Positivity of coefficients of the inverse of a certain power series Consider the unique formal power series $g(z)$ with $g(0)=0$ and $g'(0)=1$ satisfying the equation 
 $$
g(z)-g(z)^8+g(z)^{15}=z,
 $$
that is the inverse of
 $$
z-z^8+z^{15} 
 $$
in the group of formal power series of the form $z+O(z^2)$. Does this series have any non-positive coefficients of powers $z^{7k+1}$? (Other coefficients are manifestly zero).
Background.  Existence of non-positive coefficients among those would imply that a certain operad fails the Ginzburg-Kapranov test for Koszulness. (See http://arxiv.org/abs/0907.1505). 
Comments.  The first ten thousand or so of those coefficients are positive. Some elementary complex analysis approaches didn't work, but maybe I am not aware of some useful tricks.
 A: Combining Omar and fedja's comments gives a quick solution. By the Lagrange Inversion formula, we want to show that the coefficient of $t^k$ in $(1-t+t^2)^{-(7k+1)}$ is positive. Writing this as a contour integral, we want to compute
$$\frac{1}{2 \pi i} \oint \frac{dt}{t (1-t+t^2)} \left( \frac{1}{t (1-t+t^2)^7} \right)^k.$$
Set $f(t) = \tfrac{1}{t(1-t+t^2)}$ and $g(t) = \frac{1}{t(1-t+t^2)^7}$, so we want $\tfrac{1}{2 \pi i} \oint f(t) g(t)^k dt$.
The critical points of $g$ are at $t=1/5$ and $t=1/3$, so take our contour to be a circle of radius $1/5$ around the origin. We want $\tfrac{1}{10 \pi} \int_{\theta=0}^{2 \pi} f(e^{i \theta}/5) e^{i \theta} g(e^{i \theta}/5)^k d \theta$.
We have $g(1/5) = \tfrac{30517578125}{1801088541} \approx 17$, and $g(e^{i \theta}/5)$ smaller than that for all other $\theta$. We set $a=g(1/5)$. Here is a plot of $|g(e^{i \theta}/5)|$; I leave rigorous proof for others. So values of $\theta$ very near $0$ dominate the integral. The Taylor series for $g(e^{i \theta}/5)$ starts $a-b \theta^2$ where $b=\tfrac{152587890625}{37822859361}>0$. (Note the absence of an $i \theta$ term, because we chose a path through the critical point.)

Also, $f(1/5)=\tfrac{121}{25}>0$, call this $c$. So the integral is roughly $\int c (a-b \theta^2)^k d \theta \approx \int c a^k e^{-(b/a)k \theta^2} d \theta = \frac{C}{\sqrt{k}} a^k$ for some positive $C$. In particular, it is positive for large enough $k$ and, as fedja says, $10000$ is going to be way more than enough. I leave rigorous bounds to you.
