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Antidiagonal of sequence: q=5; t(n,m) = Sum((2*cos(i*Pi/q))^m*cos[(m - 2*n)*i*Pi/q), {i, 0, q - 1}]/q.
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%I #4 Dec 10 2016 17:30:36

%S 1,0,1,0,1,1,1,0,2,2,0,1,1,3,5,0,1,1,3,5,11,1,0,2,2,6,10,22,0,1,1,3,5,

%T 11,21,43,0,1,1,3,5,11,21,43,85,1,0,2,2,6,10,22,42,86,170

%N Antidiagonal of sequence: q=5; t(n,m) = Sum((2*cos(i*Pi/q))^m*cos[(m - 2*n)*i*Pi/q), {i, 0, q - 1}]/q.

%C Row sums are {1, 1, 2, 3, 5, 10, 20, 45, 100, 215, ...}.

%D J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 41.

%F q=5; t(n,m) = Sum[(2*cos(i*Pi/q))^m*cos((m - 2*n)*i*Pi/q), {i, 0, q - 1}]/q;

%F out_n,m = Antidiagonal(t(n,m)).

%e {1},

%e {0, 1},

%e {0, 1, 1},

%e {0, 0, 2, 1},

%e {0, 0, 1, 3, 1},

%e {1, 0, 0, 3, 4, 2},

%e {0, 1, 0, 1, 6, 5, 7},

%e {0, 1, 1, 0, 4, 10, 7, 22},

%e {0, 0, 2, 1, 1, 10, 15, 14, 57},

%e {0, 0, 1, 3, 1, 5, 20, 22, 36, 127}

%t t[n_, m_, q_] = Sum[(2*Cos[i*Pi/q])^m*Cos[(m - 2*n)*i*Pi/q], {i, 0, q - 1}]/q;

%t a = Table[Table[If[ Rationalize[ N[t[n, m, q]]] < 10^(-10), 0, Rationalize[N[t[n, m, q]]]], {m, 0, 10}, {n, 0, 10}], {q, 3, 10, 2}];

%t Table[Flatten[Table[Table[a[[l]][[m, n - m + 1]], {m, 1, n}], {n, 1, 10}]], {l, 1, Length[a]}]

%K nonn,tabl,uned

%O 0,9

%A _Roger L. Bagula_, Mar 24 2010