Expert: Josh - 1/23/2007

Question
Gud day Sir,
aside from 6 and 28 what are the other perfect nos. (if there's any), i tried to formulate a program for that one but still i havnt able to succeed.. tnx

Answer
G'day Leonard,

Perfect numbers (n) have the property that the sum of all divisors of n equals 2n.

Alternatively, if we write S(n) for the set of all integers which divide n, then, the sum of all elements in S(n), excluding n itself, must be n.

e.g., If n=28, S(n)={1,2,4,7,14,28}. The sum of elements in S(n)\n={1,2,4,7,14} equals 1+2+4+7+14=28.

The next few perfect numbers are 496, 8128, 33550336....

Up to 10^300, no odd perfect numbers have yet been found. It is unclear whether odd perfect numbers exist. However, for even perfect numbers, a number of properties are known:

(A) Euclid construction of perfect numbers:

Euler was amongst the first to show that all even perfect numbers P=q*(2^(p-1)), where q=(2^p)-1 is prime.
Proof: see http://mathworld.wolfram.com/PerfectNumber.html

(B) All even perfect numbers P>6 are of the form:
P=1+9*T(n), where T(n)=n(n+1)/2 is a triangular number such that n=8k+2.
e.g., For P=496, (496-1)/9=n(n+1)/2. Solving this quadratic equation, n=-11,10. Keeping with positive integers, 10=8k+2, where k=1.

(C) All even perfect numbers are hexagonal numbers, see http://mathworld.wolfram.com/HexagonalNumber.html for geometric interpretation.

(D) Even perfect numbers are always the sum of consecutive positive integers beginning from 1.
e.g., For P=6, 6=Sum n, from n=1 to n=3
     For P=28, 28=Sum n, from n=1 to n=7
     For P=496, 496=Sum n, from n=1 to n=31.
The number of terms in the sum, L (3,7,31, respectively in the above example) are Mersenne primes, q=(2^p)-1.

(E) Interestingly, all even perfect numbers (except 6) end in 16, 28, 36, 56, 76, or 96.