I am trying to solve the following recursion $$ \sum _{i=1}^{n-1} -\mu (n-i)+\sum _{i=1}^{n-1} 2 i \mu (n-i)=(n-1)^2 (\mu (n)-1) \tag{1} $$
By increasing $n$ to $n+1$, we get $$ \sum _{i=1}^n -\mu (-i+n+1)+\sum _{i=1}^n 2 i \mu (-i+n+1)=n^2 (\mu (n+1)-1) \tag{2} $$ and then by shifting index, we get $$ \sum _{i=1}^{n-1} -\mu (n-i)+\sum _{i=1}^{n-1} 2 (i+1) \mu (n-i)+\mu (n)=n^2 (\mu (n+1)-1)\tag{3} $$ Taking the difference of (1) and (3), we arrive at $$ -2 \sum _{i=1}^{n-1} \mu (n-i)-\mu (n)=(n-1)^2 (\mu (n)-1)-n^2 (\mu (n+1)-1)\tag{4} $$
Replace $n$ by $n+1$ and shift index again, we get $$ (n+1)^2 \mu (n+2)+1= \\2 \sum _{i=1}^{n-1} \mu (n-i)+n^2 \mu (n+1)+2 \mu (n)+\mu (n+1)+2 n+2 \tag{5} $$ Then taking the difference between (4) and (5), we get $$ (n+1)^2 \mu (n+2)+(n-2) n \mu (n)=\left(2 n^2+1\right) \mu (n+1)+2 \tag{6}$$ This has a solution $$ \mu(n)=\frac{2}{3} H_{n-1}:=\frac{2}{3} \sum_{i=1}^{n-1} \frac{1}{i} $$ What bothers me is that this solution does not fit in the original equation (1)! How is this possible? I don't think I have made a mistake in the computations.
The sums are $$ \sum _{i=1}^{n-1} -\mu (n-i)+\sum _{i=1}^{n-1} 2 i \mu (n-i)=(n-1)^2 (\mu (n)-1) \tag{1} $$ Your equation you have arrived $$ (n+1)^2\mu(n+2)-(2n^2+1)\mu(n+1)+n(n-2)\mu(n)-2=0\tag{eq} $$ Have indeed general solution $$ \mu(n)=-1+\frac{2}{3(n-1)}+\frac{2}{3}H_{n-2}+C_1+C_{2}\left(-\frac{1}{2}\psi^{(1)}(n-2)+\frac{\pi^2}{12}+\frac{-12+43n-46n^2+20n^3-3n^4}{4(2-3n+n^2)^2}\right)\tag{sol} $$ as long as $n\geq 3$. That is because for $n=1,2$ the solution has singularities at $n=1$ and $n=2$ ($n=2$ is also a curious point since $H_n=\sum_{1\leq n\leq n}k^{-1}$ and by convection we use $H_0=0$). It is like beeing solving a two order differential equation with no initial conditions. The solution it can be far away from what you might expecting. The Mathematica do its best. Its gives you $-1+\frac{2}{3(n-1)}+\frac{2}{3}H_{n-2}+C_1$ and not $C_1+\frac{2}{3}H_{n-1}$, to tell you that $n=1$ and $n=2$ are pathological cases. Hence for $n\geq 3$ the solution goes well. But for $n=1,2$ you have to define or give more input. The values at $n=1$ and $n=2$ are usefull for your computation of the sums you have. By demanding your sums to be in accordance to each other you can define $$ \mu(n)=\left\{\begin{array}{cc} c_1\textrm{ , for } n=1\\ c_2\textrm{ , for }n=2\\ 2/3H_{n-1}-1\textrm{ , for }n\geq 3 \end{array}\right\}\tag 2 $$
where $H_0=0$. Setting these into (eq) and (1), you take a solution: $c_1=-5/4$, $c_2=-1/4$.