6 and 28 are the start of a pattern; both are the product of a power of 2 and and less than the next power of two. In other words, they are 2

^{n}* (2

^{n+1}- 1) for n equal to 1 and 2 respectively. However, the next number in the sequence, 8 * 15, or 120, is not perfect. Its factors, 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, and 60, add up to far more than 120 (in fact, twice as much.). Why the difference?

Well, suppose (2

^{n+1}- 1) is prime. We can now enumerate the factors of 2

^{n}* (2

^{n+1}- 1) : They will be

1, 2, ..., 2Since^{n}, (2^{n+1}- 1), 2*(2^{n+1}- 1), ... 2^{n-1}* (2^{n+1}- 1)

1 + 2 + ...+ 2and^{n}= 2^{n+1}- 1

(2^{n+1}- 1) + 2*(2^{n+1}- 1) + ... 2^{n-1}* (^{2n+1}- 1) =

(2^{n}- 1)(2^{n+1}- 1) =

2^{2n+1}- 2^{n+1}- 2^{n}+ 1

if you add them together, you get

2^{n+1}- 1 + 2^{2n+1}- 2^{n+1}- 2^{n}+ 1 =

2^{2n+1}- 2^{n}=

2^{n}* (2^{n+1}- 1)

or the original number. But when (2

^{n+1}- 1) isn't prime, there will be additional factors (as with 120, which, because 15 isn't prime, has the "extra" factors 3, 5, 6, 10, etc.), so the result isn't perfect. Primes of the form (2^{n+1}- 1) are called Mersenne primes. Only 47 of them are known, the largest being 2^{43,112,609}− 1. There are also 47 known perfect numbers, one corresponding to each Mersenne prime. It can be proved that all even perfect number is of this form; it's an open question whether there are any odd perfect numbers.
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