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A product triangle sequence based on:a=1;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 1,1,1,1,2,1,1,6,6,1,1,15,90,15,1,1,40,600,600,40,1,1,104,4160,62400,

%T 4160,104,1,1,273,28392,1135680,1135680,28392,273,1,1,714,194922,

%U 20271888,810875520,20271888,194922,714,1,1,1870,1335180,364504140

%N A product triangle sequence based on:a=1;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, 4, 14, 122, 1282, 70930, 2328692, 851810570, 76548543502,

%C 189143352851092,...}

%F a=1;

%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, 2, 1},

%e {1, 6, 6, 1},

%e {1, 15, 90, 15, 1},

%e {1, 40, 600, 600, 40, 1},

%e {1, 104, 4160, 62400, 4160, 104, 1},

%e {1, 273, 28392, 1135680, 1135680, 28392, 273, 1},

%e {1, 714, 194922, 20271888, 810875520, 20271888, 194922, 714, 1},

%e {1, 1870, 1335180, 364504140, 37908430560, 37908430560, 364504140, 1335180, 1870, 1},

%e {1, 4895, 9153650, 6535706100, 1784247765300, 185561767591200, 1784247765300, 6535706100, 9153650, 4895, 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