I was fooling around, trying to come up with a pseudo-rapid way to compute $\pi$. Then I remembered that for the Riemann zeta function ($\zeta$), we always have: \begin{equation} \zeta(2n)=c\pi^{2n}, \end{equation} where $n$ is a positive integer and where $c$ is some rational number. Thus, I thought that an efficient way of computing $\pi$ would be to be simply to invert this relation (denoting $s=2n$): \begin{equation} \pi=(\zeta(s)/c)^{1/s} \end{equation} Then I thought that to make this even faster, one could find a power $s$ such that $c$ takes the form $c=1/d$, with $d$ an integer (since, numerically, multiplication is faster than division). Then we would have: \begin{equation} \pi=(d\zeta(s))^{1/s} \end{equation} I think the largest the $s$ power is, the faster one can compute $\pi$ up to some precision (by summing a finite number of terms of the zeta-function).
- However, I wasn't able to find $s>10$ such that $c=1/d$... Using Mathematica I searched up to $s=1000$.
This inspired me to ask the following question:
Is there a $s>10$ such that $c=1/d$ with $d\in\mathbb{N}$ ?
I rant too much to post comments!
Let me begin by saying that this is equivalent to proving the existence, of $2n>10$, of $$\dfrac{\pi^{2n}}{\zeta(2n)}\in\mathbb{N}$$ We know that, $B_n$ are the Bernoulli numbers, $$\dfrac{\pi^{2n}}{\zeta(2n)}=\dfrac{(2n)!}{(-1)^{n+1}B_{2n}2^{2n-1}}$$ Note that $\dfrac{\pi^{2n}}{\zeta(2n)}=\dfrac{B_n}{A_n}$, where $A_n,B_n$ are the sequences $A002432$ and $A046988$, respectively, in OEIS.
Now, if we write $\xi_n=\dfrac{B_n}{A_n}$, we see that we can define inductively, $$\xi_1=\dfrac{1}{6}\\ \xi_n=\sum_{i=1}^{n-1}(-1)^{i-1}\frac{\xi_{n-i}}{(2i+1)!}+(-1)^{n+1}\frac{n}{(2n+1)!}\\ \implies \xi_n=\dfrac{\pi^{2n}}{\sum^\infty_{k=1}\dfrac{1}{k^{2n}}}$$ Using mathematica, one gets the values of $\xi_n$. I am not sure if for all finite values of $n>5$, $\xi_n\in\mathbb{N}$. From this MO post, you can see that $\zeta(2n)$ is a rational multiple of $\pi^{2n}$, and for a proof, see here.