Like the title asks, are there any known positive integers $k$ such that there are infinitely many primes that can be written as a sum of $k$ fibonacci numbers? For $k=1$ this is a famous unsolved problem, but if we relax this to make it so that $k$ could be some other positive integer, are there any currently that are known?
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asked Apr 30, 2023 at 22:44
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$\begingroup$ This is a natural question but likely extremely hard: the Fibonacci numbers $F_m$ satisfying $F_m \leq x$ has density $O(\log x)$, and for any fixed $k$, the numbers expressible as a sum of at most $k$ Fibonacci numbers has density $O((\log x)^k)$. This is still far too thin for modern techniques to detect primes; this sequence has log density $0$. In fact several open problems involve detecting primes in sequences of log density strictly less than one but still positive, and some even involve sets with full log density! $\endgroup$– Stanley Yao XiaoCommented May 1, 2023 at 0:10
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1$\begingroup$ On the other hand, I think it might be known that for all sufficiently large $k$, there is some prime that can be written as a sum of $k$ (positive) Fibonacci numbers. It feels analogous to the question of Are there primes of every Hamming weight?. $\endgroup$– WojowuCommented May 1, 2023 at 2:55
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This paper almost answers this: Primes as sums of Fibonacci numbers, Michael Drmota, Clemens Müllner, Lukas Spiegelhofer https://arxiv.org/abs/2109.04068 Memoirs of the American Mathematical Society
The abstracts includes the following: "One of our main results says that for every sufficiently large integer k there exists a prime number that can be represented as the sum of k different and non-consecutive Fibonacci numbers."
Hence it works for all sufficiently large numbers $k\geq k_0$. On page 4 it is mentioned that an effective bound $k_0$ could in principle be worked out, (but that would be very messy).
Update: as usual in analytic number theory, if one can prove the existence of an object one can often prove that there are many. Theorem 1.2, also on page 4, gives a strong quantitative result, implying that for each sufficiently large $k$ there are many such primes, (choose $x$ such that $k-\mu \log_{\gamma} x$ is small). The question if there are infinitely many such primes seems open.
answered May 2, 2023 at 0:00
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$\begingroup$ Are there infinitely many for a given $k$, as the question asks? $\endgroup$ Commented May 2, 2023 at 4:58
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$\begingroup$ I don't think Theorem 1.2 implies infinitude - for any fixed $k$ and $x$ large enough, the RHS of the expression there is approximately $\frac{\pi(x)}{\sqrt{\log x}}(\frac{1}{x}+O((\log x)^{-1/2}))$, which doesn't imply it tends to infinity. $\endgroup$– WojowuCommented May 2, 2023 at 7:00
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$\begingroup$ @Wojowu: are you sure the exponential factor gives $1/x$ ? I did not work it out, but maybe it gives $1/x^{\alpha}$ with $0<\alpha < 1$ ?? $\endgroup$ Commented May 2, 2023 at 7:27