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An antidiagonal triangle based on: t(n,q) = If[n == 0, 1, Sum[Eulerian[n, k]*q^n, {k, 1, n, 2}]].
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%I #8 Dec 10 2016 17:28:47

%S 1,1,1,4,2,1,12,8,3,1,52,30,12,4,1,360,260,60,16,5,1,2656,2562,780,

%T 108,20,6,1,20160,23408,8568,1768,180,24,7,1,177472,262950,94752,

%U 20580,3380,282,28,8,1,1814400,3594620,1467840,277984,41160,5772,420,32,9,1

%N An antidiagonal triangle based on: t(n,q) = If[n == 0, 1, Sum[Eulerian[n, k]*q^n, {k, 1, n, 2}]].

%C Row sums are {1, 2, 7, 24, 99, 702, 6133, 54116, 559453, 7202238, ...}.

%F t(n,q) = If[n == 0, 1, Sum[Eulerian[n, k]*q^n, {k, 1, n, 2}]].

%e {1},

%e {1, 1},

%e {4, 2, 1},

%e {12, 8, 3, 1},

%e {52, 30, 12, 4, 1},

%e {360, 260, 60, 16, 5, 1},

%e {2656, 2562, 780, 108, 20, 6, 1},

%e {20160, 23408, 8568, 1768, 180, 24, 7, 1},

%e {177472, 262950, 94752, 20580, 3380, 282, 28, 8, 1},

%e {1814400, 3594620, 1467840, 277984, 41160, 5772, 420, 32, 9, 1}

%t Clear[t, n, q, a]

%t << DiscreteMath`Combinatorica`

%t t[n_, q_] = If[n == 0, 1, Sum[Eulerian[n + 1, k]*q^k, {k, 1, n, 2}]];

%t a = Table[Table[t[n, q], {n, 0, 10}], {q, 1, 11}];

%t Table[Table[a[[m, n - m + 1]], {m, 1, n}], {n, 1, 10}];

%t Flatten[%]

%K nonn,tabl,uned

%O 0,4

%A _Roger L. Bagula_, Mar 05 2010