<s>[INST] Is there any simple formula for $x \sum\limits_{k=1}^n \frac{p_{n-k}(x)}{k}$? Is there any simple formula?
$$
\begin{align}
p_0(x) &= 1 \\
p_n(x) &= x \sum\limits_{k=1}^n \frac{p_{n-k}(x)}{k}
\end{align}
$$
I am thankful for any help or hint.
Edit: The first few polynomials are
$$
\begin{align}
p_0(x) &= 1 \\
p_1(x) &= x \\
p_2(x) &= x^2 + \frac{1}{2}x \\
p_3(x) &= x^3 + x^2 + \frac{1}{3}x \\
p_4(x) &= x^4 + \frac{3}{2}x^3 + \frac{11}{12}x^2 + \frac{1}{4}x \\
p_5(x) &= x^5 + 2x^4 + \frac{7}{4}x^3 + \frac{5}{6}x^2 + \frac{1}{5}x
\end{align}
$$
I know that the first coefficient is always $1$, the second is $\frac{n-1}{2}$ and the last is $\frac{1}{n}$.
 [/INST] This is not an answer!
Just too long for a comment.
A closed form could be quite complicated...
Using Mathematica FindSequenceFunction powerful feature, I found
$$\frac{x}{n}+\frac{ (2 (\psi ^{(0)}n+\gamma ))}{n}x^2+\frac{\left(6 \psi ^{(0)}n^2+12 \gamma  \psi ^{(0)}n+6 \psi ^{(1)}n+6 \gamma ^2-\pi ^2\right)}{2 n}x^3+\ldots$$
where $\gamma$ is the Euler constant, and $\psi ^{(0)},\psi ^{(1)}$ are polygamma and its derivative.
Indeed we get
$$
\begin{array}{cr}
 1 & x \\
 2 & x^2+\frac{x}{2} \\
 3 & x^3+x^2+\frac{x}{3} \\
 4 & \ldots+\frac{3 x^3}{2}+\frac{11 x^2}{12}+\frac{x}{4} \\
 5 & \ldots\frac{7 x^3}{4}+\frac{5 x^2}{6}+\frac{x}{5} \\
\end{array}
$$
</s>