Dirichlet inverse for $\left\{1,0,-1,0,1,0,-1,0,1,0,-1,0,\ldots\right\}$

Question

I am looking for the Dirichlet inverse of $\left\{1,0,-1,0,1,0,-1,0,1,0,-1,0,\ldots\right\}$ or equivalently $$ f(n)=\frac{ {i^{n-1}+(-i)^{n-1}}}{2}. $$ It is an interesting inverse, it seems always to evaluate to $\pm1,0$, i.e. $f^{-1}(n) \in \left\{-1,0,1\right\}$. An explicit formula for $f^{-1}(n)$ would help me.

Answer 1

Note that $$ \sum\limits_{n = 1}^\infty {\frac{{f(n)}}{{n^s }}} = \sum\limits_{n = 0}^\infty {\frac{{( - 1)^n }}{{(2n + 1)^s }}} = \beta (s) = \prod\limits_{p > 2} {\left( {1 + ( - 1)^{\frac{{p + 1}}{2}} p^s } \right)^{ - 1} } , $$ where $\beta$ is the Dirichlet beta function and the Euler product is over the odd primes. Consequently, $$ \sum\limits_{n = 1}^\infty {\frac{{f^{ - 1} (n)}}{{n^s }}} = \prod\limits_{p > 2} {\left( {1 + ( - 1)^{\frac{{p + 1}}{2}} p^s } \right)} . $$ Expanding the product on the right-hand side, we see that $$ f^{ - 1} (n) = ( - 1)^{\sum\nolimits_{p|n} {\frac{{p + 1}}{2}} } $$ when $n$ is an odd, square-free integer and $f^{-1} (n)=0$ otherwise.