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). Does this mean that the $S$ is actually the $2$nd member of a series of differential operators with interesting divisibility properties?

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). Does this mean that the $S$ is actually the $2$nd member of a series of differential operators with interesting divisibility properties?

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 $x+c_1x^2+c_2x^2+c_3x^3+... \in A\left[\left[x\right]\right]$. 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). Does this mean that the $S$ is actually the $2$nd member of a series of differential operators with interesting divisibility properties?

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). Does this mean that the $S$ is actually the $2$nd member of a series of differential operators with interesting divisibility properties?