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Re: linear second order homogeneous differential equation recursions

• To: mathgroup at smc.vnet.net
• Subject: [mg69976] Re: linear second order homogeneous differential equation recursions
• From: Roger Bagula <rlbagula at sbcglobal.net>
• Date: Thu, 28 Sep 2006 06:17:06 -0400 (EDT)
• References: <200609250753.DAA11496@smc.vnet.net> <efactm$1vv$1@smc.vnet.net>

Daniel Lichtblau wrote:

>
>Out[17]= ((-2)^n*Cos[(Sqrt[5]*Pi)/2]*Gamma[(3 - Sqrt[5] + 2*n)/4]*
>   Gamma[(3 + Sqrt[5] + 2*n)/4]*(-2*Cos[(n*Pi)/2]*Gamma[(5 - Sqrt[5])/4]*
>     Gamma[(5 + Sqrt[5])/4] + Gamma[(3 - Sqrt[5])/4]*Gamma[(3 + Sqrt[5])/4]*
>     Sin[(n*Pi)/2]))/(Pi2*n!)
>
>In[22]:= FullSimplify[Table[an*n!, {n,0,12}]]
>Out[22]= {1, 1, -1, -5, 11, 95, -319, -3895, 17545, 276545, -1561505,
>     -30143405, 204557155}
>
>
>Daniel Lichtblau
>Wolfram Research
>
>  
>
Daniel Lichtblau,
Thanks for your help.

I have another more difficult problem.
I came up with a new type of sequence:
a[n]=f[n]*a[n-4]/(n*(n-1)*(n-2)*(n-3))

f[n_Integer] = Module[{a}, a[n] /. RSolve[{a[n] == 2*a[n - 1] - a[n
- 2] + 1, a[0] == 1, a[1] == 2}, a[n], n][[1]] // FullSimplify]
(1/2)(2 + n + n^2)
Rationalize[N[Table[f[n], {n, 0, 25}], 100], 0]
{1, 2, 4, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79, 92, 106, 121, 137, 154, 
172,
191, 211, 232, 254, 277, 301, 326}
Clear[a]
a[n_] := a[n] = f[n]*a[n - 2]/(n*(n - 1)*(n - 2)*(n - 3)); a[0] = 1; 
a[1] = 1; a[2] = 1; a[3] = 1;
Table[a[n + 2]*Gamma[n + 3]*Gamma[n + 1]/2, {n, 0, 30}]
{1, 3, 11, 48, 242, 1392, 8954, 64032, 501424, 4290144, 39612496, 394693248,
4198924576, 47757883008, 575252666912, 7354713983232, 98943458708864,
1404750370797312, 20877069787570304, 325902086024976384, 
5302775726042857216,
90274877828918458368, 1596135493538900022016, 29429610172227417427968,
561839693725692807749632, 11153822255274191205199872,
228668755346356972754100224, 4863066503299547365467144192,
106559639991402349303410704384, 2416944052139875040637170663424,
56370049555451842781504262619136}
I had to move it up two terms or the Gammas give it Infinities.

Your solution before when I get the powers right does work,
but I can't compute it on my machine:
it doesn't work. It gives an alternative method for my A123025 sequence!

A123025 <http://www.research.att.com/%7Enjas/sequences/A123025> 	
	Simple Scaled differential equation recursion RF: 
a[n+2]=-(n^2-n+1)*(a(n))/((n+2)*(n+1)). 	
	+20
1


	*1, 1, -1, -5, 11, 95, -319, -3895, 17545, 276545, -1561505, 
-30143405*, 204557155, 4672227775, -37024845055, -976495604975, 
8848937968145, 264630308948225, -2698926080284225, -90238935351344725, 
1022892984427721275, 37810113912213439775, -471553665821179507775 (list 
<http://www.research.att.com/%7Enjas/sequences/table?a=123025&fmt=4>; 
graph <http://www.research.att.com/%7Enjas/sequences/table?a=123025&fmt=5>)


	OFFSET 	

1,5

	

	REFERENCES 	

Richard Bronson,Schaum's Ouline of Modern Introductory Differential 
Equations, MacGraw-Hill, New York,1973,page 107, solved problem 19.17


	FORMULA 	

a(n) = -(n^2-n+1)*(a(n))/((n+2)*(n+1)) output= a(n)*n!


	MATHEMATICA 	

a[n_] := a[n] = -(n^2 - n - 1)*a[n - 2]/(n*(n - 1)); a[0] = 1; a[1] = 1; 
Table[a[n]*n!, {n, 0, 30}]


	KEYWORD 	

nonn,uned,new


	AUTHOR 	

Roger Bagula (rlbagula(AT)sbcglobal.net), Sep 24 2006


Roger Bagula

• References:
  • linear secod order homogeneous differential equation recursions
    • From: Roger Bagula <rlbagula@sbcglobal.net>