Being (2^13) – 1 a prime number, find the fifth perfect number in order of magnitude.

This is an Euclides question that has been intriguing me for some time.

Hope you can help.

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1

I don’t understand the question. The first perfect numbers are 6,28,496,8128,33550336,8589869056,137438691328,2305843008139952128,…6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128, \ldots, so the fifth is 3355033633550336. Was this your question?

– Dietrich Burde

2 days ago

1

(213−1)⋅(213−1)(2^{13}-1)\cdot(2^{13-1})

– barak manos

2 days ago

1

33550336=212⋅(213−1)33550336=2^{12}\cdot (2^{13}-1).

– Dietrich Burde

2 days ago

1

In order for (2n−1)⋅(2n−1)(2^{n}-1)\cdot(2^{n-1}) to be perfect, 2n−12^{n}-1 must be prime. In order for 2n−12^{n}-1 to be prime, nn must be prime.

– barak manos

2 days ago

2

Yes, divide it by 22 as many times as you can, then add 11, then take log2\log_2 on the result.

– barak manos

2 days ago

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