Attempting to solve the recurrence relation $$a_{n+1} = 2na_n+a_n+2$$
with $a_0=1$ (solution in summation form, closed form not necessary)
Let $E(x)=\sum^{\infty}_{n=0}a_n\frac{x^n}{n!}$, and multiply each time in the relation by $ \frac{x^n}{n!}$ and take the sum.
$$\sum^{\infty}_{n=0}a_{n+1}\frac{x^n}{n!}=\sum^{\infty}_{n=0}(2n+1)a_n\frac{x^n}{n!}+2\sum^{\infty}_{n=0}\frac{x^n}{n!}$$
notice that $\sum^{\infty}_{n=0}a_{n+1}\frac{x^n}{n!}=E'(x)$, and $\sum^{\infty}_{n=0}\frac{x^n}{n!}=e^x$. But I'm not sure what I can do with the middle sum, differentiating/integrating it doesn't seem to help, nor does substitutions like $m=2n+1$
Best Answer
Hint: The middle term is $2 \sum a_n\frac {x^{n}} {(n-1)!}+E(x)$. You can write this as $2x\sum a_{n+1}\frac {x^{n}} {n!} +E(x)=2xE'(x)+E(x)$