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let $a_{1}=1,a_{2}=3$, and $a_{n+2}=(n+1)a_{n+1}+a_{n}$

find the close form $a_{n}$

my try:

let $$\dfrac{a_{n+2}}{(n+1)!}=\dfrac{a_{n+1}}{n!}+\dfrac{a_{n}}{(n+1)!}$$ so $$\dfrac{a_{n+2}}{(n+1)!}-\dfrac{a_{2}}{1!}=\sum_{i=1}^{n}\dfrac{a_{i}}{(i+1)!}$$

But we can't find the $a_{n}$ Thank you

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this does not seem to have closed form. –  Santosh Linkha Sep 25 '13 at 11:50
@experimentX: yes it has a closed form: it is:$$c_1I_n(-2)+c_2K_n(2)$$ where $I_n$ and $K_n$ are the modified Bessel functions of the first and second kind –  Riccardo.Alestra Sep 25 '13 at 11:56
Hello, can you post your solution? –  china math Sep 25 '13 at 12:20
Well, the obvious way to solve this is to rewrite the problem in terms of differential equation by collecting all the terms of the sequence into a generating function. Then just solve the differential equation (this will probably give the Bessel functions that Wolfram Alpha provided) and finally plug in initial conditions to pin down the exact solution. –  Marek Sep 25 '13 at 12:27
Well, some would think (like...say, me) that the Bessel functions, leave alone modified of this or that kind, hardly fit into what many would call "closed form" for a simple, straightforward sequence defined recursively. –  DonAntonio Sep 25 '13 at 12:40

1 Answer 1

up vote 3 down vote accepted

Hint: Use the following recurrence equations: $$I_n(z)=\frac{2(n+1)}{z}I_{n+1}(z)+I_{n+2}(z)$$ and: $$K_n(z)=\frac{2(n+1)}{z}K_{n+1}(z)-K_{n+2}(z)$$ If you put $z=2$ and $a_n=I_n(-2)+K_n(2)$ you get: your recurrence equation in $a_n$ You can find what you need here

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