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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?

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

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  • $\begingroup$ 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
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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 $\eta(q) = q^{1/24} \prod_{k=1}^\infty (1-q^k)$.

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