A277363 Self-convolution of a(n)/4^n gives fibonorials (A003266).

1, 2, 6, 52, 646, 13756, 458780, 24525352, 2094232006, 287618113900, 63647556127412, 22739228686869592, 13126310109506278556, 12250085882856201785816, 18488349380363585366790264, 45134497176992058331312333648, 178246891228174428563552421395782

Offset: 0

Author

Vladimir Reshetnikov, Oct 10 2016

Keywords

• nonn

Comments

Self-convolution of a(n) gives A003266(n)*4^n.

Crossrefs

Cf. A000045, A003266, A277362.

Programs

Maple

a:= proc(n) option remember; `if`(n=0, 1, (4^n
      *mul((<<0|1>, <1|1>>^i)[1, 2], i=1..n)-
      add(a(k)*a(n-k), k=1..n-1))/2)
    end:
seq(a(n), n=0...20);  # Alois P. Heinz, Oct 12 2016

Mathematica

With[{n = 20}, Sqrt[Sum[Fibonorial[k] (4 x)^k, {k, 0, n - 1}] + O[x]^n][[3]]] (* before version 10.0 define Fibonorial[n_] := Product[Fibonacci[k], {k, 1, n}] *)

Formula

Sum_{k=0..n} a(k)/4^k * a(n-k)/4^(n-k) = A003266(n).