Dirichlet series of the $k$-th divisor of $n^2$

Question

I know that the Dirichlet series of the $k$-th divisor of n is $$\sum_{n=0}^\infty \frac{d_k(n)}{n^s}= \zeta^k(s).$$ where $d_k(n)$ is the number of $k$-tuples of integers $(a_1,a_2,\cdots,a_k)$ with product $n$, i.e. $$d_k(n)= \sum_{a_1\cdots a_k=n}1.$$

I would like to know how or where to find a formula for the Dirichlet series of $$\sum_{n=0}^\infty \frac{d_k(n^2)}{n^s}.$$

I computed when $k=2$, which is $$\sum_{n=0}^\infty \frac{d_2(n^2)}{n^s}=\frac{\zeta^3(s)}{\zeta(2s)}.$$

For $k\ge 2,$ I used the Euler product and stopped with

$$ \sum_{n=0}^\infty \frac{d_k(n^2)}{n^s} = \zeta^k(s) \prod_P \frac{1}{2} \left( \left(1-P^{-s/2}\right)^k+\left(1+P^{-s/2}\right)^k\right)$$

Answer 1

Given any $m$ multiplicative arithmetic functions $f_1,f_2,\cdots f_m:\mathbb{N}\to \mathbb{C}$ and any $n\in \mathbb{N}$ we have:

$$\sum_{\substack{k_1k_2k_3\cdots k_m=n\\(k_1,\ldots k_m)\in \mathbb{N}^m}}\prod_{j=1}^mf_j(k_j)=\prod_{p\mid n}\left(\sum_{\substack{e_1+e_2+e_3+\cdots e_m=v_p(n)\\(e_1,\ldots e_m)\in (\mathbb{N}\cup \{0\})^m}}\prod_{j=1}^mf_j(p^{e_j})\right)$$

With the product on the right ranging over primes $p\mid n$ and where $v_p(n)$ is the $p{\small -}$adic order of $n$. For instance when $m=2$ we have $\sum_{d\mid n}f_1(d)f_2\left(\frac{n}{d}\right)=\prod_{p\mid n}\left(\sum_{k=0}^{v_p(n)} f_1(p^k)f_2(p^{v_p(n)-k})\right)$.

Now as a result of this we get the following Dirichlet series generating function identity:

$$\mathfrak{D}(s;f_1*f_2*f_3*\cdots *f_m)=\sum_{n=1}^{\infty}\frac{(f_1*f_2*f_3*\cdots *f_m)(n)}{n^s}\\=\sum_{n=1}^{\infty}\frac{1}{n^s}\left({\small \sum_{\substack{k_1k_2k_3\cdots k_m=n\\(k_1,\ldots k_m)\in \mathbb{N}^m}}\prod_{j=1}^mf_j(k_j)}\right)=\prod_{p}\left(\sum_{n=0}^{\infty}\sum_{\substack{e_1+e_2+e_3+\cdots e_m=n\\(e_1,\ldots e_m)\in (\mathbb{N}\cup \{0\})^m}}\prod_{j=1}^mf_j(p^{e_j})\right)$$

Also since the Dirichlet convolution of multiplictive arithmetic functions is multiplictive. We see that $d_k(n)=\sum_{\substack{k_1k_2k_3\cdots k_m=n\\(k_1,\ldots k_m)\in \mathbb{N}^m}}1$ is also multiplicative. Further since the product of two multiplictive functions is also multiplictive and we have that $(\lambda*\mu)(n)$ is the indicator function for squares of natural numbers, we see using iverson brackets that $\alpha(n)=[\exists k\in \mathbb{N}:k^2=n]$ is multiplictive. Now combining this we find that $d_k(n)\alpha(n)$ is multiplictive and since we have:

$$\sum_{n=1}^{\infty}\frac{d_k(n)\alpha(n)}{n^{s/2}}=\sum_{n=1}^{\infty}\frac{d_k(n^2)}{n^s}$$

You can now use the previous identities at the start to express your series as an Euler product.

Also more generally if $\zeta$ is a primitive $m^\text{th}$ root of unity then we have that:

$$[\exists k\in \mathbb{N}:k^m=n]=\left(1*\zeta*\zeta^{2\Omega}*\zeta^{3\Omega}*\cdots *\zeta^{(m-1)\Omega}\right)(n)\\=\prod_{p\mid n}[m\mid v_p(n)]=[m\mid \gcd(v_2(n),v_3(n),v_5(n),v_7(n),v_{11}(n),\cdots)]$$

Where $\Omega(n)=\sum_{p}v_p(n)$ counts the number of (not neccisarly distinct prime factors of $n$) and where the function $\zeta^{k\Omega(n)}$ is completely multiplicative. Lastly for the special case (as used earlier) when $k$ is odd and $\zeta=-1$ this map $(-1)^{\Omega(n)}=\lambda(n)$ is known as the Liouville function.