Perfect numbers: Difference between revisions
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Revision as of 14:31, 22 March 2015
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).
See also
- Rational Arithmetic
- Perfect numbers on OEIS
- Odd Perfect showing the current status of bounds on odd perfect numbers.
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
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>
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 ;
bool is_perfect( int ) ;
int main( ) {
cout << "Perfect numbers from 1 to 33550337:\n" ;
for ( int num = 1 ; num < 33550337 ; num++ ) {
if ( is_perfect( num ) )
cout << num << '\n' ;
}
return 0 ;
}
bool is_perfect( int number ) {
int sum = 0 ;
for ( int i = 1 ; i < number ; i++ )
if ( number % i == 0 )
sum += i ;
return sum == number ;
}</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>
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
Elixir
<lang elixir>def is_perfect(x) do
[1 | lc x inlist :lists.seq(2, div(n, 2)), rem(n, x) == 0, do: x] |> :lists.sum() == n
end </lang>
Erlang
<lang erlang>is_perfect(X) ->
X == lists:sum([N || N <- lists:seq(1,X-1), X rem N == 0]).</lang>
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>
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"
// 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)
}
}
}
func computePerfect(n int64) bool {
var sum int64
for i := int64(1); i < n; i++ {
if n%i == 0 {
sum += i
}
}
return sum == 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 = map (\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 = do mapM_ print $ take 10 perfect mapM_ print $ map (\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
<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
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>function isperfect(n)
n == sum([n % i == 0 ? i : 0 for i = 1:n-1])
end</lang>
- Output:
julia> filter(isperfect, 1:10000)
4-element Int64 Array:
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>
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>
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>
Nim
<lang nim>import math
proc isperfect(n: int): bool =
let max: int = (sqrt(n.toFloat)).toInt
var sum: int = 1
for i in 2..max:
if n mod i == 0:
let q: int = n div i
sum += (i + q)
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>func: isPerfect(n) { | i | 0 n 2 / loop: i [ n i rem ifZero: [ i + ] ] n == } </lang>
- Output:
10000 seq filter(#isPerfect) println [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*$n; }</lang> Use this naive method to show the first 5. Takes about 30 seconds (half the time Pari/GP takes with this method): <lang perl>use ntheory qw/divisor_sum/; for (1..33550336) {
print "$_\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 17 are found in ~1 second, with the first 20 taking a bit under 10 seconds. <lang perl>use ntheory qw/forprimes is_prime/; use bigint; forprimes {
my $n = 2**$_ - 1; print "$_\t", $n * 2**($_-1),"\n" if is_prime($n);
} 2, 4500;</lang>
- Output:
2 6 3 28 5 496 7 8128 13 33550336 17 8589869056 19 137438691328 31 2305843008139952128 61 2658455991569831744654692615953842176 89 191561942608236107294793378084303638130997321548169216 ... 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423 ...
Perl 6
<lang perl6>sub perf($n) { $n == [+] grep $n %% *, 1 .. $n div 2 }</lang>
PHP
<lang php>function is_perfect($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 $num . PHP_EOL;
}</lang>
PicoLisp
<lang PicoLisp>(de perfect (N)
(let C 0
(for I (/ N 2)
(and (=0 (% N I)) (inc 'C I)) )
(= C N) ) )</lang>
PL/I
<lang PL/I>perfect: procedure (n) returns (bit(1));
declare n fixed; declare sum fixed; declare i fixed binary;
sum = 0;
do i = 1 to n-1;
if mod(n, i) = 0 then sum = sum + i;
end;
return (sum=n);
end perfect;</lang>
PowerShell
<lang powershell>Function IsPerfect($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>
Functionnal 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>
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 (define (perfect? n)
(= n
(for/fold ((sum 0))
((i (in-range 1 (add1 (floor (/ n 2))))))
(if (= (remainder n i) 0)
(+ sum i)
sum))))
(filter perfect? (build-list 1000 values))
- -> '(0 6 28 496)</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 14Sep2013. <lang rexx>/*REXX version of the ooRexx pgm (code was modified for 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:
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 14Sep2013, a REXX say statement was added to display the perfect numbers. <lang rexx>/*REXX version of the PL/I program (code was modified for Classic REXX).*/ parse arg low high . /*obtain the specified number(s).*/ if high== & low== then high=34000000 /*if no args, 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 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
- a corresponding factor is used when a factor is found
- testing is stopped if the sum of the factors exceeds X
<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 args, 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 output. */ numeric digits max(9,w+2) /*ensure enough digits to handle#*/
do i=low to high /*process the single # or range. */
if isPerfect(i) then say right(i,w) 'is a perfect number.'
end /*i*/
exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────ISPERFECT subroutine────────────────*/ isPerfect: procedure; parse arg x /*get 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 factors ≤√X*/
if x//j\==0 then iterate /*J isn't a factor of X, so skip.*/
s = s + j + x%j /*··· add it and the other factor*/
if s>x then return 0 /*Sum too big? It ain't perfect.*/
end /*j*/ /*(above) is marginally faster. */
return s==x /*if the sum matches X, perfect! */</lang>
- Output:
using the defaults
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 args, 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 output. */ numeric digits max(9,w+2) /*ensure enough digits to handle#*/
do i=low to high /*process the single # or range. */
if isPerfect(i) then say right(i,w) 'is a perfect number.'
end /*i*/
exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────ISPERFECT subroutine────────────────*/ isPerfect: procedure; parse arg x 1 y /*get the number to be tested. */ if x==6 then return 1 /*handle special case of six. */
/*[↓] perfect #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
y=r /*find digital root of dig root. */
end /*DO 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 factors ≤√X*/
if x//j\==0 then iterate /*J isn't a factor of X, so skip.*/
s = s + j + x%j /*··· add it and the other factor*/
if s>x then return 0 /*Sum too big? It ain't perfect.*/
end /*j*/ /*(above) is marginally faster. */
return s==x /*if the sum matches X, perfect! */</lang> output is the same as the traditional version and is about 5.3 times faster (testing 100,000 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 the specified number(s).*/ if high== & low== then high=34000000 /*if no args, use a range.*/ if low== then low=1 /*if no LOW, then assume unity.*/ if low//2 then low=low+1 /*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 output. */ numeric digits max(9,w+2) /*ensure enough digits to handle#*/
do i=low to high by 2 /*process the single # or range. */
if isPerfect(i) then say right(i,w) 'is a perfect number.'
end /*i*/
exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────ISPERFECT subroutine────────────────*/ isPerfect: procedure; parse arg x 1 y /*get the number to be tested. */ if x==6 then return 1 /*handle special case of six. */
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
y=r /*find digital root of dig root. */
end /*DO until*/ /*wash, rinse, repeat ··· */
if r\==1 then return 0 /*is dig root ¬1? Then ¬perfect.*/
s = 3 + x%2 /*the first three factors of X. */
do j=3 while j*j<=x /*starting at 3, find factors ≤√X*/
if x//j\==0 then iterate /*J isn't a factor of X, so skip.*/
s = s + j + x%j /*··· add it and the other factor*/
if s>x then return 0 /*Sum too big? It ain't perfect.*/
end /*j*/ /*(above) is marginally faster. */
return s==x /*if the sum matches X, perfect! */</lang> output is the same as the traditional version and is about 11.5 times faster (testing 100,000 numbers).
Lucas-Lehmer method
This version uses memoization to implement a fast version of Lucas-Lehmer test,
it is faster than the traditional method by an order of magnitude.
<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 args, use a range.*/
if low== then low=1 /*if no LOW, then assume unity.*/
if low//2 then low=low+1 /*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 output. */
numeric digits max(9,w+2) /*ensure enough digits to handle#*/
@.=0; @.1=2 /*highest magic # and its index.*/
do i=low to high by 2 /*process the single # or range. */
if isPerfect(i) then say right(i,w) 'is a perfect number.'
end /*i*/
exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────ISPERFECT subroutine────────────────*/ isPerfect: procedure expose @.; parse arg x /*get the # 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 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 factors ≤√X*/
if x//j\==0 then iterate /*J divides X evenly, so ... */
s = s + j + x%j /*··· add it and the other factor*/
if s>x then return 0 /*Sum too big? It ain't perfect.*/
end /*j*/ /*(above) is marginally faster. */
return s==x /*if the sum matches X, perfect! */</lang> output is the same as the traditional version and is about 75 times faster (testing 100,000 numbers).
Lucas-Lehmer + other optimizations
This version uses the Lucas-Lemer method, digital root, and restricts itself to even numbers, and
also utilizes a check for the last-two-digits as per François Édouard Anatole Lucas (in 1891), in
conjuction with this, the i DO loop is fast advanced according to the last number tested.
<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 args, 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 output. */
numeric digits max(9,w+2) /*ensure enough digits to handle#*/
@. =0; @.1=2; !.=2; _=' 6' /*highest magic # and its index.*/
!._=22; !.16=12; !.28=8; !.36=20; !.56=20; !.76=20; !.96=20
/* [↑] "Lucas' numbers, 1891. */
do i=low to high by 0 /*process the single # or range. */
if isPerfect(i) then say right(i,w) 'is a perfect number.'
i=i+!.? /*use a fast advance for DO loop.*/
end /*i*/
exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────ISPERFECT subroutine────────────────*/ isPerfect: procedure expose @. !. ? /*expose (make global) some vars.*/ parse arg x 1 y -2 ? /*#, another copy, and last digit*/ if x==6 then return 1 /*handle special case of six. */ if !.?==2 then return 0 /*test last 2digs: 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 test? */
/*[↓] perfect #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
y=r /*find digital root of dig root. */
end /*until*/ /*wash, rinse, repeat ··· */
if r\==1 then return 0 /*Digital root ¬1? Then ¬perfect.*/ 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 factors ≤√X*/
if x//j\==0 then iterate /*J divides X evenly, so ... */
s = s + j + x%j /*··· add it and the other factor*/
if s>x then return 0 /*Sum too big? It ain't perfect.*/
end /*j*/ /*(above) is marginally faster. */
return s==x /*if the sum matches X, perfect! */</lang>
output is the same as the traditional version and is about 483 times faster (testing 100,000 numbers).
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
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
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>
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)
- Programming Tasks
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