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Self-convolution of a(n)/4^n gives fibonorials (A003266).
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%I #8 Oct 12 2016 18:55:13

%S 1,2,6,52,646,13756,458780,24525352,2094232006,287618113900,

%T 63647556127412,22739228686869592,13126310109506278556,

%U 12250085882856201785816,18488349380363585366790264,45134497176992058331312333648,178246891228174428563552421395782

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

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

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

%p a:= proc(n) option remember; `if`(n=0, 1, (4^n

%p *mul((<<0|1>, <1|1>>^i)[1, 2], i=1..n)-

%p add(a(k)*a(n-k), k=1..n-1))/2)

%p end:

%p seq(a(n), n=0...20); # _Alois P. Heinz_, Oct 12 2016

%t 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}] *)

%Y Cf. A000045, A003266, A277362.

%K nonn

%O 0,2

%A _Vladimir Reshetnikov_, Oct 10 2016