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%I #2 Oct 12 2012 14:54:52
%S 3,3,46,6,46,347,532,532,347,1932,14505,740,14505,1932,9199,203925,
%T 152405,152405,203925,9199,40250,2087884,6882086,-86372,6882086,
%U 2087884,40250,168318,17968725,152844537,78623775,78623775,152844537,17968725,168318
%N The nonzero difference between the Pascal {1,8,1} level triangle sequence and the {1,8,1} Catalan generalized triangle: t0(n,m)=Binomial[n, m]*Product[k!*(n + k)!/((m + k)!*(n - m + k)!), {k, 1, 7}]. A(n,k)=(3*n - 3*k + 1)A(n - 1, k - 1) + (3*k - 2)A(n - 1, k); t(n,m)=A(n,m)-t0(n,m). The first three levels and the external columns are zero and extracted.
%C Row sums are:
%C {6, 98, 1758, 33614, 731058, 17934068, 499210710}.
%F t0(n,m)=Binomial[n, m]*Product[k!*(n + k)!/((m + k)!*(n - m + k)!), {k, 1, 7}]. A(n,k)=(3*n - 3*k + 1)A(n - 1, k - 1) + (3*k - 2)A(n - 1, k); t(n,m)=A(n,m)-t0(n,m).
%e {3, 3},
%e {46, 6, 46},
%e {347, 532, 532, 347},
%e {1932, 14505, 740, 14505, 1932},
%e {9199, 203925, 152405, 152405, 203925, 9199},
%e {40250, 2087884, 6882086, -86372, 6882086, 2087884, 40250},
%e {168318, 17968725, 152844537, 78623775, 78623775, 152844537, 17968725, 168318}
%t f[0, 0] = 1; f[n_, m_] := f[n, m] = Binomial[n, m]*Product[k!*(n + k)!/((m + k)!*(n - m + k)!), {k, 1, 6}]; A[n_, 1] := 1 A[n_, n_] := 1 A[n_, k_] := (3*n - 3*k + 1)A[n - 1, k - 1] + (3*k - 2)A[n - 1, k]; a = Table[If[f[n - 1, k - 1] - A[n, k] == 0, {}, -f[n - 1, k - 1] + A[n, k]], {n, 10}, {k,n}]; c = Delete[Union[Table[Flatten[a[[n]]], {n, 1, 10}]], 1]; Flatten[a]
%K sign,uned
%O 1,1
%A _Roger L. Bagula_ and _Gary W. Adamson_, Sep 20 2008
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