I want to determine all solutions to the recurrence relation

$$ a_n−2na_{n−1}+n(n−1)a_{n−2} = 2n·n!, \ \ \ n≥2, a_0 = a_1 = 1. $$

by using exponential generating functions.

My idea was something like that: $$ \sum_{n\geq2} a_n \frac{z^n}{n!}−\sum_{n\geq2}2na_{n−1} \frac{z^n}{n!}+\sum_{n\geq2}n(n−1)a_{n−2} \frac{z^n}{n!} - \sum_{n\geq2}2n\frac{z^n}{n!} = 0 \\ $$


$$Â(z) =\sum_{n\geq0} a_n \frac{z^n}{n!}$$

which is the exponential generating function, I can derive that $\sum_{n\geq2}2na_{n−1} \frac{z^n}{n!}$ is $z^2 Â'(z)$ and for $\sum_{n\geq2}2n\frac{z^n}{n!}$, that $zÂ(2z)$ (which I am also not 100% sure of). However, I get stuck for the third summation term. How would I continue, am I even on the right path?

  • $\begingroup$ You're definitely on the right path! Don't forget the $2$ on the 2nd term. The last term should be $\sum {2n\cdot n!\cdot\frac{z^n}{n!}}$, which will be related to a geometric series (it has nothing to do with $A(z)$). As for the 3rd term, see if you can interpret it in terms of $A''(z)$. $\endgroup$ Nov 24, 2020 at 17:20
  • $\begingroup$ Seems like you're on the right path, but check the term $z^2A'(z)$ it should equal $\Sigma na_{n}\frac{z^{(n+1)}}{n!}$ which isn't exactly what you want, I got $2(zA(z)-z)$ for the first term, you can check it for yourself. $\endgroup$
    – Mars
    Nov 24, 2020 at 17:34
  • $\begingroup$ @Mars If you multiply by $z^n$ and divide by $n!$ then last term should be $2nz^n$ whose sum is $$\sum _{n=2}^{\infty } 2 n z^n=\frac{2 (z-2) z^2}{(z-1)^2}$$ $\endgroup$
    – Raffaele
    Nov 24, 2020 at 19:21
  • $\begingroup$ @Raffaele Ah, I see now. Thank you for catching that mistake, I updated my answer. $\endgroup$
    – Mars
    Nov 24, 2020 at 21:48

1 Answer 1


For the first term $$\sum_{n\geq2} a_n \frac{z^n}{n!}$$ I get $Â(z)-1-z$ by taking care of the first two initial conditions. For the second term $$\sum_{n\geq2}2na_{n−1} \frac{z^n}{n!}$$ I get $2(zÂ(z)-z)$, notice $\frac{n}{n!}=\frac{1}{(n-1)!}$ and that taking derivatives shift the index of $a_n$ up, multiplying by z shifts the index of $a_n$ down. The -z is obtained by accounting for initial conditions.

For the third term $$\sum_{n\geq2}n(n−1)a_{n−2} \frac{z^n}{n!}$$ I get $z^2Â(z)$. Notice that $\frac{n(n-1)}{n!}=\frac{1}{(n-2)!}$.

For the fourth term is actually $$ \sum_{n\geq 2}2nz^n$$ I get $\frac{2(z-2)z^2}{(z-1)^2}$. Because $\sum_{n\geq 1}nz^n=\frac{z}{(1-z)^2}$ and taking out the first term we get $\sum_{n\geq 2}nz^n=\frac{z}{(1-z)^2}-z=\frac{z(1-(1-z)^2)}{(1-z)^2}=\frac{z^2(2-z)}{(1-z)^2}$.

  • $\begingroup$ Can you maybe explain how you came up with the solution for the 4th term? $\endgroup$ Nov 24, 2020 at 18:43
  • $\begingroup$ @NikollosXerxes Yeah I'll edit my post now. $\endgroup$
    – Mars
    Nov 24, 2020 at 18:59

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