Euler product over a finite subset of composite numbers

Question

The product $$\prod_{p\text{ prime}}\frac{p^n}{p^n-1}=\zeta(n)$$ is well-known. This works over primes, and only over the entire set of them.

Is any similar expression known for composites? For a finite subset of them?

e.g. $$\prod_{c\text{ composite, c}=c_{i}}^{c=c_{j}}\frac{c^n}{c^n-1}=f(m)$$ for some $f$.

This is generic. If such an expression is not known, is anything known for the case $i=1$ and $n=1$? If not, then for $i=n=1$ and $j=\infty$? If that is also not known, is anything known on $\lim \inf f(m)$ or $\lim \sup f(m)$?

Answer 1

Surely we can say something in the case of the set of all composite numbers and for an integer exponent. From the following representation of Gamma function $$\Gamma\left(z\right)=\lim_{n\rightarrow+\infty}\frac{n^{z-1}n!}{z\left(z+1\right)\cdots\left(z+n-1\right)}$$ it is possible to prove that $$\prod_{n\geq0}\frac{\left(n+a_{1}\right)\cdots\left(n+a_{k}\right)}{\left(n+b_{1}\right)\cdots\left(n+b_{k}\right)}=\frac{\Gamma\left(b_{1}\right)\cdots\Gamma\left(b_{k}\right)}{\Gamma\left(a_{1}\right)\cdots\Gamma\left(a_{k}\right)},\,a_{1}+\dots+a_{k}=b_{1}+\dots+b_{k}$$ for $a_{i},b_{j}\in\mathbb{C}\setminus\mathbb{Z}^{-},$ (see Whittaker and Watson, A course of modern analysis, $4$th edition, Cambridge University Press, Cambridge, $1927$, section $12.13$) then if $k>1$ is a positive integer we have $$\prod_{n\geq2}\frac{n^{k}}{n^{k}-1}=\prod_{n\geq0}\frac{\left(n+2\right)^{k}}{\left(n+2-\omega_{0}^{k}\right)\cdots\left(n+2-\omega_{k-1}^{k}\right)}=\frac{\Gamma\left(2-\omega_{0}^{k}\right)\cdots\Gamma\left(2-\omega_{k-1}^{k}\right)}{\Gamma\left(2\right)^{k}}$$ where $\omega_{j}^{k}=\exp\left(\frac{2\pi ij}{k}\right)$. Then, since $\Gamma(2)=\Gamma(1)=1$, $$\prod_{c\text{ composite}}\frac{c^{k}}{c^{k}-1}=\color{red}{\frac{\Gamma\left(2-\omega_{1}^{k}\right)\cdots\Gamma\left(2-\omega_{k-1}^{k}\right)}{\zeta\left(k\right)}}.$$ ADDENDUM: I think it is interesting to see how the answer of Daniel D. is linked to mine; from the power series $$\sum_{n\geq2}\frac{\left(-1\right)^{n}x^{n}\zeta\left(n\right)}{n}=\gamma x+\log\left(\Gamma\left(x+1\right)\right),\,-1<x\leq1$$ we have $$\sum_{n\geq2}\frac{\left(-1\right)^{n}x^{n}\left(\zeta\left(n\right)-1\right)}{n}=\left(\gamma-1\right)x+\log\left(\Gamma\left(x+1\right)\right)+\log\left(\left(x+1\right)\right)$$ which can be extended to $x=-1$ and where $\gamma$ is the Euler-Mascheroni constant. Now recalling that if $f(x)=\sum_{n\geq0}a_{n}x^{n}$ then we have the multisection formula $$\sum_{n\geq0}a_{qn+p}x^{qn+p}=\frac{1}{q}\sum_{j=0}^{q-1}\omega_{q}^{-jp}f\left(\omega_{q}^{j}x\right)$$ where $\omega_{q}^{j}$ is a primitive $q$-root of the unity, we get, assuming that $f(x):=\left(\gamma-1\right)x+\log\left(\Gamma\left(x+1\right)\right)+\log\left(\left(x+1\right)\right)$ and $m>1$, that $$\sum_{n\geq1}\frac{\zeta\left(mn\right)-1}{n}=\lim_{x\rightarrow-1^{+}}\sum_{j=0}^{m-1}f\left(\omega_{m}^{j}x\right)=\color{red}{\sum_{j=1}^{m-1}\log\left(\Gamma\left(2-\omega_{m}^{j}\right)\right)}.$$ Notice that we used the relations $\sum_{j=0}^{q-1}\omega_{q}^{j}=0$ and $$\lim_{x\rightarrow-1^{+}}\left[\log\left(\Gamma\left(x+1\right)\right)+\log\left(\left(x+1\right)\right)\right]$$ $$=\lim_{x\rightarrow-1^{+}}\left[\log\left(\Gamma\left(x\omega_{q}^{0}+1\right)\right)+\log\left(\left(x\omega_{q}^{0}+1\right)\right)\right]=0.$$

Answer 2

My bit, with some manipulations one can relate the zeta function $\zeta(m)=f_p(m)$ with the prime zeta function $P$

$f_p(m) =\prod_p\frac{1}{1-p^{-m}} =\exp\ln\prod_p\frac{1}{1-p^{-m}} =\exp\sum_p\ln(\frac{1}{1-p^{-m}}) =\exp\sum_p\sum_k\frac{p^{-mk}}{k} =\exp\sum_k\frac{1}{k}\sum_p\frac{1}{p^{mk}} =\exp\sum_k\frac{1}{k}P(mk) $

Now call $f_n(m)$ the product over all the naturals and let's do the same set of manipulation than before

$f_n(m) =\prod_n\frac{1}{1-n^{-m}} =\exp\ln\prod_n\frac{1}{1-n^{-m}} =\exp\sum_n\ln(\frac{1}{1-n^{-m}}) =\exp\sum_n\sum_k\frac{n^{-mk}}{k} =\exp\sum_k\frac{1}{k}\sum_n\frac{1}{n^{mk}} =\exp\sum_k\frac{1}{k}(\zeta(mk)-1) $

Finally the product over all composites $f_c(m)$ can be obtained as the quotient of the two other expressions since $f_n(m)=\prod_n\frac{1}{1-n^{-m}}=\prod_c\frac{1}{1-c^{-m}}\prod_p\frac{1}{1-p^{-m}}=f_c(m)f_p(m)$ so

$f_c(m)=f_n(m)/f_p(m)=\exp\sum_k\frac{1}{k}(\zeta(mk)-1-P(mk))$

if $i\neq1$ and $j\neq\infty$ instead of $\zeta(mk)$ and $P(mk)$ we should get a partial sum of those.

Answer 3

This Answer addresses the coefficients of the Dirichlet-Series of $f(s) = \prod_{c \text{ comp}} \frac 1{1-c^{-s}}$.

We introduce $$g(s) = \prod_{n=2}^\infty (1 + n^{-s} + n^{-2s} + \dots ),$$ which is a holomorpic function on $\Re s > 1$, as $\left | 1 - \frac 1{1-n^{-s}}\right | \ll n^{-s}$. This can be re-written as a dirichlet-series $$g(s) = \sum_{n=1}^\infty c(n) n^{-s},$$ where $c(n) = \#\{(e_2, e_3, \dots) \in \mathbb N_{\geq 0}^* | 2^{e_2} 3^{e_3} 4^{e_4} \dots = n \}$ is the amount of multiplicative partitions of $n$. With $\mu$ (the möbius-function) we obtain $$\prod_{c \text{ comp}} \frac 1{1-c^{-s}} = \zeta(s)^{-1} g(s) = \sum_{n=1}^\infty a(n) n^{-s},$$ where $a(n) = (c \star \mu)(n) = \sum_{d \mid n} \mu(d) c(\frac nd)$.