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How do I solve a recurrence of the form $$nd^{n-1}a_n+a_{n+1}d^{n+1}=b_n$$ for $a_n$, where $b_n$ is another (known) sequence and $d$ is a constant? My only idea was to use a generating function and try to separate the added terms on the left and hammer them into a form where they can go together, but I can't make it work.

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  • $\begingroup$ what is the objective ? a definition for $a_n$ ? $\endgroup$ – Cardinal Jul 16 '15 at 19:27
  • $\begingroup$ Rewriting this we get $a_{n+1}= a_n \frac{-n}{d^2} + \frac{b_n}{d^{n+1}}$ $\endgroup$ – Jan Jul 16 '15 at 19:32
  • $\begingroup$ @Cardinal Yes, in terms of $d$ and $b_n$. $\endgroup$ – Laertes Jul 16 '15 at 19:32
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I guess that you would like to express $a_n$ in terms of $b_n$.

From $$ a_{n+1}d^{n+1}+nd^{n-1}a_n=b_n \tag1 $$you have, assuming $d \neq0$, $$ a_{n+1}+nd^{-2}a_n=d^{-n-1}b_n \tag2 $$ and you obtain $$ \frac{(-1)^n}{\prod_{k=1}^n(kd^{-2})}a_{n+1}-\frac{(-1)^{n-1}nd^{-2}}{\prod_{k=1}^n(kd^{-2})}a_n=\frac{(-1)^n}{\prod_{k=1}^n(kd^{-2})}d^{-n-1}b_n \tag3 $$ giving $$ \frac{(-1)^n}{n!\:d^{-2n}}a_{n+1}-\frac{(-1)^{n-1}}{(n-1)!\:d^{-2(n-1)}}a_n=\frac{(-1)^n}{n!}d^{n-1}b_n. \tag4 $$ Then, summing from $n=1$ to $n=N$, you get by telescoping, $$ \frac{(-1)^N}{N!\:d^{-2N}}a_{N+1}-a_1=\sum_{n=1}^N\frac{(-1)^n}{n!}d^{n-1}b_n \tag5 $$ and

$$ a_{N+1}=(-1)^N \frac{N!}{d^{2N}}\sum_{n=1}^N\frac{(-1)^n}{n!}d^{n-1}b_n +(-1)^N \frac{N!}{d^{2N}}a_1. \tag6 $$

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