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MathGroup Archive 2006

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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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