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A324140 Define b(n) and c(n) by b(0)=0, b(1)=c(0)=c(1)=1; b(n)=c(n-1)*c(n-2), c(n) = (b(n-1)+c(n-1)*(b(n-2)+c(n-2)); sequence gives b(n). 1

%I

%S 0,1,1,2,12,144,6912,4478976,150923575296,3091996642249801728,

%T 2121885128576356504259996418048,

%U 30493397665311618727858440951711886636410099204096

%N Define b(n) and c(n) by b(0)=0, b(1)=c(0)=c(1)=1; b(n)=c(n-1)*c(n-2), c(n) = (b(n-1)+c(n-1)*(b(n-2)+c(n-2)); sequence gives b(n).

%H S. G. Wagner, <a href="https://www.fq.math.ca/Papers1/45-3/wagner.pdf">The Fibonacci number of Fibonacci trees and a related family of polynomial recurrence systems</a>, Fib. Q., 45 (2007), 247-253.

%p f:=proc(n) option remember;

%p if n=0 then [0,1];

%p elif n=1 then [1,1]

%p else

%p [f(n-1)[2]*f(n-2)[2],(f(n-1)[1]+f(n-1)[2])*(f(n-2)[1]+f(n-2)[2])];

%p fi; end;

%p [seq(f(n),n=0..11)];

%Y Cf. A324141.

%K nonn

%O 0,4

%A _N. J. A. Sloane_, Feb 20 2019

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Last modified February 3 11:32 EST 2023. Contains 360034 sequences. (Running on oeis4.)