I have two questions. The second question is really about showing my motives to ask the first question but it would be great if someone could verify that my argument works.

Let $a_k(n)$ be the $n$th largest product of $k$ distinct primes. Then $a_1(n)$ enumerates set of primes and $a_2(n)$ enumerates the set of square free semiprimes, $a_3(n)$ enumerates the set of sphenic numbers and so on.

Question 1: Is there an established name in the literature for this $\zeta$ like function $\zeta_k(s)=\sum_n^\infty {a_k(n)^{-s}}$?


We should get a couple of things from these nice series: Among other things there should be a corresponding theorem to theorem 1 of this paper which explores "almost prime" zeta functions. These "almost" prime zeta functions are described as the sum of reciprocals which have $k$ prime factors but these factors needn't be distinct. We will need these factors to be distinct for the argument below to work.

Almost Prime Zeta Function (Not what I am asking for ... but interesting)

This is defined on wikipedia as $$P_k(s)\equiv \sum_{n: \Omega(n)=k} \frac 1 {n^s}$$ where $\Omega$ is the total number of prime factors. The way that k-almost prime is defined we see that $3$-almost prime numbers include stuff like $7\times 7 \times 13$ where as the sphenic numbers do not.

Theorem 1 of Mathar gives a strategy for computing $P_k$ recursively. Note that $P_2(s)= \frac{P_1(2s)+P_1(s)^2}{2!}$. This can be seen quickly. It is a symmetry argument. This can be compared with

$\zeta_2(s) = \frac{\zeta_1 ^2(s)-\zeta_1(2s)}{2!}$. Same symmetry argument but now we want to make sure the diagonal of the array is NOT included.

Here is why I am convinced people must study these things: It seems to me that


$$\zeta(s)^{-1}=1+\sum_{k=1}(-1)^k \zeta_k(s)$$


We will exploit $$ \prod_{n=1}^m (1-x_n)=\sum_{S\subset [m]}{(-1)^{|s|}}\prod_{i\in S} x_i $$ where $[m]=\{1\dots m\}$ and we can make an argument by taking $m\to \infty$.

Take $x_n$ to be the reciprocal of the $n$th prime raised to the $s$. That is, $x_n= {p_n^{-s}} \implies 1-x_n={1-p_n^{-s}}$.

$$ \zeta(s)^{-1}$$ $$=\prod_{n=1}^\infty (1-p_n^{-s})$$ $$=\sum_{S\subset \mathbb{N}}{(-1)^{|S|}}\prod_{n\in S} p_n^{-s}$$ $$=\sum_{k=0}^\infty\sum_{|S|=k}{(-1)^{k}}\prod_{n\in S} p_n^{-s}$$ $$=1+\sum_{k=1}^\infty (-1)^k\zeta_k(s)$$

The first equality is Euler's product. The second equality makes me nervous because of this expression $S \subset \mathbb{N}$ but we may elect to ignore all the infinite subsets of $\mathbb{N}$ as this product with evaluate to zero in these cases. So how will we add up all of these cases? We have a different product for each finite subset of the natural numbers. So what we can do is throw them in different classes. We will consider the case where all of the subsets have cardinality $k$ but then this would mean that this is sum of all the reciprocals of the products of $k$ distinct primes raised to the $s$ power but this was the definition of $\zeta_k(s)$.


I checked this for a few numbers and it looks fine but I am not confident about the math. I also don't know the name of this function which makes it harder to confirm this using numerical tools.

Question 2: Does everything look ok?

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    $\begingroup$ Your claim is correct. I'd prove it via $$\begin{aligned} \frac{1}{\zeta(s)} &= \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} \\ &= \sum_{k = 0}^{\infty} \sum_{\Omega(n) = k} \frac{\mu(n)}{n^s} \\ &= 1 + \sum_{k = 1}^{\infty} (-1)^k \sum_{\Omega(n) = k} \frac{\lvert\mu(n)\rvert}{n^s} \\ &= 1 + \sum_{k = 1}^{\infty} (-1)^k \zeta_k(s)\end{aligned}$$ for $\operatorname{Re} s > 1$. $\endgroup$ – Daniel Fischer Jul 16 '18 at 12:34
  • $\begingroup$ Yay! I better do more reading about this Mobius Inversion that everyone seems to think is all the rage. $\endgroup$ – Mason Jul 16 '18 at 12:43
  • $\begingroup$ Wow. What a powerful tool: I thought youtube.com/watch?v=Vsib1v5vfkc was a pretty good intro to this topic. $\endgroup$ – Mason Jul 16 '18 at 13:02
  • $\begingroup$ I would still love to know if my argument has any hiccups but considering this construction is no more than slapping some absolute bars on a commonly seen tool I suspect that my question 1 is answered in the negative: This thing doesn't get it's own name. @DanielFischer. You might consider posting your comment as an answer. $\endgroup$ – Mason Jul 16 '18 at 13:18
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    $\begingroup$ @Mason, it seems to me that a similar expression for $\zeta_k(s)$ also holds true:$$\frac{\zeta(s)}{\zeta(2s)}=1 + \sum_{k = 1}^{\infty} \zeta_k(s)$$ $\endgroup$ – Aleksey Druggist Mar 11 at 2:17

Concerning question 1, apart from the case $k = 1$, the famous prime zeta function (often denoted $P(s)$), I'm not aware of these functions having special names. Calling them "squarefree $k$-almost prime zeta functions" would of course make sense, even if it's a little cumbersome.

Concerning question 2, apart from a couple of trivial typos - you forgot the minus sign of the exponent in $\prod (1 - p_n^s)$, in the lines below that you index the primes by $n$ and $k$ respectively, but the product index is called $i$ - there is one thing causing concern, which you already pointed out yourself:

The second equality makes me nervous because of this expression $S \subset \mathbb{N}$ but we may elect to ignore all the infinite subsets of $\mathbb{N}$ as this product with evaluate to zero in these cases.

For an infinite $S$, the $(-1)^{\lvert S\rvert}$ doesn't make sense. But, as you note, in that case the product diverges to $0$, so we can justify interpreting $$(-1)^{\lvert S\rvert}\prod_{i\in S} p_i^{-s}$$ as $0$. Then we have a sum of uncountably many terms, but only countably many of them are nonzero. Still one needs to give an argument that the value of that sum is in fact equal to the value of the product. That's mildly tedious, but not technically difficult. (The most difficult part is avoiding the "it's obvious" trap.) Everything works out, though. (Always restricting to $\operatorname{Re} s > 1$ of course.)

We get a simpler proof of the claim by reordering the known Dirichlet series of the reciprocal zeta function: \begin{align} \frac{1}{\zeta(s)} &= \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} \\ &= \sum_{k = 0}^{\infty} \sum_{\Omega(n) = k} \frac{\mu(n)}{n^s} \\ &= 1 + \sum_{k = 1}^{\infty} (-1)^k \sum_{\Omega(n) = k} \frac{\lvert\mu(n)\rvert}{n^s} \\ &= 1 + \sum_{k = 1}^{\infty} (-1)^k\zeta_k(s)\,. \end{align}


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