A closed form for $\sum_{i\cdot j^k=n}(-1)^i$?

Question

$$\alpha_k(n) \stackrel{\text{def.}}{=} \sum_{i\cdot j^k=n}(-1)^i.$$ Does a closed form exist for $\alpha_k(n)$? For low values of $k$: $$\alpha_0(n)=(-1)^n$$ $$\alpha_1(n)=\begin{cases} 2\nu_2(n)-\sigma_0\left(\frac{n}{2^{\nu_2(n)}}\right) & \text{ if n is even, n}\ne2^{m} \\ \nu_2(n)-1 & \text{ if n is even, n}=2^{m} \\ -\sigma_0(n) & \text{ if n is odd} \end{cases}$$ And with $a,b$ odd, $a\ne1$ having no repeated prime factors: $$\alpha_2(n)=\begin{cases} \sigma_0(b)p & \text{ if n=}ab^22^{2p+1} \\ \sigma_0(b)(p-2) & \text{ if n =}ab^22^{2p} \end{cases}.$$

Answer 1

Yes, define $t_k(n)$ for fixed $k\ge 1$ to be the indicator function for the integer powers of $k$.

By definition $t_k(n)$ is multiplicative. In addition let the dirichlet character modulo two be : $\chi_2(n)$.

$$\sum_{ij^k=n}(-1)^i=\sum_{ij^k=n}1-2\chi_2(i)=\sum_{ia=n}t_k(a)(1-2\chi_2(i))=\sum_{d\mid n}t_k(d)-2\sum_{d\mid n}t_k(d)\chi_2(\frac{n}{d})$$ $$=\prod_{p\mid n}(1+\lfloor\frac{v_p(n)}{k}\rfloor)-2\prod_{p\mid n}(\sum_{j=0}^{\lfloor{\frac{n}{k}\rfloor}}\chi_2(p^{n-jk}))$$

Where $v_p(n)$ denotes the p-adic order of $n$, which finally gives us that $a_k(n)=$

$$ \begin{cases} -\prod\limits_{p\mid n}\left(1+\left\lfloor\frac{v_p(n)}{k}\right\rfloor\right) & 2 \nmid n\\ \prod\limits_{p\mid n}\left(1+\left\lfloor\frac{v_p(n)}{k}\right\rfloor\right) & 2 \mid n, k\nmid n \\ \left(1-{2}{\left(1+\left\lfloor\frac{v_p(n)}{k}\right\rfloor\right)^{-1} }\right)\prod\limits_{p\mid n}\left(1+\left\lfloor\frac{v_p(n)}{k}\right\rfloor\right) & 2 \mid n,k\mid n \end{cases}$$