Conditions for a Maclaurin series to have (positive) integer coefficients Take for instance
$$
\frac{1}{\sqrt{z^2-6z+1}}=\sum_{\ell=0}^\infty c_\ell z^\ell\ .
$$
The coefficient $c_\ell$ of the Maclaurin series are all positive integers
$$
\vec{c}=\{1,3,13,63,321,\ldots\}
$$
given by the formula $c_\ell=p_\ell(3)$, where $p_\ell(t)$ is a Legendre polynomial. 
I am wondering if there exist necessary/sufficient conditions for a function $f(z)$ to ensure that its Maclaurin expansion has (positive) integer coefficients. I have never heard of anything like that, but it seems like a very curious property for a function.
Many thanks for your help.
 A: For sure, if a function is a combination of terms of the form $\frac{1}{(\alpha-x)^m}$ with $\alpha\in\mathbb{R}^+$ and $m\in\mathbb{N}$ it has the wanted property. In other cases, such property can be derived from a differential equation. In the present case, it is probably faster to notice that the coefficient are integers by a differential equation, then notice that
$$ \frac{d}{dx}\left(\log\frac{1}{1-6x+x^2}\right)=\frac{6-2x}{1-6x+x^2} $$
has a positive Taylor series at the origin, so does the function $\frac{1}{\sqrt{1-6x+x^2}}$.
A: This is a very nice observation. Like you said, we have
$$\frac{1}{\sqrt{1 - 2 x z + z^2}}= \sum_{l\ge 0} p_l(x) z^l$$
where $p_l(x)$ is the $l$-th Legendre polynomial. One of the explicit formulas for the Legendre polynomials is
$$p_l(2x+1) = \sum_{k=0}^l \binom{l}{k}\binom{l+k}{k} x^k$$
Therefore, we have
$$\frac{1}{\sqrt{1 - 2 (2x+1) z + z^2}}= \sum_{l\ge 0} p_l(2x+1) z^l$$
where $p_l(2x+1)$ are polynomials in $x$ with positive integral coefficients.
Note: For any $a>0$ we have
$$\frac{1}{(1-2(2x+1) z + z^2)^a} = \sum_{l\ge 0} q_l(x) z^l$$
where $q_l$ are polynomials in $x$ with positive coefficients. We can show the positivity by factoring $(1-2(2x+1) z + z^2)$.
