TITLE: Theta functions, re-expressed
QUESTION [17 upvotes]: 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?

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

REPLY [22 votes]: 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 $\eta(q) = q^{1/24} \prod_{k=1}^\infty (1-q^k)$.