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It follows from a general theorem of Honda that the formal group with the logarithm $$ x+x^{2^s}/2+x^{3^s}/3+x^{4^s}/4+\cdots $$ has integer coefficients. I became interested in it because its $p$-typizations give the formal groups of the $s$th Morava K-theories (after reducing modulo $p$).

In particular I wonder whether the series $$ \sum_{n\geqslant1}\frac{x^{n^2}}n $$ which one obtains for $s=2$ is related in any way to modular forms and elliptic curves.

Does anybody know where to find information about this?

P.S. - Decided to add a picture: here is the color-coded modulus of the derivative of the above series as a function of a complex variable $x$ in the unit disk, where its ``modular-like'' behavior is especially apparent.

enter image description here

P.P.S. - ...and for some further suspense, here are the first few terms of the formal group itself. Notation: $s$ is the sum of the two variables and $p$ is their product. Note the reappearing factors. \begin{align*} s\\ -p&(2s^2-p)\\ +2s^3p&(2s^2-p)\\ -sp&(3s^6-9s^4p+10s^2p^2-3p^3)\\ -s^2p&(2s^2-p)(4s^4+6s^2p-3p^2)\\ +s^4p&(12s^6-21s^4p+20s^2p^2-6p^3)\\ +2sp&(2s^2-p)(4s^8+18s^6p-5s^4p^2-4s^2p^3+p^4)\\ -2s^3p&(18s^{10}+18s^8p-67s^6p^2+87s^4p^3-48s^2p^4+9p^5)\\ -s^2p&(36s^{12}+246s^{10}p+72s^8p^2-493s^6p^3+356s^4p^4-106s^2p^5+12p^6)\\ +3s^9p&(3s^6-9s^4p+10s^2p^2-3p^3)\\ +... \end{align*}

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  • $\begingroup$ I don't know. $\sum x^{n^2}$ is a theta-function, and I know that one is very closely related to modular forms. $\endgroup$ – Gerry Myerson Nov 3 '13 at 5:12
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    $\begingroup$ I'm not an expert, but the Eichler integral can be defined, purely formally, as the operator that sends, with your notation, $x^n \mapsto x^n/n$ (recall that for modular forms $x=exp(2\pi i z)$, so this is integration $dz$). Your series is then a formal ``half-integral'' of the standard theta function. I've seen such things arise in talks, though I don't know anything about them myself (and I trust your ability to google as much as my own). I'd recommend searching for "half-derivative" rather than "half-integral", since the latter appears quite frequently with another meaning. $\endgroup$ – rlo Nov 3 '13 at 18:34
  • $\begingroup$ @rlo Wow! Many thanks, it seems to be the key! $\endgroup$ – მამუკა ჯიბლაძე Nov 3 '13 at 18:48
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    $\begingroup$ I still do not have anything definite to say; however, thanks to @rlo I found a really amazing identity: in link Andrews, Jiménez Urroz and Ono have found, generalizing previous amazing discoveries of Andrews and Zagier, among other things, an identity equivalent to the following: $$ 4\sum_{n\geqslant1}nq^{n^2}=\left(\sum_{n\in\mathbb Z}q^{n^2}\right)\left(2\sum_{n\geqslant1}\frac{q^{2n-1}}{1+q^{2n-1}}+ \sum_{n\geqslant1}\left(1-\prod_{k\geqslant n}\left(\frac{1+q^k}{1-q^k}\right)^{(-1)^k}\right)\right). $$ Isn't this crazy!? $\endgroup$ – მამუკა ჯიბლაძე Nov 9 '13 at 17:26
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    $\begingroup$ Here is another very interesting version. It does not come any closer to the goal for me, but still... Let $$ \theta_{\geqslant n}(q):=\prod_{k\geqslant n}\left(\frac{1-q^k}{1+q^k}\right)^{(-1)^k}; $$ in particular $\theta_{\geqslant1}=\theta$ is the "usual" theta function. Then, $$ 2\sum_{n\geqslant1}nq^{n^2}=\theta(q)\sum_{n\geqslant1}\frac{\theta_{\geqslant n}(q)-\theta_{\geqslant n}(q)^{-1}}2. $$ Cool, isn't it? :) $\endgroup$ – მამუკა ჯიბლაძე Nov 10 '13 at 10:07
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The following answer is not really satisfactory for me; however it seems to be the analog of current results on similar phenomena like partial, mock and quantum modular forms, so I decided to post it here in hope that somebody will contribute further improvements.

Using help from another question I posted later on, I can now claim this:

let $\tilde\theta(\tau):=\sum_{n\geqslant1}ne^{n^2\pi i\tau}$ be (up to a constant) the derivative wrt $\tau$ of the series in question with $x=e^{\pi i\tau}$; then in the upper half-plane, $$ \tilde\theta(-1/\tau)=(i\tau)^\frac32\tilde\theta(\tau)-\frac{i\tau}\pi\int\limits_0^\infty t\coth(\sqrt{\pi i\tau}t)e^{-t^2}dt. $$ The last term must be closely related to the Mordell integral; for large $z=i\tau/\pi$ its asymptotic behavior is given by the (divergent) series $$ \sum_{n\ge0}\frac{B_{2n}}{2n!}z^{1-n}=\frac z2+\frac1{12}-\frac1{120z}+\frac1{504z^2}-\frac1{1440z^3}+\frac1{3168z^4}-\frac{691}{3931200z^5}+... $$ which somehow explains the near-modular features of $\tilde\theta$. I think I will post a followup question to clarify relationship with some recent work mentioned by @rlo in a comment above.

Another thing I do not understand well: it seems that I cannot extend the first equality analytically simultaneously to both branches of the square root.

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Again in no sense a complete answer - I just found a partial explanation for my very last remark about reappearing factors in the homogeneous summands of the formal group.

Let $L_s(x)$ be the above logarithm, i. e. $\sum_{n\geqslant1}\frac{x^{n^s}}n$, and let $F_s(x_1,x_2,...)$ be the (many variable variant) of the corresponding formal group, i. e. $F_s=E_s(\sum_iL_s(x_i))$ where $E_s$ is the compositional inverse of $L_s$).

Adapting to our situation the argument for Theorem 4.3.9 (page 141, or page 121 of the online version) of Ravanel's Green Book one finds$$F_s(x_1,x_2,...)=F_s(w_1,w_2^{2^{s-1}},...,w_n^{n^{s-1}},...)$$where $w_n$ are the Witt symmetric functions (determined by $\sum_ix_i^n=\sum_{d|n}dw_d^{\frac nd}$).

Thus in particular for $s=2$ we get$$F_2(x,y)=F_2(w_1(x,y),w_2(x,y)^2,w_3(x,y)^3,...)=F_2(x+y,(-xy)^2,(-xy(x+y))^3,(-xy(x+y)^2)^4,(-xy(x^3+x^2y+xy^2+y^3))^5,...)$$which tells us that there will be lots of reappearing factors in the expansion...

I've asked a new question Efficiently computing (plethysm-like?)substitutions of symmetric functions about handling tricky expressions like that for $F_s$ above.

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