Let $H(t) = \sum_{n=1} ^{\infty} \pi(n)t^n$ where $\pi(n)$ is the prime counting function. This is the Hilbert series of some $\mathbb{Q}$-vector space. By the prime number theorem, the radius of convergence is $1$. Observing $\pi(n) = \pi(n-1)+1$ if $n$ is prime and $\pi(n) = \pi(n-1)$ if $n$ is composite, we might rewrite this as $H(t) = f(t)/(1-t)$ where $f(t) =\sum_{p \text{ prime }} t^p$. Define the sequence $b_n$ for $n=-1,0,1,2,\cdots$ as $ f'(t)/f(t) = \sum_{n=-1} b_n t^n$. Then we can recover the primes from this sequence:
$$ p = 2 + \sum_{q <p, q \text{ prime }} b_{p-1-q}$$
For example:
For example the first coefficients are given by the series:
$$2*t^{(-1)} + 1 + (-1)*t + 4*t^2 + (-5)*t^3 + 11*t^4 + (-16)*t^5 + 22*t^6 + (-37)*t^7 + 67*t^8 + (-101)*t^9 + 166*t^{10} + (-260)*t^{11} + 404*t^{12} + (-652)*t^{13} + \cdots $$
so $b_{-1} = 2, b_0 = 1, b_1 = -1 , b_2 = 4$ etc.
We have for example:
$$3 = 2+b_0 = 2+1$$
$$5 = 2+b_2+b_1 = 2+4-1$$
$$7 = 2+b_4+b_3+b_1 = 2+11-5-1$$ Let $a_{n,k}$ denote the number of ordered ways of writing $n$ as a sum of $k$ primes. Then after some calculation, one finds that: $$a_{n,k} = \frac{k}{n-2k} \sum_{v=0}^{n-1} {a_{v,k} b_{n-1-v}}$$ which is a recurence relation. Furthermore if $\alpha_n$ $n=0,1,2,3,\cdots$ are all roots of $f(t)$ not equal to zero, then for $n\ge 0$ $$ b_n = - \sum_{k=0} ^ \infty \frac{1}{\alpha_k^{n+1}}$$
The numbers $b_n$ might be computed inductively using: $$ n a_{n,1} = \sum_{v=0}^n b_{v-1} a_{n-v,1}$$ from which one sees, that $b_n \in \mathbb{Z}$.
One real root of $f(t)$ seems to be the number,
$$ \gamma = -0.62923 \cdots $$
(OEIS: http://oeis.org/A078756 )
Since everything related to primes has something to do with the Riemann Zeta function, I wonder, what is the relation to the stuff above to the Riemann Zeta function?
If someone knows of any reference or has any idea, that would be great.
Is there for example a way to compute the real root $\gamma$? What is the relation of $\gamma$ to the other complex roots? What other properties do the numbers $b_n$ have? etc.
Thanks for your help.
Edit: I found a conjectural way to compute $\gamma$ and using the Euler Product a link to Riemann Zeta function:
$$\zeta(s) = \prod_{p} {\frac{1}{1-(2+\sum_{q<p} b_{p-1-q})^{-s}}}$$
and for $\gamma$ numerical coincidences suggest that:
$$\lim_{n \rightarrow \infty} \frac{b_n}{b_{n+1}} = \gamma = -0.629233\cdots$$
which could be one way to define $\gamma$. Then one has to show, that this limit exists and that $f(\gamma) = 0$.
If someone has an idea in this direction,that would be nice.
Second edit: Here is the computation for the recurence relation of $a_{n,k}$: For $k\ge 1$ we have on the one hand: $$\log(f(t)^k)' = \frac{k f(t)^{k-1} f'(t)}{f(t)^k} = k \frac{f'(t)}{f(t)} = \sum_{n=-1}^\infty k b_n t^n$$
On the other hand it is $$f(t)^k = \sum_{n=0}^\infty a_{n,k}t^n$$ which means that $$\log(f(t)^k)' = \frac{\sum_{n=0}^\infty n a_{n,k} t^{n-1}}{\sum_{n=0}^\infty a_{n,k} t^{n}}$$
Hence it follows, that (by multiplying with the denominator): $$\sum_{n=0}^\infty n a_{n,k} t^{n-1} = k (\sum_{n=0}^\infty a_{n,k} t^{n}) (\sum_{n=0} b_{n-1}t^n) \frac{1}{t}$$
After multiplying with $t$ and using the Cauchy product formula we get:
$$\sum_{n=0}^{\infty} n a_{n,k} t^{n} = \sum_{n=0}^\infty k( \sum_{v=0}^n a_{v,k}b_{n-1-v})t^n$$
and comparing coefficients we find that:
$$n a_{n,k} = k ( \sum_{v=0}^n a_{v,k} b_{n-1-v})$$
and with $b_{-1} = 2$ it follows after solving this equation for $a_{n,k}$ that:
$$ a_{n,k} = \frac{k}{n-2k} \sum_{v=0}^{n-1} a_{v,k} b_{n-1-v}$$
Especially for $k=1$ and $n=p$ prime we find that:
$$ p = 2+ \sum_{q<p} b_{p-1-q}$$
