Recall the classical $\theta(q):=\prod_{k=1}^{\infty}(1-q^k)$ and define the sequences $a_n$ and $b_n$ by

$$\frac{\theta^3(q)}{\theta(q^3)}=\sum_{n=0}^{\infty}a_nq^n \qquad \text{and} \qquad F(q):=\sum_{i,j\in\Bbb{Z}}q^{i^2+ij+j^2}=\sum_{n=0}^{\infty}b_nq^n.$$

Edit. In accord with Noam's commentary, we may replace $\theta$ by $\eta$.

Question. Is the following true? If so, any proof?

$$b_n=\begin{cases} \,\,\,\,\,\,\, a_n\qquad \text{if $a_n\geq0$} \\ -2a_n \qquad \text{if $a_n<0$}. \end{cases}$$

  • $\begingroup$ Out of idle curiosity: is it a generating function for something interesting? It's kinda noteworthy that "reciprocal" function $\eta (q^p)^p / \eta (q)$ counts projective irreps of $S_n$ over $\mathbb F_p$. $\endgroup$ – Denis T. Apr 10 '17 at 20:13
  • $\begingroup$ $F(q)$ is one of the theta functions in Borweins' cubic theta-function identity, see B. Berndt, "Ramanujan's Notebooks", part 5, chapter 33. $\endgroup$ – Nemo Dec 1 '17 at 21:05

Yes, it is true.

This generating function $\sum_n a_n q^n$ turns out to be the same as $(3F(q^3)-F(q))/2$: they coincide through the $q^{100}$ term, which is more than enough to prove equality between modular forms of weight $1$ for a congruence group of such low index in ${\rm SL}_2({\bf Z})$. Your conjecture then follows from $b_{3n} = b_n$ for all $n$ (for which an explicit bijection is $(i,j) \mapsto (i+2j,i-j)$).

P.S. I think the generating function for the $a_n$ would be "classically" called not $\theta^3(q) / \theta(q^3)$ but $\eta^3(q) / \eta(q^3)$, where $\eta(q) = q^{1/24} \prod_{k=1}^\infty (1-q^k)$.


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.