Help me manipulating with exponential generating function (recurrence relation) I have recurrence relation $f_0=0, f_1=1$
$$f_n = \frac{2n-1}{n}f_{n-1} - \frac{n-1}{n}f_{n-2} + 1$$
$$nf_n = nf_{n-1} + (n-1)f_{n-1} - (n-1)f_{n-2} + n$$
I tried to solve it using ordinary generating functions and it turned out to be close to impossible to solve:
$$ln(F(x))' = \frac{(1-x^3)(1-x)^2+x^2}{(x-x^3-x^4)(1-x)^2}$$ where F(x) was generating function for sequence $\langle f_n \rangle$
So i thought exponential generating functions will do the job here, but i can't see the simple trick to solve it... I'd really really appreciate some help on this, because i'm stuck with this problem for like 3 days and i'd really like to solve it by generating functions. Thanks in advance!
 A: I hope I am not wrong.
I started discarding the $1$ in the rhs; writing the first terms, it seems to me that the general formula is $$f_n=f_0+(f_1-f_0) H_n$$ in which appears the harmonic number. Knowing now that the $H_n$ has to appear, I found (almost emprically) that for the recurrence you posted should be
$$f_n=f_0+\frac{1}{4} n (n+3)+(f_1-f_0-1) H_n $$
A: Here are the steps which are quite simple, if we know the answer to look for.
Making use of Claude Leibovici's empirical result we put 
$$ g_n = n (f_n-f_{n-1})$$
to obtain
$$ g_n = g_{n-1} + n$$
with $g_1 = f_1-f_0$ and $g_0 = 0.$

This gives $$g_n = f_1-f_0 + \sum_{k=2}^n k$$
or $$g_n = f_1-f_0-1 + \frac{1}{2} n (n+1).$$
Now $$\frac{g_n}{n} = f_n - f_{n-1}$$
so that
$$\sum_{k=1}^n \frac{g_k}{k} = f_n - f_0.$$
But $$\sum_{k=1}^n \frac{g_k}{k}
= (f_1-f_0-1) H_n + \frac{1}{2} \sum_{k=1}^n (k+1)
\\= (f_1-f_0-1) H_n + \frac{1}{2} 
\left(-1+ \frac{1}{2}(n+1) (n+2)\right)$$
which is
$$(f_1-f_0-1) H_n + \frac{1}{4} n (n+3)$$
so that
$$f_n = f_0 +
(f_1-f_0-1) H_n + \frac{1}{4} n (n+3).$$
