# Efficiently summing reciprocal of polynomial over primes.

For fun, I have been trying to calculate the sum of the reciprocals of the cube-full numbers. I have managed to show that the limit is equal to $$\frac{\zeta(3)\zeta(4)\zeta(5)}{\zeta(8)\zeta(10)}\prod_{p\ \mbox{prime}}\left(1-\frac1{(p^4+1)(p^5+1)}\right)\approx1.3397841535.$$ This product converges pretty fast, because of the order $$9$$ polynomial $$f(x)=(x^4+1)(x^5+1)$$ in the numerator. By simply taking the primes up to $$10^8$$, I already got $$64$$ digits of precision. $$1.3397841535743472465991525865148860527752422497881828066630150676$$ However, this method requires exponential time to calculate more digits. I was wondering if there is a faster, or even a polynomial time algorithm to calculate more digits.

One thing I tried is to take the logarithm of the product. $$\log\left(\prod_{p\ \mbox{prime}}\left(1-\frac1{f(p)}\right)\right)=\sum_{p\ \mbox{prime}}\log\left(1-\frac1{f(p)}\right)$$ By taking the Taylor series of the natural logarithm, we get $$\log\left(1-\frac1{f(p)}\right)=\sum_{n=1}^\infty\frac{-1}{n(f(p))^n}.$$ By absolute convergence, we can interchange the sums to obtain $$\sum_{n=1}^\infty\frac{-1}n\sum_{p\ \mbox{prime}}\frac1{(f(p))^n}.$$ For all $$n$$, of course $$(f(p))^n$$ is a polynomial, so the question becomes how we can efficiently sum the reciprocal of a polynomial over the primes. Is there some sort of analog for the Euler-Maclaurin formula for primes?

• nice question. what a cute limit at the start! – mathworker21 Dec 29 '19 at 1:58

$$\log(1-\frac{x^{9}}{(1+x^4)(1+x^5)}) = \sum_{m\ge 0} a_m x^{9+m}$$
$$\log\prod_p (1-\frac{p^{-9}}{(1+p^{-4})(1+p^{-5})}) =\sum_p \sum_m a_m p^{-9-m}=\sum_m a_m \sum_p p^{-9-m}$$ $$=\sum_m a_m \sum_l \frac{\mu(l)}{l} \log\zeta(l(9+m)) = \sum_{n\le N}(\sum_{l|n}\frac{ a_{n/l-9} \mu(l)}{l} )\log\zeta(n)+O(2^{-N})$$
• This looks promising, but I am not experienced enough to fill in the gaps. Are we supposed to calculate the terms $a_m$ from the Taylor series? Is that possible in polynomial time? And how do we know it converges for $|x|<1$? And most importantly, how do you go from the sum over $p$ to the sum over $l$? Suddenly $\mu$ and $\log\zeta$ pop up, so something must be done with inclusion exclusion and the Euler product, but this step still seems like a pretty big leap. Finally, do you have a proof that the partial sums converge exponentially in the final sum? – SmileyCraft Dec 27 '19 at 20:54