## Perfect numbers, interlude (Challenge #11)

Recall that the first three perfect numbers are 6, 28, and 496. And if you tried computing $\sigma(8128)$ as I suggested near the end of part I, you might have noticed that 8128 is perfect as well! Do you see any interesting patterns here? (Hint: try factoring…)

Post your discoveries as comments here (unless you already know the answer, in which case you should refrain from spoiling other people’s fun =).

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### 4 Responses to Perfect numbers, interlude (Challenge #11)

1. DB says:

So far, these perfect numbers are of the form 2^(n-1) * (2^n – 1), but only when (2^n – 1) is prime. Works for n=2, 3, 5, and 7, so maybe n itself should be prime?

2. I figured out the same thing as DB while I was on Thanksgiving break and away from the computer. Nice way to pass the time.

But it is not required that n be prime. I’m pretty sure that it works when n is 9.

3. Brent says:

DB: right. As it turns out, n does have to be prime, but that’s not a sufficient condition.

Steve: It is a nice way to pass the time, indeed. =) I don’t think 9 works, though. 2^8 * 511 = 130816, the sum of the proper divisors of which is 171696, not 130816. That’s because 511 is not prime; 2^(3*3) – 1 = (2^3 – 1)(2^6 + 2^3 + 1) = 7 * 73.