A173005 A product triangle sequence based on recursion:a=4; f(n,a)=(2*a+1)*f(n-1,a)+f(n-2,a).
1, 1, 1, 1, 9, 1, 1, 80, 80, 1, 1, 711, 6320, 711, 1, 1, 6319, 499201, 499201, 6319, 1, 1, 56160, 39430560, 350439102, 39430560, 56160, 1, 1, 499121, 3114515040, 246007756722, 246007756722, 3114515040, 499121, 1, 1, 4435929, 246007257601
Offset: 0
Examples
{1}, {1, 1}, {1, 9, 1}, {1, 80, 80, 1}, {1, 711, 6320, 711, 1}, {1, 6319, 499201, 499201, 6319, 1}, {1, 56160, 39430560, 350439102, 39430560, 56160, 1}, {1, 499121, 3114515040, 246007756722, 246007756722, 3114515040, 499121, 1}, {1, 4435929, 246007257601, 172697094835902, 1534842394188558, 172697094835902, 246007257601, 4435929, 1}, {1, 39424240, 19431458835440, 121233114567545603, 9575881454449171680, 9575881454449171680, 121233114567545603, 19431458835440, 39424240, 1}, {1, 350382231, 1534839240742160, 85105473729326613333, 59743922859711995180563, 530973050767752120484320, 59743922859711995180563, 85105473729326613333, 1534839240742160, 350382231, 1}
Programs
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Mathematica
Clear[f, c, a, t]; f[0, a_] := 0; f[1, a_] := 1; f[n_, a_] := f[n, a] = (2*a + 1)*f[n - 1, a] - f[n - 2, a]; c[n_, a_] := If[n == 0, 1, Product[f[i, a], {i, 1, n}]]; t[n_, m_, a_] := c[n, a]/(c[m, a]*c[n - m, a]); Table[Flatten[Table[Table[t[n, m, a], {m, 0, n}], {n, 0, 10}]], {a, 1, 10}]
Formula
a=4; f(n,a)=(2*a+1)*f(n-1,a)+f(n-2,a);
c(n)=If[n == 0, 1, Product[f(i, a), {i, 1, n}]];
t(n,m)=c(n)/(c(m)*c(n-m)
Comments