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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 c(n).
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%I #11 Jan 05 2025 19:51:41

%S 1,1,2,6,24,288,15552,9704448,318616436736,6659685075614957568,

%T 4578804751138453509192623849472,

%U 65342994997096325845410944896525471363735926865920

%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 c(n).

%H S. G. Wagner, <a href="https://web.archive.org/web/2024*/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. A324140.

%K nonn

%O 0,3

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