Re: linear secod order homogeneous differential equation recursions

• To: mathgroup at smc.vnet.net
• Subject: [mg69870] Re: linear secod order homogeneous differential equation recursions
• From: Paul Abbott <paul at physics.uwa.edu.au>
• Date: Tue, 26 Sep 2006 00:59:31 -0400 (EDT)
• Organization: The University of Western Australia
• References: <ef834t\$ggg\$1@smc.vnet.net>

In article <ef834t\$ggg\$1 at smc.vnet.net>,
Roger Bagula <rlbagula at sbcglobal.net> wrote:

> I have this factorial based recursion:
> a[n] = (a0*n^2 + b0*n + c0)*a[n - 2]/(n*(n - 1))
>
> a[n]*n!=Integer
> I want to get a form:
> b[n]=a[n]*n!

The solution is

a[n] n! == (a0^(n/2) (2^n C[1] + (-2)^n C[2]) *
Gamma[(b0 + 2 a0 (n + 2) - Sqrt[b0^2 - 4 a0 c0])/(4 a0)] *
Gamma[(b0 + 2 a0 (n + 2) + Sqrt[b0^2 - 4 a0 c0])/(4 a0)])/
(Gamma[(4 a0 + b0 - Sqrt[b0^2 - 4 a0 c0])/(4 a0)] *
Gamma[(4 a0 + b0 + Sqrt[b0^2 - 4 a0 c0])/(4 a0)])

Cheers,
Paul

_______________________________________________________________________
Paul Abbott                                      Phone:  61 8 6488 2734
School of Physics, M013                            Fax: +61 8 6488 1014
The University of Western Australia         (CRICOS Provider No 00126G)
AUSTRALIA                               http://physics.uwa.edu.au/~paul

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