[seqfan] Re: Cassini transformation of linear recursion of order 3

[Vladimir Shevelev]

The following step on the way of solution of the general problem: if a(n)=La(n-1)+Ma(n-2)+Na(n-3)+Pa(n-4) with arbitrary
initials, then for g(n)=a^2(n)-a(n-1)a(n+1) we have
g(n)=-Mg(n-1)-(LN+P)g(n-2)-(L^2P+2MP-N^2)g(n-3)+P(LN+P)g(n-4)-MP^2g(n-5)+P^3g(n-6).
 
In general, the problem reduced to the following interesting algebraic problem:
Suppose we know a decomposition of a polynomial x^K+b_1x^(K-1)+...+b_K=(x-x_1)(x-x_2)...(x-x_K). 
Consider polynomial of order s=K(K-1)/2 with roots x_1x_2, x_1x_3,...,x_(K-1)x_K. It is required to find its coefficients, i.e.,
(x-x_1x_2)(x-x_1x_3)...(x-x_(K-1)x_K)=x^s+c_1x^(s -1)+...+c_s;   c_i=?
It is easy to see that c_1=-b_2 and c_s=(b_K)^(K-1), but other coefficients are calculated much difficulter.
 
Best regards,
Vladimir


----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Monday, March 4, 2013 19:15
Subject: [seqfan] Cassini transformation of linear recursion of order 3
To: seqfan at list.seqfan.eu

> Dear SeqFans,
>  
> I proved the following: 
> If a(n+3)=L*a(n+2)+M*a(n+1)+N*a(n), n>=0, a(0)=b, a(1)=c, a(2)=d
> and for n>=1, g=a^2(n)-a(n-1)*a(n+1),
> then g(n+3)=-M*g(n+2)-L*N*g(n+1)+N^2*g(n).
> This identity we call the Cassini transformation of linear 
> recursion of order 3.
> In case |g(n)|<=|a(n)|, this allows, using  telesopic 
> summation, to obtain a fast convergent series for 
> lim(a(n+1)/a(n)) as n goes to infinity. Could anyone say whether 
> such a transformation of linear recursion of order 3 is known?
>  
> Best regards,
> Vladimir
> 
>  Shevelev Vladimir‎
> 
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 Shevelev Vladimir‎