5
$\begingroup$

Is it possible to write the following double product in terms of the zeta function?

\begin{align} &\prod_{i=1}^{\infty}\prod_{j=1}^{\infty} \frac{1}{1-(p_i\ p_j)^{-s}} \end{align}

Extending the zeta function to semiprimes with

\begin{align} &\prod _{i=1}^{\infty}\dfrac{1}{1 -\ q_{i}^{\ \ -s}}\\ \end{align}

where $q$ runs through the semiprimes is rewritable as

\begin{align} &\left( \prod_{i=1}^{\infty}\prod_{j=1}^{\infty} \frac{1}{1-(p_i\ p_j)^{-s}}\right)^{1/2}\left( \prod_{i=1}^{\infty} \frac{1}{1-(p_i{^2})^{-s}}\right)^{1/2} \end{align}

The last part can clearly be rewritten as $\zeta(2s)^{1/2}$, but I don't know how I can rewrite the first part in terms of the zeta function.

In addition, I believe it may be that the triprime zeta function

\begin{align} &\prod_{\Omega(q)=3}^{}\dfrac{1}{1 -\ q^{\ -s}}\\ \end{align}

can be written

\begin{align} &\left(\prod_{i=1}^{\infty} \prod_{j=1}^{\infty}\prod_{k=1}^{\infty} \frac{1}{1-(p_i\ p_j\ p_k)^{-s}}\right)^{1/6}\left( \prod_{i=1}^{\infty}\prod_{j=1}^{\infty} \frac{1}{1-(p_i^{\ 2}\ p_j)^{-s}}\right)^{1/2}\left( \prod_{i=1}^{\infty} \frac{1}{1-(p_i{^3})^{-s}}\right)^{1/3} \end{align}

and the function

\begin{align} &\prod_{\Omega(q)=4}^{}\dfrac{1}{1 -\ q^{\ -s}}\\ \end{align}

can be written

\begin{align} &\left(\prod_{i=1}^{\infty} \prod_{j=1}^{\infty}\prod_{k=1}^{\infty} \prod_{l=1}^{\infty}\frac{1}{1-(p_i\ p_j\ p_k\ p_l)^{-s}}\right)^{1/24} \left(\prod_{i=1}^{\infty} \prod_{j=1}^{\infty}\prod_{k=1}^{\infty} \frac{1}{1-(p_i^{\ 2}\ p_j\ p_k)^{-s}}\right)^{1/4}\\ &\left( \prod_{i=1}^{\infty}\prod_{j=1}^{\infty} \frac{1}{1-(p_i^{\ 3}\ p_j)^{-s}}\right)^{1/3} \left( \prod_{i=1}^{\infty}\prod_{j=1}^{\infty} \frac{1}{1-(p_i^{\ 2}\ p_j^{\ 2})^{-s}}\right)^{1/8} \left( \prod_{i=1}^{\infty} \frac{1}{1-(p_i{^4})^{-s}}\right)^{1/4} \end{align}

which of course presents the greater problem of defining the further multiple products in terms of the zeta function.

$\endgroup$
  • 3
    $\begingroup$ Any particular reason why this product is being considered? $\endgroup$ – Terry Tao Dec 16 '14 at 21:38
  • 1
    $\begingroup$ @TerryTao I am studying the semiprime counting function. I will update the post accordingly. $\endgroup$ – martin Dec 16 '14 at 21:53
  • 6
    $\begingroup$ Ah, I see. Doesn't look very promising, as there does not appear to be an Euler product factorisation. I think you're better off working with the second von Mangoldt function $\Lambda_2(n) = \Lambda(n) \log n + \Lambda*\Lambda(n)$ to count semiprimes (note that $\sum_n \frac{\Lambda_2(n)}{n^s} = \frac{\zeta''(s)}{\zeta(s)}$). $\endgroup$ – Terry Tao Dec 17 '14 at 3:58
  • $\begingroup$ @TerryTao If you have time, I would be really grateful if you could take a look at the products post revisions. I am fairly sure I have removed the errors now. The initial confusion on my part was due, I think, to an error converting the semiprime reciprocal sum \begin{align} &\sum_{\Omega(n)=2}^{}\dfrac{1}{n^{s}}=\frac{1}{2} \sum _{j=1}^{\infty}\sum _{i=1}^{\infty} \frac{1}{(p_i\ p_j)^{s}}+\frac{1}{2} \sum _{k=1}^{\infty} \frac{1}{(p_k{^2})^{s}} \end{align} (which I am almost certain is correct) to a product. I am fairly sure things are fixed now. $\endgroup$ – martin Dec 18 '14 at 8:33
  • $\begingroup$ @martin, just to clarify, are you implying that $\sum_{\Omega(n)=2}\frac{1}{n^s}=\prod_{i=1}^{\infty}\prod_{j=1}^{\infty}\frac{1}{1-(p_{i}p_{j})^{-s}}$? $\endgroup$ – A.Neves Dec 24 '14 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.