Perfect numbers

Write a function which says whether a number is perfect.

A number is perfect if the sum of its factors is equal to twice the number. An equivalent condition is that n is perfect if the sum of n's factors that are less than n is equal to n.

Note: The faster Lucas-Lehmer_test is used to find primes of the form 2ⁿ-1, all known perfect numbers can derived from these primes using the formula (2ⁿ - 1) × 2^{n - 1}. It is not known if there are any odd perfect numbers.

Ada

<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; </Ada>

ALGOL 68

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

Output:

BASIC

Works with: QuickBasic version 4.5

<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</qbasic>

Forth

: perfect? ( n -- ? )
  1
  over 2/ 1+ 2 ?do
    over i mod 0= if i + then
  loop
  = ;

Haskell

perf n = n == sum [i | i <- [1..n-1], n `mod` i == 0]

J

is_perfect=: = [: +/ ((0=]|[)i.) # i.

The program defined above, like programs found here in other languages, assumes that the input will be a scalar positive integer.

Examples of use, including extensions beyond those assumptions:

   is_perfect 33550336
1
   }.I. is_perfect"0 i. 10000
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_pos_int=: 0&< *. ]=>.
   (is_perfect"0 *. is_pos_int) 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

Java

<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; }</java> Or for arbitrary precision: <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).compareTo(BigInteger.ZERO) == 0){ sum= sum.add(i); } } return sum.compareTo(n) == 0; }</java>

Python

<python>def perf(n):

   sum = 0
   for i in xrange(1,n):
       if n % i == 0:
           sum += i
   return sum == n</python>

Using functional form <python>def perf(n):

  return n == sum( i for i in xrange(1,n) if n % i == 0)</python>