# [seqfan] Re: 3 nonlinear recurrences (Georgi Guninski)

Jaume Oliver i Lafont joliverlafont at gmail.com
Mon Sep 21 19:59:04 CEST 2009

```Hello, Georgi

> 
> A059480 A recurrence equation.
> a(n) = a(n - 1) + (n + 1)*a(n - 2)
>
> a[i+1]=( a[i-2]*a[i-1]-a[i-1]^2+a[i-2]*a[i]+a[i-1]*a[i])/a[i-2]
> (i think i proved this by substituting the definition and simplifying to
> 0)

The result is not needed in order to proof it, it can be derived from
the definition.

a(n) = a(n - 1) + (n + 1)*a(n - 2)    
so
a(n+1)=a(n)+(n+2)*a(n-1)             

from ,
n+1=(a(n)-a(n-1))/a(n-2)                                                  

from ,
n+1=(a(n+1)-a(n))/a(n-1)-1=(a(n+1)-a(n)-a(n-1))/a(n-1)        

Now equating the right terms of  and  and isolating a(n+1)
yields your result, without the "n" in the formula.

Jaume

```
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