[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
> [1]
> 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) [1]
so
a(n+1)=a(n)+(n+2)*a(n-1) [2]
from [1],
n+1=(a(n)-a(n-1))/a(n-2) [3]
from [2],
n+1=(a(n+1)-a(n))/a(n-1)-1=(a(n+1)-a(n)-a(n-1))/a(n-1) [4]
Now equating the right terms of [3] and [4] and isolating a(n+1)
yields your result, without the "n" in the formula.
Jaume
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