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.

Let $f\left( x \right) = \sum\limits_{n = 0}^\infty  {a_n x^n } $ be the generating function of the solution of the recurrence.
$$
\sum\limits_{n = 0}^\infty  {na_n x^n }  = \sum\limits_{n = 0}^\infty  {\left( {n + 1} \right)a_n x^{n + 1}  - 4\sum\limits_{n = 0}^\infty  {a_{n - 1} x^n } }  + 
12\sum\limits_{n = 5}^\infty  {na_n x^n } H_n 
$$
which gives the following DE
$$
xf'\left( x \right) = x^2 f'\left( x \right) - 3xf\left( x \right)-\frac{2 x \left(50 x^5-67 x^4+2 x^3+3 x^2+6 x+6 \log (1-x)\right)}{(x-1)^2}$$
Whose solution is
$$f(x)=\frac{2}{75 (x-1)^2} \left(-7553 x^5+3750 x^5 \log (1-x)+\\+24040 x^4-18750 x^4 \log (1-x)-29405 x^3+37500 x^3 \log (1-x)+\\+16830 x^2-37500 x^2 \log (1-x)-3840 x+18750 x \log (1-x)-3840 \log (1-x)\right)$$
Coefficients of the MacLaurin expansion of $f(x)$ are
$$0,0,0,0,0,\frac{137}{5},\frac{578}{15},\frac{1667}{35},\frac{395}{7},\frac{41137}{630},\frac{13007}{175},\frac{964849}{11550},\ldots$$
There is no closed form for $a_n$, though.
