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.

Given by a CAS.
Isuppose that is missing the condition $a_3=0$.
Starting from $n=5$, the sequence
$$\left\{\frac{137}{25},\frac{1009}{150},\frac{17953}{2450},\frac{151717}{19600},\frac{
   170875}{21168},\frac{1474379}{176400},\frac{3751927}{435600},\frac{20228477}{22869
   00}\right\}$$ is not recognized by $OEIS$.
Being totally stuck, I used a CAS without conditions and got, after a lot of simplifications,
$$a_n=\frac{(-3 n^4+22 n^3-69 n^2+194 n+288+96 C) } {4 (n-3) (n-2) (n-1) n }+$$ $$\frac{12 (n-4) (n+1) \left(n^2-3 n+6\right) H_{n+1} } {4 (n-3) (n-2) (n-1) n }$$
For $n=0,1,2,3$, this leads to indeterminate forms. For $n=4$
$$a_4=C+\frac{25}{4}=0 \implies C=-\frac{25}{4}$$ leading to
$$a_n=(n-4)\frac{12 (n+1) \left(n^2-3 n+6\right) H_{n+1}-(n-3) \left(3 n^2-n+26\right) } {4 (n-3) (n-2) (n-1) n }$$ which is identical to what @Raffaele wrote in a comment.
Asymptotically,
$$a_n=3 \left(\gamma -\frac{1}{4}\right)+3 \log (n)+\frac{11}{2 n}-\frac{25}{4
   n^2}+O\left(\frac{1}{n^3}\right)$$
I would really like to know how, starting from scratch, we could arrive to the result.
