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

[Vladimir Shevelev]

Thank you, Andy!
 
I understand that there are general results, but my concrete aim is to find a linear recurrence of order R(N) with constant coefficients for g=a^2(n)-a(n-1)*a(n+1), if a(n) satisfies a linear  recursion of order  N with constant coefficients. Thus my result yields that R(3)=3. I think that even function R(N) is unknown and it is, in my opinion,  an important conctete open problem.  As a continuation of this topic, I proved that R(4)=6. For example, for tetranacci numbers (A000078)  a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) with a(0)=a(1)=a(2)=0, a(3)=1, g(n) satisfies a  recursion
g(n) = -g(n-1) - 2*g(n-2) - 2*g(n-3) + 2*g(n-4) - g(n-5) + g(n-6). 
I am sure that a concrete recursion for g(n) if a(n) satisfies a linear recursion of order N with constant coefficients is a very interesting and difficult problem. Namely in this sense I asked what results are known, beginning with my one for N=3.
 
Best regards,
Vladimir

----- Original Message -----
From: Andrew N W Hone <A.N.W.Hone at kent.ac.uk>
Date: Tuesday, March 5, 2013 2:01
Subject: [seqfan] Re: Cassini transformation of linear recursion of order 3
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Dear Vladimir, 
> 
> I don't know to what extent this is "known" but such 
> transformations involving polynomial 
> functions of the terms of a linear recurrence have nice 
> properties. 
> 
> For instance, if the sequence (a_n) satisfies a linear 
> recurrence of order N with constant coefficients, then 
> 
> g_n = P(a_n, a_{n+1}, ... , a_{n+k-1}) 
> 
> also satisfies a linear recurrence with constant coefficients, 
> for any polynomial P in k variables (where w.l.o.g. 
> we can assume that k is at most N). With a bit of work one could 
> also get an upper bound on 
> the order of the recurrence for (g_n). 
> 
> One can easily prove this for sequences defined over the complex 
> numbers, since (generically) one 
> can write the terms of the first sequence as a sum 
>  
> a_n = \sum_j   C_j x_j^n 
> 
> over the roots x_j of the characteristic polynomial of the 
> recurrence for a_n, and then from the polynomial 
> P one has an expression of the form 
> 
> g_n = \sum_k D_k y_k^n, 
> 
> where the y_k are formed from certain products of x_j. Then g_n 
> satisfies 
> 
> F(S) g_n = 0 
> 
> where F(S) = \prod _k (S-y_k) and S is the shift operator 
> (sending n -> n+1). 
> 
> With a bit more work one could come up with a more general proof 
> that would work 
> with recurrences defined over any field, or over a ring. 
> 
> Your particular choice of g_n is very nice: it is a 2 x 2 
> determinant (a discrete Wronskian, or Casorati 
> determinant). 
> 
> Best wishes,
> Andy   
> 
> ________________________________________
> From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of 
> Vladimir Shevelev [shevelev at bgu.ac.il]
> Sent: 05 March 2013 03:21
> To: seqfan at list.seqfan.eu
> Subject: [seqfan] Cassini transformation of linear recursion of 
> order 3
> 
> 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/
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/
> 

 Shevelev Vladimir‎