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.

I have one different idea, variation of constant.
If one has a recurrence relation of the form
$$ a_{n+1} = \alpha_n a_n + \beta_n,\quad n\geq 0, $$
then we can make the ansatz
$$ a_n^{(p)} = C_n \prod_{j=}^{n-1} \alpha_j $$
and get
$$ C_{n+1} \prod_{j=0}^n \alpha_j = \alpha_n C_n \prod_{j=0}^{n-1} \alpha_j + \beta_n $$
$$ \Longrightarrow C_{n+1} = C_n + \frac{\beta_n}{\prod_{j=0}^n \alpha_j} $$
$$ \Longrightarrow C_n = \sum_{l=0}^{n-1} \frac{\beta_l}{\prod_{j=0}^l \alpha_j} + C_0$$
and therefore
$$ a_n = \sum_{l=0}^{n-1} \frac{\beta_l}{\prod_{j=0}^l \alpha_j} \prod_{j=0}^{n-1} \alpha_j + C \prod_{j=0}^{n-1} \alpha_j = \sum_{l=0}^{n-1} \frac{12 H_{l+1}}{\prod_{j=0}^l \frac{j-3}{j+1}} \prod_{j=0}^{n-1} \frac{j-3}{j+1} + C \prod_{j=0}^{n-1} \frac{j-3}{j+1} \\= \sum_{l=0}^{n-1} 12 H_{l+1} \prod_{j=l+1}^{n-1} \frac{j-3}{j+1} + C \prod_{j=0}^{n-1} \frac{j-3}{j+1},$$
if we define
$$ \alpha_n = \frac{n-3}{n+1},\qquad \beta_n = 12 H_{n+1}.$$
I do not know how to continue from here, maybe someone knows how to simplify this terms to get to the solution?
