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Re: linear second order homogeneous differential equation recursions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69896] Re: linear second order homogeneous differential equation recursions
  • From: Roger Bagula <rlbagula at sbcglobal.net>
  • Date: Wed, 27 Sep 2006 06:03:51 -0400 (EDT)
  • References: <ef834t$ggg$1@smc.vnet.net> <efad1i$20h$1@smc.vnet.net>

Paul Abbott wrote:

>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
>
>  
>
Paul Abbott,
Thank you for your help. Both you and Daniel Lichtblau are better at 
getting solution out of Mathematica than I am;
with five conmstants  ( C[1] and C[2] added to {a0,b0,c0}) and  even 
with your simplified form a very complex answer!
Not exactly a simple Binet type...
I had continued my investigation and had found that the form
seemed related to Hypergeometrics which your solution seems to confirm.
Is there any way to convert it in Mathematica to an hypergeometric form?
I had further generalized the form by
f[n]-> Binet of the Fibonacci :
f[n_Integer] = Module[{a}, a[n] /. RSolve[{a[n] == a[ n - 1] + a[n
- 2], a[0] == 0, a[1] == 1}, a[n], n][[1]] // FullSimplify]
Clear[a]
a[n_] := a[n] = f[n]*a[n
- 2]/(n*(n - 1)); a[0] = 1; a[1] = 1;
Table[ExpandAll[a[n]*n!], {n, 0, 30}]
{1, 1, 1, 2, 3, 10, 24, 130, 504, 4420, 27720, 393380, 3991680, 91657540,
1504863360, 55911099400, 1485300136320, 89290025741800, 3838015552250880,
373321597626465800, 25964175210977203200, 4086378207619294646800,
459851507161617245875200, 117103340295746126693347600,
21322394684069868456741273600, 8785678105688353155168403690000,
2588389457883293541569193426124800, 1725665322163094950031867515982420000,
822618641999347403739646931950148812800,
887387152950606153059937200876123854180000,
684451614889137013807535833259801818202112000}

That idea was inspired by the Fibonacci Polynomials and the Bernoulli 
/Polynomials/ constants
which I found:
g[n_]=Gamma[n+1]*BernoulliB[n-1]...= (n!*Bernoulli[n-1])
Table[g[n],{n,1,30}]
{1, -1, 1, 0, -4, 0, 120, 0, -12096, 0, 3024000, 0, -1576143360, 0,
1525620096000, 0, -2522591034163200,
   0, 6686974460694528000, 0, -27033456071346536448000, 0,
160078872315904478576640000, 0, -1342964491649083924630732800000, 0,
15522270327163593186886877184000000, 0,
-241364461951740682229320388129587200000, 0}
Also:
h[n_] = (n + 1)*n!*BernoulliB[n];
Table[h[n], {n, 0, 30}]


I figure this is an even wider range of solutions that I had first thought,
with a wide range of applications in
different areas.
Roger Bagula


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