I speculated in 2008 that the modified Neretin polynomials presented in [A145900](http://oeis.org/A145900) of the On-line Encyclopedia of Integer Sequences, which can be summed to give a normalized Schwarzian derivative for a complex function and are related to a representation of the Virasoro algebra, all have integer coefficients. Definitions, references, and links are provided in the entry. Can anyone prove or disprove this conjecture?



Conjecture: "Neretin polynomials" for a normalized Schwarzian have integer coefficients
Before I give the answer, let me try to formulate the question in the way *I* would have asked it here. Fortunately I am not bound by the OEIS requirements of brevity and ASCII, and there is LaTeX here...


**Question.** Let $A$ be the polynomial ring $\mathbb Z\left[c\_1,c\_2,c\_3,...\right]$ in infinitely many commuting indeterminates $c\_1$, $c\_2$, $c\_3$, .... Let $g$ be the formal power series $x+c\_1x^2+c\_2x^3+c\_3x^4+... \in A\left[\left[x\right]\right]$. By considering $A=\mathbb Z\left[c\_1,c\_2,c\_3,...\right]$ as a subring of $\mathbb Q\left[c\_1,c\_2,c\_3,...\right]$, we can define a power series $S = \dfrac{x^2}{6} \left(\dfrac{g^{\prime\prime\prime}}{g^{\prime}} - \dfrac32 \left(\dfrac{g^{\prime\prime}}{g^{\prime}}\right)^2\right) \in \left(\mathbb Q\left[c\_1,c\_2,c\_3,...\right]\right)\left[\left[x\right]\right]$. To prove that this $S$ actually lies in $A\left[\left[x\right]\right]$.


**Remark.** If we let $f=g^{\prime}$, then $S$ can also be written as $\dfrac{x^2}{6}\left(D^2\left(\ln f\right) - \dfrac12 \left(D\left( \ln f\right)\right)^2\right)$. We will not need this, however.


**Answer to the question.** Since the constant term of the power series $g^{\prime} \in A\left[\left[x\right]\right]$ is $1$, the power series $g^{\prime}$ has a multiplicative inverse $\dfrac{1}{g^{\prime}}$ in $A\left[\left[x\right]\right]$.


For every $k\in\mathbb N$, the power series $\dfrac{1}{k!}g^{(k)}$ lies in $A\left[\left[x\right]\right]$ (because for every $n\in\mathbb N$, the coefficient of this power series $\dfrac{1}{k!}g^{(k)}$ before $x^n$ is


$\dfrac{1}{k!}\left(n+k\right)\left(n+k-1\right)...\left(n+1\right)c\_{n+k} = \dbinom{n+k}{k} c\_{n+k} \in A$ (where $c\_0$ denotes $1$)


). Applied to $k=2$, this yields $\dfrac{g^{\prime\prime}}{2} \in A\left[\left[x\right]\right]$. On the other hand, applied to $k=3$, it yields $\dfrac{g^{\prime\prime\prime}}{6} \in A\left[\left[x\right]\right]$.


Now,


$S = \dfrac{x^2}{6} \left(\dfrac{g^{\prime\prime\prime}}{g^{\prime}} - \dfrac32 \left(\dfrac{g^{\prime\prime}}{g^{\prime}}\right)^2\right) = x^2 \left(\dfrac{g^{\prime\prime\prime}}{6}\cdot\dfrac{1}{g^{\prime}} - \left(\dfrac{g^{\prime\prime}}{2}\right)^2\cdot\left(\dfrac{1}{g^{\prime}}\right)^2\right)$


is in $A\left[\left[x\right]\right]$ (because each of $x^2$, $\dfrac{g^{\prime\prime\prime}}{6}$, $\dfrac{1}{g^{\prime}}$ and $\dfrac{g^{\prime\prime}}{2}$ is in $A\left[\left[x\right]\right]$).


**Meta-Question.** The formula $S = \dfrac{x^2}{6}\left(D^2\left(\ln f\right) - \dfrac12 \left(D\left( \ln f\right)\right)^2\right)$ reminds me of $p\_2 = e\_1^2 - 2e\_2$ (one of the [formulae for power sums in terms of elementary symmetric functions](http://en.wikipedia.org/wiki/Newton%2527s_identities#Expressing_power_sums_in_terms_of_elementary_symmetric_polynomials)). Does this mean that the $S$ is actually the $2$nd member of a series of differential operators with interesting divisibility properties?


