A series related to prime numbers

Solution 1:

• 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 $\frac{f(t)}{1-t} = \sum_{n \ge 2} \pi(n) t^n \approx\sum_{n \ge 2} \frac{n}{\log n} t^n$