# What is known about the pattern for $\zeta(2n+1)$?

Related to the question Does $\zeta(3)$ have a connection with $\pi$?:

It is well known that

$$\zeta(2n) = f(2n) \pi^{2n}$$

where $f(n)$ is an function in rationals: (the denominator = OEIS A002432: 6, 90, 945, 9450).

Apery proved that $\zeta(3)$ is irrational, and presumably $f(2n+1)$ is so for all (but the latter is open).

But what else is known about $f(2n+1)$? What sort of pattern does it share (if any) with $f(2n)$? Numerically does it fit well? Is there any clue/progress since Apery as to its form? Of course, $f(2n+1)$ may in effect divide out the $\pi$, meaning that somehow $\pi$ is not involved in $\zeta(2n+1)$ in any essential way.

• An answer on MO (mathoverflow.net/questions/38190/…) says that the standard conjecture is that these numbers are all algebraically independent of each other and $\pi$ over $\mathbb{Q}$. I think an answer to this question is likely to be quite high-level so you might want to ask on MO. – Qiaochu Yuan Apr 28 '11 at 3:10
• For now, the following looks helpful: math.jussieu.fr/~miw/articles/pdf/MZV2011IMSc1.pdf – Qiaochu Yuan Apr 28 '11 at 3:15
• This question is closely related to math.stackexchange.com/questions/12815/…. I'm not sure if it's close enough to be considered a duplicate. – Alex Becker Apr 28 '11 at 3:15
• The main reason why we cannot prove too much about $f(2n+1)$ is because we don't now any pattern for them.... – N. S. Feb 1 '16 at 22:32

## 1 Answer

As Qiaochu writes in his comment, $\zeta(2n+1)$ (for $n \geq 1$) is expected to be algebraically independent of $\pi$, and furthermore for different values of $n$ these numbers are also expected to be algebraically independent of one another.

These expectations are far from being proved (as far as I know), but are not idle speculations: contemporary number theorists have a very tightly woven web of conjectures about values of $\zeta$-functions, and $L$-functions, which is supported by large amounts of theoretical and experimental evidence, and I don't think there is any reason to doubt that the expectations are correct.

Given this, I don't expect that the numbers $f(2n+1)$ will follow any significant pattern.

• I gave a +1 just for "very tightly woven web of conjectures"... – J. M. is a poor mathematician Apr 28 '11 at 3:18