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%I #2 Mar 30 2012 17:34:29
%S 2,7,7,1,110,1,1,503,503,1,1,576,8926,576,1,1,667,54772,54772,667,1,1,
%T 778,118799,1091404,118799,778,1,1,911,207621,8440107,8440107,207621,
%U 911,1,1,1068,329900,27180372,187139238,27180372,329900,1068,1,1,1251
%N Adjusted recursive triangular sequence with row sums 2*(n+5)!/6!: A(n,k)= A(n - 1, k - 1) + A(n - 1, k) + (n + 4)*(n + 3)*A(n - 2, k - 1).
%C Row sums are:
%C {2, 14, 112, 1008, 10080, 110880, 1330560, 17297280, 242161920, 3632428800,...}.
%C The division by 6! and changing the first element gives a nicer looking result.
%F A(n,k)= A(n - 1, k - 1) + A(n - 1, k) + (n + 4)*(n + 3)*A(n - 2, k - 1).
%e {2},
%e {7, 7},
%e {1, 110, 1},
%e {1, 503, 503, 1},
%e {1, 576, 8926, 576, 1},
%e {1, 667, 54772, 54772, 667, 1},
%e {1, 778, 118799, 1091404, 118799, 778, 1},
%e {1, 911, 207621, 8440107, 8440107, 207621, 911, 1},
%e {1, 1068, 329900, 27180372, 187139238, 27180372, 329900, 1068, 1},
%e {1, 1251, 496770, 65297294, 1750419084, 1750419084, 65297294, 496770, 1251, 1}
%t Clear[A] A[1, 1] = 2*6!/720; A[2, 1] := A[2, 2] = 7!/720;
%t A[3, 2] = (2*8! - 2*6!)/720; A[4, 2] = A[4, 3] = ( 9! - 6!)/720;
%t A[n_, 1] := 6!/720; A[n_, n_] := 6!/720;
%t A[n_, k_] := A[n - 1, k - 1] + A[n - 1, k] + (n + 4)*(n + 3)*A[n - 2, k - 1];
%t a = Table[A[n, k], {n, 10}, {k, n}];
%t Flatten[a]
%t Table[Apply[Plus, a[[n]]], {n, 1, 10}];
%t Table[Apply[Plus, 720*a[[n]]]/(2*(n + 5)!), {n, 1, 10}];
%K nonn,tabl
%O 1,1
%A _Roger L. Bagula_, Dec 31 2008
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