TITLE: Sufficient conditions for the coefficients of a generating function to dominate those of its square
QUESTION [6 upvotes]: Let $f(z)$ be a generating function (so in particular, its power series coefficients are nonnegative). I am interested in conditions which would ensure that for every $n$, the coefficient of $z^n$ in $f$ is greater than  or equal to the coefficient of $z^n$ in $f^2$. Equivalently, I would like the generating function $f-f^2$ to have nonnegative coefficients.
One necessary condition is that $f$ have no constant term, so $f(0)=0$. Another necessary condition is that $f$ have infinitely many terms; polynomials will never satisfy this condition. But these necessary conditions are very far from sufficient.
For an example of a generating function that does satisfy this condition, let $f$ denote the generating function for the shifted Catalan numbers,
$$
f=\frac{1-\sqrt{1-4z}}{2}=z+z^2+2z^3+5z^4+14z^5+\cdots.
$$
Then we have that $f-f^2=z$, which satisfies the condition.
However the condition seems to be quite fragile. If we instead take
$$
f=\frac{1-2z-\sqrt{1-4z}}{2z}=z+2z^2+5z^3+14z^4+42z^5+\cdots
$$
then we have $f-f^2=z+z^2+z^3-6z^5-\cdots$, which does not satisfy it.
The only other necessary condition I know of is that the dominant singularity of $f$ cannot be a pole, because if it were a pole then (by Pringsheim's Theorem) there would be a real number $z_0>0$ such that $f(z_0)>1$, and thus we would have $f^2(z_0)>f(z_0)$, so the condition could not hold.
Other than a few specific examples (e.g., the shifted Catalan numbers above), I know of no sufficient conditions.

REPLY [6 votes]: Take any series $g$ with zero free term and all other coefficients being nonnegative, and solve $f-f^2=g$ with respect to $f$, i.e.,
$$f=\frac{1-\sqrt{1-4g}}2.$$
This is the general form of all such $f$, and it can be easily seen that the coefficients of $f$ are nonnegative as well.