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Let $\sigma_m (n)=\sum_{d|n}d^m$, and $d(n)=\sigma_0(n)$ as stated in the title.

My question is: Does the q-series $f(q)=\sum_{n\ge 1} d(n)q^n$ gives something like a modular form (i.e. with some kind of identity under the action of $SL_2$) when $q=e^{2\pi i z}$?

When $d(n)$ is replaced by $\sigma_{2m-1} (n)$ the q-series gives out Eisenstein series, which are exactly modular forms. One way to obtain such a relation is to use Hecke's converse theorem for the L-function $\zeta(s) \zeta(s-2m+1)$. So I feel natural to think about $\zeta(s)^2$ for properties of $\sum_{n\ge 1} d(n)q^n$.

I tried to modify the argument of Hecke's converse theorem but it does not really give me a neat relation.

Any help will be appreciated. Thanks in advance.

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up vote 7 down vote accepted

The coefficient $d(n)$ is the limit of the $n$-th Hecke eigenvalue of the (nonholomorphic) Eisenstein series $E(z,s)$ as $s\to 1/2$. The limit of the Eisenstein series itself is zero, hence it is more natural to consider the $(\partial/\partial s)E(z,s)$ at $s=1/2$, which is precisely $$ \sqrt{y}\log y+4\sqrt{y}\sum_{n=1}^\infty d(n)K_0(2\pi ny)\cos(2\pi nx). $$

In short, you have to leave the realm of holomorphic modular forms to find the object you are looking for: $d(n)$ is "morally" the $n$-th Hecke eigenvalue of the Eisenstein series with Laplace eigenvalue $1/4$ (which is zero eventually, so we pass to the derivative). This is another reason why Maass forms are so natural, something that not every arithmetician appreciates!

You can read more about this fascinating story in Iwaniec: Introduction to the spectral theory of automorphic forms. The limit Eisenstein series above is discussed at the end of Chapter 3.

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@GH: Ahh, I really appreciate it! I did not think about the Maass form case. – Ted Mao Apr 23 '13 at 20:45
Glad you like it! BTW Hecke also tried hesperately to find certain things among holomorphic modular forms when Maass, his student, found them among Maass forms. – GH from MO Apr 23 '13 at 21:01

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