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A product triangle sequence based on:a=3;f(n, a) = f(n - 1, a) + a*f(n - 2, a); c(n,a)=If[n == 0, 1, Product[f(i, a)*f(i + 1, a), {i, 1, n}]]; t(n,m,a)=If[Floor[n/2] >= m, c(n, a)/c(n - m, a), c(n, a)/c(m, a)]
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%I #2 Mar 30 2012 17:34:39

%S 1,1,1,1,4,1,1,28,28,1,1,133,3724,133,1,1,760,101080,101080,760,1,1,

%T 3880,2948800,392190400,2948800,3880,1,1,21049,81670120,62069291200,

%U 62069291200,81670120,21049,1,1,110236,2320357564,9002987348320

%N A product triangle sequence based on:a=3;f(n, a) = f(n - 1, a) + a*f(n - 2, a); c(n,a)=If[n == 0, 1, Product[f(i, a)*f(i + 1, a), {i, 1, n}]]; t(n,m,a)=If[Floor[n/2] >= m, c(n, a)/c(n - m, a), c(n, a)/c(m, a)]

%C Row sums are:

%C {1, 2, 6, 58, 3992, 203682, 398095762, 124301964740, 6860281000355442,

%C 10604146187026386826, 16491558999095964500974244,...}

%F a=3;

%F f(n, a) = f(n - 1, a) + a*f(n - 2, a);

%F c(n,a)=If[n == 0, 1, Product[f(i, a)*f(i + 1, a), {i, 1, n}]];

%F t(n,m,a)=If[Floor[n/2] >= m, c(n, a)/c(n - m, a), c(n, a)/c(m, a)]

%e {1},

%e {1, 1},

%e {1, 4, 1},

%e {1, 28, 28, 1},

%e {1, 133, 3724, 133, 1},

%e {1, 760, 101080, 101080, 760, 1},

%e {1, 3880, 2948800, 392190400, 2948800, 3880, 1},

%e {1, 21049, 81670120, 62069291200, 62069291200, 81670120, 21049, 1},

%e {1, 110236, 2320357564, 9002987348320, 6842270384723200, 9002987348320, 2320357564, 110236, 1},

%e {1, 588772, 64903870192, 1366161563671408, 5300706867045063040, 5300706867045063040, 1366161563671408, 64903870192, 588772, 1},

%e {1, 3109597, 1830843644884, 201824880037432624, 4248211899907919302576, 16483062171642726893994880, 4248211899907919302576, 201824880037432624, 1830843644884, 3109597, 1}

%t f[0, a_] := 0; f[1, a_] := 1;

%t f[n_, a_] := f[n, a] = f[n - 1, a] + a*f[n - 2, a];

%t c[n_, a_] := If[n == 0, 1, Product[f[i, a]*f[i + 1, a], {i, 1, n}]];

%t t[n_, m_, q_] = If[Floor[n/2] >= m, c[n, q]/c[n - m, q], c[n, q]/c[m, q]];

%t Table[Flatten[Table[Table[t[n, m, q], {m, 0, n}], {n, 0, 10}]], {q, 1, 10}]

%K nonn,tabl,uned

%O 0,5

%A _Roger L. Bagula_, Mar 19 2010