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

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

  • 2
    $\begingroup$ @MaxAlekseyev: The OP means the number of ways to write $n$ as a product of $k$ factors, and $\zeta^k(s)$ is the $k$-th power of the Riemann zeta function. Anyways, the OP's Dirichlet series for $k\geq 3$ is not meromorphic on $\mathbb{C}$, so there is no such pretty formula as for $k=2$. This is the so-called Estermann phenomenon: www-users.math.umn.edu/~garrett/m/mfms/notes_2013-14/… $\endgroup$
    – GH from MO
    Mar 12, 2018 at 20:59
  • $\begingroup$ @GH from MO : does it mean that the coefficient $ a_{n} $ of the Dirichlet series is not $ \ll_{\varepsilon}n^{\varepsilon} $ for all $ \varepsilon>0 $? $\endgroup$ Mar 12, 2018 at 21:51
  • 2
    $\begingroup$ @SylvainJULIEN: No, it does not mean that. They are $\ll_\epsilon n^\epsilon$, but this only buys you (locally uniform) absolute convergence (hence holomorphicity) for $\Re(s)>1$. Apriori it could happen that the series cannot be continued beyond this half-plane. The above series can be, but only up to $\Re(s)>1/2$ (omitting $s=1$), where they hit a natural wall of (non-isolated) singularities. $\endgroup$
    – GH from MO
    Mar 12, 2018 at 21:56
  • $\begingroup$ That looks mysterious to me. Can one define a finer measure of the growth speed of $ a_n $ with $ n $ to give a clear picture of this phenomenon ? $\endgroup$ Mar 12, 2018 at 22:00
  • 2
    $\begingroup$ @SylvainJULIEN: It is not the growth, but the arithmetic regularity (or arithmetic global structure) that matters. Most Dirichlet series, even with tiny coefficients do not extend meromorphically to $\mathbb{C}$. Analytic continuation is the exception, not the rule. This is why we love automorphic $L$-functions. They are very rare and precious objects (with analytic continuation, functional equation etc.). At any rate, this site is not for informal discussions like this one, but for formal questions and answers. $\endgroup$
    – GH from MO
    Mar 12, 2018 at 22:03

1 Answer 1


Given any $m$ multiplictive 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:


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.