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:

$$ p = 2+ \sum_{q<p} b_{p-1-q}$$

  • First of all $f(t) = \sum_p t^p$ means $f'(t)/f(t) = \sum_n b_n t^n$ where $b_n=0$ for $n < -1$ and $\sum_p b_{n-p} = \cases{n \ if\ n+1\ is\ prime\\ 0\ otherwise}$.

    $f(e^{-x})$ is the inverse Mellin transform of $\Gamma(s)\sum_p p^{-s}$ while things like $1/f(t), \log f(t), f'(t)/f(t)$ are quite inaccessible, in the same way that $\zeta(s)$ is accessible from only the integers while $1/\zeta(s),\log \zeta(s),\zeta'(s)/\zeta(s)$ need the primes.

  • The number of ordered ways to write $n$ as a sum of $m$ primes are the coefficients of $f(t)^m$ and the number of ordered ways to write $n$ as a sum of primes are the coefficients of $\frac{1}{1-f(t)}-1 = \sum_{m=1}^\infty f(t)^m$.

    $\frac{1}{f(t)}=\frac{1}{t^2(1-(1-\frac{f(t)}{t^2}))}= t^{-2}\sum_{m=0}^\infty (-1)^m(\frac{f(t)}{t^2}-1)^m$ where the coefficients of $(\frac{f(t)}{t^2}-1)^m$ are the number of ordered ways to write $n+2m$ as a sum of $m$ primes $\ge 3$. $\frac{1}{t^2-f(t)}=t^{-2}\sum_{m=0}^\infty (\frac{f(t)}{t^2}-1)^m$.

  • Let $g(x) = \sum_{p^k} e^{-p^k x} \log p$ the inverse Mellin transform of $\Gamma(s) \frac{-\zeta'(s)}{\zeta(s)}$. We have the explicit formula $g(x) = \sum Res(\Gamma(s) \frac{-\zeta'(s)}{\zeta(s)} x^{-s}) = x^{-1}- \sum_\rho \Gamma(\rho) x^{-\rho}-\sum_{k=0}^\infty (a_k+b_k \log(x))x^k$. A corresponding explicit formula exists for $f(t)$ but it will be messy because $\Gamma(s)\sum_p p^{-s}$ has many branch points and a naturaly boundary $\Re(s)=0$.

  • We don't know anything on the zeros and particular values of $f(t)$ and since it is analytic only for $|t|<1$ then $f'(t)/f(t)$ won't be equal to a sum over $f$'s zeros. The most of what we know is the asymptotic of $f(t)$ as $t \to 1$, the error terms depends on the RH. It will give the asymptotic of $\log f(t)$ and possibly for $f'(t)/f(t)$, as $t \to 1$

    If $f(t)$ has a zero on $|t|< 1$, let $z_0$ be one with minimal absolute value, assume there is no other zero on $|z_0|$, then $\frac{f'(t)}{f(t)}-\frac{1}{t-z_0}$ is analytic for $|t|\le |z_0|+\epsilon$ so that $b_n = z_0^{-n}+O(|z_0|+\epsilon)^{-n}$ and $\lim_{n \to \infty} b_n/b_{n+1} = z_0$. By numerical approximation you can show such a $z_0$ exists in which case $z_0 \in (-1,0)$. So your claim $z_0= -\gamma$ is that $f'(t)/f(t)$ is analytic for $|t| < \gamma$. It is plausible there are some heuristics for such a thing given $\frac{f(t)}{1-t} = \sum_{n \ge 2} \pi(n) t^n \approx\sum_{n \ge 2} \frac{n}{\log n} t^n$


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