perfect numbers. Also, an epilog was written for the re-worked function.
<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! */</lang> output when using the input defaults of: 1 10000
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
<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! */</lang>
output when using the default inputs:
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.
For 10,000 numbers tested, this version is 19.6 times faster than the ooRexx program logic.
For 10,000 numbers tested, this version is 25.6 times faster than the PL/I program logic.
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. <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! */</lang>
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. <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!*/</lang>
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. <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!*/</lang>
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
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 i is fast advanced according to the last number tested.
An integer square root function was added to limit the factorization of a number. <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! */</lang>
output is the same as the traditional version and is about 500 times faster (testing 34,000,000 numbers).
Ring
<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 </lang>
Ruby
<lang ruby>def perf(n)
sum = 0 for i in 1...n sum += i if n % i == 0 end sum == n
end</lang> Functional style: <lang ruby>def perf(n)
n == (1...n).select {|i| n % i == 0}.inject(:+)
end</lang> Faster version: <lang ruby>def perf(n)
divisors = [] for i in 1..Math.sqrt(n) divisors << i << n/i if n % i == 0 end divisors.uniq.inject(:+) == 2*n
end</lang> Test: <lang ruby>for n in 1..10000
puts n if perf(n)
end</lang>
- Output:
6 28 496 8128
Fast (Lucas-Lehmer)
Generate and memoize perfect numbers as needed. <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 </lang>
- Output:
[6, 28, 496, 8128] true 0.001053954
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.
Run BASIC
<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</lang>
- Output:
6 28 496 8128
Rust
<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);
} </lang>
SASL
Copied from the SASL manual, page 22: <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) } </lang>
Scala
<lang scala>def perfectInt(input: Int) = ((2 to sqrt(input).toInt).collect {case x if input % x == 0 => x + input / x}).sum == input - 1</lang>
or
<lang scala>def perfect(n: Int) =
(for (x <- 2 to n/2 if n % x == 0) yield x).sum + 1 == n
</lang>
Scheme
<lang scheme>(define (perf n)
(let loop ((i 1)
(sum 0))
(cond ((= i n)
(= sum n))
((= 0 (modulo n i))
(loop (+ i 1) (+ sum i)))
(else
(loop (+ i 1) sum)))))</lang>
Seed7
<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;</lang>
- Output:
6 28 496 8128 33550336
Sidef
<lang ruby>func is_perfect(n) {
var sum = 0;
for i in (1 ..^ n) {
i.divides(n) && (sum += i);
}
sum == n;
}
10000.times { |i|
is_perfect(i) && say i;
}</lang>
A much faster check for even perfect numbers: <lang ruby>func is_even_perfect(n) {
var square = (8*n + 1) square.is_sqr || return false
var tp = ((square.isqrt + 1) / 2) tp.is_pow || return false
(tp-1).is_prime ? true : false
}
for i in range(0, 10000, 2) {
is_even_perfect(i) && say i
}</lang>
- Output:
6 28 496 8128
Simula
<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;</lang>
Slate
<lang slate>n@(Integer traits) isPerfect [
(((2 to: n // 2 + 1) select: [| :m | (n rem: m) isZero]) inject: 1 into: #+ `er) = n
].</lang>
Smalltalk
<lang smalltalk>Integer extend [
"Translation of the C version; this is faster..."
isPerfectC [ |tot| tot := 1.
(2 to: (self sqrt) + 1) do: [ :i |
(self rem: i) = 0
ifTrue: [ |q|
tot := tot + i.
q := self // i.
q > i ifTrue: [ tot := tot + q ]
]
].
^ tot = self
]
"... but this seems more idiomatic"
isPerfect [
^ ( ( ( 2 to: self // 2 + 1) select: [ :a | (self rem: a) = 0 ] )
inject: 1 into: [ :a :b | a + b ] ) = self
]
].</lang>
<lang smalltalk>1 to: 9000 do: [ :p | (p isPerfect) ifTrue: [ p printNl ] ]</lang>
Swift
<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)
}
}</lang>
- Output:
6 28 496 8128
Tcl
<lang tcl>proc perfect n {
set sum 0
for {set i 1} {$i <= $n} {incr i} {
if {$n % $i == 0} {incr sum $i}
}
expr {$sum == 2*$n}
}</lang>
Ursala
<lang Ursala>#import std
- import nat
is_perfect = ~&itB&& ^(~&,~&t+ iota); ^E/~&l sum:-0+ ~| not remainder</lang> This test program applies the function to a list of the first five hundred natural numbers and deletes the imperfect ones. <lang Ursala>#cast %nL
examples = is_perfect*~ iota 500</lang>
- Output:
<6,28,496>
VBScript
<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</lang>
- Output:
C:\>cscript /nologo perfnum.vbs 6 True C:\>cscript /nologo perfnum.vbs 29 False C:\>
XPL0
<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)];
];
]</lang>
- Output:
6 28 496 8128 33550336
zkl
<lang zkl>fcn isPerfectNumber1(n)
{ n == [1..n-1].filter('wrap(i){ n % i == 0 }).sum(); }</lang>
- Output:
[1..0d10_000].filter(isPerfectNumber1).println(); L(6,28,496,8128)
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