Product over the primes with relation to the Dirichlet series

Question

What is the value of $\displaystyle \prod_p\left(1+\frac{p^s}{(p^s-1)^2}\right)$
I got this product by defining a function $a(n)$ such that $a(n)=a(p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_n^{a_n})=a_1a_2a_3...a_n$ and using that function in a Dirichelt series
$\sum_{n=1}^\infty \frac{a(n)}{n^s}=\prod_p(1+\sum_{n=1}^\infty\frac{a(p^n)}{p^{ns}})=\prod_p(1+\sum_{n=1}^\infty\frac{n}{p^{ns}})=\prod_p\left(1+\frac{p^s}{(p^s-1)^2}\right)$
I defined $a(1)=1$

Answer 1

This is the Dirichlet generating function (g.f.) of OEIS sequence A005361 "Product of exponents of prime factorization of n" which states "Dirichlet g.f.: ζ(s)ζ(2s)ζ(3s)/ζ(6s)". The proof is simple. Recall that the definition of the Riemann zeta function is $$ \zeta(s) := \sum_{n=1}^\infty n^{-s} = \prod_p (1-p^{-s})^{-1}.$$ This implies that $$ \zeta(ks) = \prod_p (1-X^k)^{-1} $$ where $\, X := p^{-s}.\,$ Now our Dirichlet g.f. is ($X\,$ or $\,X^{-1}\,$ the same here)

$$ G(s) := \prod_p \left(1 + \frac{p^s}{(p^s-1)^2} \right) = \prod_p \left(1 + \frac{X}{(1-X)^2}\right).$$

Use some algebra to get $$ 1 + \frac{x}{(1-x)^2} = \frac{1-x+x^2}{(1-x)^2} = \frac{1-x^6}{(1-x)(1-x^2)(1-x^3)}.$$ Combine previous equations to get $$ G(s) = ζ(s)ζ(2s)ζ(3s)/ζ(6s). $$