Let us denote by $S_k$ the set of values of the sequence $a(n)=\sum_{i=1}^n i^k$.
Does $\bigcup_{m=1}^{\infty}S_m$ contain infinitely many primes?
Does the set of all sums of all positive powers of the first $n$ natural numbers contain infinitely many primes?
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sequences-and-series
prime-numbers
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1I suspect the answer is "no"; however a proof of this would (among other things) prove that there are only finitely many [Fermat primes](https://en.wikipedia.org/wiki/Fermat_number), which is a significant open problem. – 2017-02-20
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0@vadim123 But sequence of Fermat numbers is only one sequence, here you have an infinite number of sequences in the question. – 2017-02-20
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0exactly. So the only realistic hope for an answer is if the answer is "yes". But, as I mentioned, I suspect it is "no". – 2017-02-20
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0@vadim123 In all those sequences we also have a sequence $2^n + 1$. Is it known does this sequence has an infinite number of primes as its values? – 2017-02-20
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0Faulhaber's formula grants that there is a $n(n+1)$ factor in the expression for $\sum_{m=1}^{n}m^k$. There also is a $(2n+1)$ factor if $k$ is even. I think this (maybe together with the Von-Staudt-Clausen theorem) should be enough to lead to a negative answer. – 2017-02-20
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0@user49640 No. n takes every natural value in every of these sequences. – 2017-02-20
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0@user49640 Yes. – 2017-02-20
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0@Nea, if $1+2^n$ is prime, then it is a Fermat prime. See the wikipedia link I sent. – 2017-02-20