# Multiplication of coefficients in Dirichlet series

This appears to be a relationship:

$\sum\limits_{p\;\text{prime}} \frac{1}{p^s} = \log\zeta (s) - \sum\limits_{n=1}^{\infty}\frac{\sqrt{a_{n}b_{n}}}{n^{s}}$

where $a_{n}$ is a sequence of fractions beginning $\frac{0}{1}, \frac{1}{1}, \frac{1}{1}, \frac{1}{2}, \frac{1}{1}, \frac{0}{1}, \frac{1}{1}, \frac{1}{3}, \frac{1}{2}, \frac{0}{1}, \frac{1}{1}, \frac{0}{1}...$ and $b_{n}$ is another sequence of fractions $\frac{0}{1}, \frac{0}{1}, \frac{0}{1}, \frac{1}{2}, \frac{0}{1}, \frac{1}{1}, \frac{0}{1}, \frac{1}{3}, \frac{1}{2}, \frac{1}{1}, \frac{0}{1}, \frac{0}{1}...$

for which the Dirichlet series are defined by:

$\sum\limits_{n=1}^{\infty}\frac{a_{n}}{n^{s}} = +\frac{1}{1}(\zeta (s)-1)^1 -\frac{1}{2}(\zeta (s)-1)^2 +\frac{1}{3}(\zeta (s)-1)^3 - \frac{1}{4}(\zeta (s)-1)^4 +\frac{1}{5}(\zeta (s)-1)^5 -...$

$\sum\limits_{n=1}^{\infty}\frac{b_{n}}{n^{s}} = -\frac{0}{1}(\zeta (s)-1)^1 +\frac{1}{2}(\zeta (s)-1)^2 -\frac{2}{3}(\zeta (s)-1)^3 + \frac{3}{4}(\zeta (s)-1)^4 -\frac{4}{5}(\zeta (s)-1)^5 +...$

The square root of the elementwise multiplication of $a_{n}$ times $b_{n}$,$\;$ $\sqrt{a_{n}b_{n}}$ is then: $\frac{0}{1},\frac{0}{1},\frac{0}{1},\frac{1}{2},\frac{0}{1},\frac{0}{1},\frac{0}{1},\frac{1}{3},\frac{1}{2},\frac{0}{1},\frac{0}{1},\frac{0}{1}...$

Is there any literature describing multiplication of coefficients of Dirichlet series?

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After a quick glance at your sequences, it appears that $$a_n =\frac{\Lambda(n)}{\log n}$$ and $$\sqrt{a_n b_n}=\frac{\Lambda(n)}{\log n} -\chi_p (n)$$ where $\chi_p(n)$ is the indicator function for the primes.
The sequence $b_n$ doesn't really matter too much, except the fact that it is $$\frac{\Lambda(n)}{\log n} -\chi_p(n)$$ on prime powers. Since $a_n$ is zero elsewhere, the sequence $b_n$ can be whatever we want elsewhere. (Similarly the sequence $a_n$ can be whatever we want on the primes, and you still have the relation in the first line.)
So, why does this hold? Since $\sqrt{a_nb_n}=\frac{\Lambda(n)}{\log n} -\chi_p (n)$ it follows that $$\sum_{n=1}^\infty \frac{\sqrt{a_n b_n}}{n^s}=\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n} n^{-s} +\sum_p p^{-s}.$$ Then the prime zeta functions cancel, and we recall that $$\sum_{n=1}^\infty \frac{\Lambda(n)}{\log n} n^{-s}=\log \zeta(s).$$ Hope that helps.