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?
> 
> 
> 


$$b\_n=\begin{cases} 
\,\,\,\,\,\,\, a\_n\qquad \text{if $a\_n\geq0$} \\ -2a\_n \qquad \text{if $a\_n<0$}. \end{cases}$$



Theta functions, re-expressed
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 $\eta(q) = q^{1/24} \prod\_{k=1}^\infty (1-q^k)$.


