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