# Sum of almost-prime zeta functions

Wiki defines an almost-prime zeta function as a sum of inverse powers of the k-primes (the integers which are a products of $$k$$ not necessarily distinct primes): $$P_k(s)=\sum_{n: \Omega(n)=k} \frac{1}{n^s}$$ There is also an obvious identity that decomposes the Riemann zeta function into an infinite sum of the $$P_{k}$$:

$$\zeta(s)=1+\sum_{k=1}^{\infty} P_k(s)$$ My question is about partial (finite) sums of the $$P_k(s)$$: $$\zeta(s)_N=1+\sum_{k=1}^{N} P_k(s)$$ These $$\zeta(s)_N$$ are a sort of intermediate function beetwin prime zeta function ($$N=1$$) and Riemann zeta function ($$N= \infty$$). If there is a lot of literature on the prime zeta function, then I could not find anything about the sums of $$N>1$$. What can be said about the zeros and poles of $$\zeta(s)_N$$? How does their location change with $$N$$?

• Note with $\omega(n)$ instead of $\Omega(n)$ there are coefficients such that $P_k(s) =\sum_{m=0}^{k} c_{m,k} P(s)^m$. And in both case the main idea is that in term of asymptotic and the dominating singularity at $s=1$ : $\zeta(s) \approx e^{P(s)}, P(s) \approx -\log(s-1)$, $\zeta(s) \approx \frac{1}{s-1}= \sum_{k=0}^\infty \frac{(-\log (s-1))^k}{k!},P_k(s) \approx \frac{P(s)^k}{k!}$ $\sum_{n\le x, \Omega(n) = k} 1 \approx \frac{x}{\log x} \frac{(\log \log x)^k}{k!} \approx \frac1k \sum_{p \le x} P_{k-1}(x/p) \approx \frac1k\sum_{2\le n\le x} P_{k-1}(x/n)\frac{1}{\log n}$. – reuns Mar 10 at 23:53
• The singularities encode the RH, the zeros of $\log \zeta(s), P(s),P_k(s)$ are less interesting and there are probably plenty of them as $\Re(s) \to \infty$ – reuns Mar 10 at 23:55
• I thought that the sum of $P_k$ that is $\zeta_N(s)$, with an increase in $N$, should, by properties, approach $\zeta(s)$, for example, its zeros should be lined up in a straight line $Res=1/2$ – Aleksey Druggist Mar 11 at 0:10
• They are not holomorphic at the non-trivial zeros. And I meant with $\omega(n)$ there are coefficients such that $P_k(s) = \sum_{m \le k, l \le k} c_{k,m,l} P(ls)^m$. For $\Re(s) > 1/3$, $P(s) = \log \zeta(s) - \frac{\log \zeta(2s)}2+h(s)$ with $h(s)$ analytic. From this you can relate it with the zeros of $\zeta(s)$. – reuns Mar 11 at 0:13

Here goes (my solution using symmetric polynomials and residues): First, observe that by an arithmetic argument, $$P_k(s) = [u^k] \prod_p \left(1 + \sum_{r \geq 1} p^{-rs} u^r\right) = \prod_p \left(1-\frac{u}{p^s}\right)^{-1},$$ where setting $$s \mapsto 1$$ gives the usual (non-split) Euler product for $$\zeta(s)$$. We can interpret the RHS product in terms of the complete homogeneous symmetric polynomials for $$x_j := p_j^{-s}$$: $$P_k(s) = h_k(x_1,x_2,\ldots).$$ Now by generating function expansions for these polynomials, we have that $$h_k = [z^k] \exp\left(-\sum_{k \geq 1} \frac{p_k}{k} z^k\right),$$ where $$p_k = \sum_p p^{-ks} = P(sk).$$ Here, $$P(s)$$ is the prime zeta function. So by generating function arithmetic, we have that $$\zeta(s)_N = [z^N] \frac{1}{1-z} \exp\left(-\sum_{k \geq 1} \frac{p_k}{k} z^k\right).$$ The RHS has poles at $$s := 1/k$$ for $$k \geq 1$$ by the nature of singularities of $$P(s)$$. From an inverse Z-transform, we can recover these coefficients as \begin{align} \zeta(s)_N & = \frac{1}{2\pi\imath} \oint_C \frac{s^{N-1}}{1-s} \exp\left(-\sum_{k \geq 1} \frac{P(sk)}{k} s^k\right) ds \\ & = \sum_{k \geq 1} \operatorname{Res}_{s=\frac{1}{k}}\left[\frac{s^{N-1}}{1-s} \exp\left(-\sum_{k \geq 1} \frac{P(sk)}{k} s^k\right)\right]. \end{align} Since we can expand $$P(s)$$ near $$s \rightarrow 1^{+}$$ by $$P(1+\varepsilon) \sim -\log(\varepsilon) + B + O(\varepsilon),$$ my back-of-the-envelope calculations suggest that these residues correspond to (NOTE: I haven't checked these out numerically) $$R_k = \begin{cases} -e^{-B} \exp\left(-\sum\limits_{j \geq 2} \frac{P(j)}{j}\right), & k = 1 \\ \frac{1}{k^N (k-1)} \exp(-B / k^{k+1}) \exp\left(-\sum\limits_{\substack{j \geq 1 \\ j \neq k}} \frac{P(kj)}{j \cdot k^j}\right), & k \geq 2. \end{cases}$$ The constant $$B$$ in the previous equations is defined by $$B = \sum_{n \geq 2} \frac{\mu(n)}{n} \log \zeta(n) \approx -0.315718452.$$ Now you should be able to proceed with evaluating zeros for the function.
N.b.: With respect to there not being much in the way of literature on $$P(s)$$, my friend has pointed out to me offline that the prime zeta function is essentially meaningless due to the essential and messy, complicated structure of its poles. It cannot be analytically continued past zero, and hence, is not nicely modular, nor does it have a nice functional equation like most zeta functions do. It's easier just not to work with the function unless you are forced.
• $P(s) = \sum_m \frac{\mu(m)}{m} \log \zeta(ms)$ has $\log$-singularities not poles, and $P_2(s) =\sum_{\omega(n)=2} n^{-s}=\sum_{l \ge j \ge 1} \frac12 P(sl)P(sj) - P(s(l+j))$ has no Euler product. The asymptotic of $\sum_{n \le x, \omega(n)=2} 1$ is given by the terms singular at $s=1$ ie. $P_2(s) \approx \frac12 P(s)^2 + P(s) R(s), R(s)= \sum_{l \ge 2} P(sl)$. – reuns Jun 3 at 20:50
• The $P_k(s)$ are defined in my answer as coefficients of a formal variable $u$ in what is otherwise a typical Euler product, NOT multiplicative factors of the original Euler product. The $u$ are clothes hangers where each unit exponent signifies that we have included another prime factor in the resulting $n^{-s}$ series terms. Standard generating function techniques apply from this perspective. – mds Jun 3 at 21:38