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A triangle of q factorial type based on Stirling first polynomials: t(n,k)=If[m == 0, n!, Product[Sum[(-1)^(i + k)*StirlingS1[k - 1, i]*(m + 1)^i, {i, 0, k - 1}], {k, 1, n}]].
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%I #5 Dec 09 2016 06:16:07

%S 1,1,1,1,-1,2,1,-1,2,6,1,-1,3,-12,24,1,-1,4,-36,288,120,1,-1,5,-80,

%T 2160,-34560,720,1,-1,6,-150,9600,-777600,24883200,5040,1,-1,7,-252,

%U 31500,-8064000,1959552000,-125411328000,40320,1,-1,8,-392,84672,-52920000

%N A triangle of q factorial type based on Stirling first polynomials: t(n,k)=If[m == 0, n!, Product[Sum[(-1)^(i + k)*StirlingS1[k - 1, i]*(m + 1)^i, {i, 0, k - 1}], {k, 1, n}]].

%C Row sums are:

%C {1, 2, 2, 8, 15, 376, -31755, 24120096, -123459768425, 5017134314247168,

%C -1827769039991244222327,...}.

%F t(n,k)=If[m == 0, n!, Product[Sum[(-1)^(i + k)*StirlingS1[k - 1, i]*(m + 1)^i, {i, 0, k - 1}], {k, 1, n}]];

%F out_(n,k)=Antidiagonal(t(n,k)).

%e {1},

%e {1, 1},

%e {1, -1, 2},

%e {1, -1, 2, 6},

%e {1, -1, 3, -12, 24},

%e {1, -1, 4, -36, 288, 120},

%e {1, -1, 5, -80, 2160, -34560, 720},

%e {1, -1, 6, -150, 9600, -777600, 24883200, 5040},

%e {1, -1, 7, -252, 31500, -8064000, 1959552000, -125411328000, 40320},

%e {1, -1, 8, -392, 84672, -52920000, 54190080000, -39504568320000, 5056584744960000, 362880},

%e {1, -1, 9, -576, 197568, -256048128, 800150400000, -3277416038400000, 7167708875980800000, -1834933472251084800000, 3628800}

%t Clear[t, n, m, i, k, a, b];

%t t[n_, m_] = If[m == 0, n!, Product[Sum[(-1)^(i + k)*StirlingS1[k - 1, i]*(m + 1)^i, {i, 0, k - 1}], {k, 1, n}]];

%t a = Table[Table[t[n, m], {n, 0, 10}], {m, 0, 10}];

%t b = Table[Table[a[[m, n - m + 1]], {m, n, 1, -1}], {n, 1, Length[a]}];

%t Flatten[%]

%Y A009963

%K sign,tabl,uned

%O 0,6

%A _Roger L. Bagula_, Feb 10 2009