One can show that all coefficients with sufficiently large index are positive. Indeed, using Maple, the pole of $f$ closest to the origin is:
$a:=0.543689...,$
and the residue at this pole is $c:=-0.3115580216$. So 
$$\frac{c}{z-a}=-\frac{c}{a}\sum_{n=0}^\infty \left(\frac{z}{a}\right)^n$$
has positive coefficients.

It is not difficult to estimate the integer $n_0$ such that for $n>n_0$
this part of the partial fraction decomposition will dominate the rest,
and the first $n_0$ coefficients can be computed using Maple.

(The second pole closest to the origin is $a_1=0.56984...$ and the residue
at it $c_1=0.3383$. The other 4 poles are two complex conjugate pairs,
and their absolute values are $>1$, and the residues less than 9 by absolute value, so they have no influence. On the other hand Maple computes the first 100 or 200
coefficients in no time, and they are positive.)