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

[L. Edson Jeffery]

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