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

[israel at math.ubc.ca]

Oops, I seem to have made a typo, as well as not simplifying sufficiently. 
Let's try again:

If x = s + 1/s, then 
S(N,x) = (s^(N+1)-s^(-N-1))/(s-1/s) = Sum_{k=0..N} s^{-N+2k}
Thus for x = 2*cos(t), s = exp(i*t), 
S(N,x) = sin((N+1)*t)/sin(t).
If t = k*Pi/n, then S(N,x) is periodic in N with period 2*n.

Cheers,
Robert

On Sep 1 2016, L. Edson Jeffery wrote:

>Hello Robert,
>
>Thank you for replying to my message (http://list.seqfan.eu/
>pipermail/seqfan/2016-August/016685.html). I have a couple of questions.
>
>You wrote:
>
>"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."
>
>For your first statement, is your sum
>
>S(N,x) = Sum_{k=0..N} s^{-N+2k}
>
>missing a coefficient, and could it instead be written as
>
>S(N,x) = Sum_{k=0..N} A(N-k,k)*s^{-N+2k},
>
>where A(N-k,k) is the upward antidiagonal entry in row N-k and column k of
>the array A in which row N has generating function (2-z)^N/(1-2z)^(N+1)? It
>appears that this array is symmetric and starts as
>
> 1    2     4      8      16      32       64  ...
> 2    7    20     52     128     304      704  ...
> 4   20    73    230     664    1808     4720  ...
> 8   52   230    847    2792    8536    24704  ...
>16  128   664   2792   10321   34922   110716  ...
>32  304  1808   8536   34922  129367   445244  ...
>64  704  4720  24704  110716  445244  1651609  ...
>...
>
>Row or column 0 is http://oeis.org/A000079 and row or column 1 is
>http://oeis.org/A066373, and the array is not in OEIS (not that it should
>be). The following Mathematica code (replacing N with n and A with a)
>displays the array:
>
>
>a[n_, k_] := SeriesCoefficient[Series[(2 - z)^n/(1 - 2 z)^(n + 1), {z, 0,
>10}], k]; Table[a[n, k], {n, 0, 6}, {k, 0, 6}] // Grid
>
>
>For your second statement, I understand the connection to 2n-th roots of
>unity which you gave (nice, by the way) but, since I have been trying to
>work out this proof on my own for a long time without success, could you
>please give the details of how the result simply follows from that
>connection?
>
>Thank you for your time,
>
>Ed Jeffery
>
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>Seqfan Mailing list - http://list.seqfan.eu/
>
>