[seqfan] Re: A049310 (by way of moderator)

[israel at math.ubc.ca]

If x = 2*(s+1/s), then S(N,x) = Sum_{k=0..N} s^{-N+2k}. In particular, for 
x = 2*cos(k*Pi/n) with integer k not divisible by n, s = exp(i*k*Pi/n) is a 
2n'th root of unity, and the periodicity follows from that.

Cheers,
Robert

On Aug 27 2016, Olivier Gerard wrote:

>Here is the original message By L. Edson Jeffery:
>
>
>Hello,
>
>The so-called "Chebyshev S-polynomials" are defined by the recurrence
>
>S(N,x) = x*S(N-1,x) - S(N-2,x)  (N>1),
>
>with initial conditions S(0,x) = 1, S(1,x) = x. The coefficients for S(N,x)
>appear in row N of triangle http://oeis.org/A049310.
>
>These polynomials appear to be periodic with period 2n when putting x =
>2*cos(k*Pi/n), for any nonzero integer k and any integer n>1. If true, then
>the indices can be reduced modulo 2n, a property I have found to be quite
>useful for rhombus substitution tilings over the past thirty years.
>
>So, first: What are the necessary and sufficient conditions for the
>sequence {S(N,x)} to be periodic?
>
>Second: for the product of any two S-polynomials, I conjectured that
>
>S(a,x)*S(b,x) = Sum_{k=0..min(a,b)} S(|a-b|+2k,x),
>
>for any integers a>=0, b>=0 and any x real or complex). Can someone prove
>this conjecture?
>
>Ed Jeffery
>
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>
>