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

[Vladimir Shevelev]

Here is a representation of the desired polynomial R(x)=x^(K(K-1)/2)+... in the posed algebraic problem.
Let P(x)= x^K+b_1x^(K-1)+...+b_K=(x-x_1)(x-x_2)...(x-x_K).  Then P_1(x)=P(-x)(-1)^K=(x+x_1)(x+x_2)...(x+x_K).
Put Q(x)=P(x)P_1(x)|_(x^2:=x). Let A be the companion matrix of P(x) and A*A be the Kronecker's square of A.
Then R^2(x) = det(xI-A*A)/Q(x),         (1)
where I is the identity matrix of order K^2.    
In particular, if a(n)=La(n-1)+Ma(n-2), then P(x)=x^2-Lx-M. Then Q(x)=(x^2-Lx-M)(x^2+Lx-M)|_(x^2:=x)=
x^4-(2M+L^2)x^2+M^2|_(x^2:=x)=x^2-(2M+L^2)x+M^2.
Further, A is matrix
 0  M 
 1   L
with characteristic polynomial P(x) and A*A is matrix 
0  0    0   M^2
0  0   M   LM  
0  M  0    LM
1  L   L    L^2
with characteristic polynomial   x^4-L^2x^3-2M(L^2+M)x^2-L^2M^2x+M^4=Q(x)(x^2+2Mx+M^2)                 
and   R^2(x)=(x+M)^2,  R(x)=x+M, i.e., g(n)=(-M)g(n-1).  For Fibonacci numbers L=M=1 and it is easy to see that g(n)=(-1)^(n-1).
Thus (1) is a very wide generalization of Cassini's identity for Fibonacci numbers.
 
Best regards,
Vladimir


----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Saturday, March 9, 2013 2:53
Subject: [seqfan] Re: Cassini transformation of linear recursion of order 3
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> 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
> > 
> > _______________________________________________
> > 
> > Seqfan Mailing list - http://list.seqfan.eu/
> > 
> 
>  Shevelev Vladimir‎
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/
> 

 Shevelev Vladimir‎