[seqfan] Re: 3 nonlinear recurrences
maximilian.hasler at gmail.com
Sun Sep 20 19:57:27 CEST 2009
Georgi Guninski wrote:
> A059480 A recurrence equation.
> a(n) = a(n - 1) + (n + 1)*a(n - 2)
This is not a nonlinear, but a linear recurrence:
The sequence obtained for a(0)=alpha and a(1)=beta is
alpha a0 + beta a1, where a0 and a1 are the sequences obtained for
(alpha,beta)=(1,0) resp. (0,1),
> 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
But what is the purpose of this more complicated formula?
> A135686 Two sequence type recursion of a generalized Stirling number
> type using the Fibonacci sequence
This also is a linear recurrence.
>w0,w1,w2,w3,w4,w5,w6,w7= -1, 1, -1, -1, -1, 1, 1, 1
> a[i+2] =(w0*a[i-1]^2*a[i] - (w1*a[i-2]*a[i]^2+w2*a[i-1]*a[i]^2+w3*a[i-2]*a[i-1]*a[i+1]+w4*a[i-2]*a[i]*a[i+1]))/(w5*a[i-2]*a[i-1])
This also is more complicated that the original formula.
In particular, why do you use w0,w2,...,w7 instead of just putting the
minus sign for the negative ones ?
> a[n]=a[n-1]+2^n * a[n-2]
again a linear recurrence
You get this if you write 2^(n+1)=2*2^n in the equation for a(n+1)
and substitute 2^n by its expression in terms of (a(n)-a(n-1))/a(n-2).
You can of course "disguise" any linear recurrence in this manner, but
I don't think this is very useful.
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