<s>[INST] Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$ I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$
Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.
 [/INST] Here are more details for @Phicar's suggested approach.
Let $A(z)=\sum_{n \ge 0} a_n z^n$ be the ordinary generating function of $(a_n)$.
Then $z A'(z)=\sum_{n \ge 0} n a_n z^n$, and the recurrence relation implies that
\begin{align}
z A'(z) 
&= \sum_{n \ge 5} \left((n-4)a_{n-1} + 12n H_n\right) z^n \\
&= z \sum_{n \ge 5} (n-1) a_{n-1} z^{n-1} - 3 z \sum_{n \ge 5} a_{n-1} z^{n-1} + 12z \sum_{n \ge 5} n H_n z^{n-1} \\
&= z \cdot z A'(z) - 3z A(z) + 12z\left(\frac{d}{dz}\left(\frac{-\log(1-z)}{1-z}\right) -\sum_{n=1}^4 n H_n z^{n-1}\right) \\
&= z^2 A'(z) - 3z A(z) + 12z\left(\frac{1-\log(1-z)}{(1-z)^2} -1-3z-\frac{11}{2}z^2-\frac{25}{3}z^4\right)
\end{align}
The resulting solution for $A(z)$ is messy, so there must be a better way.
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