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a(n) = 4 * a(n-1) * (3^(2^(n-1))-a(n-1)) with a(0)=1.
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%I #2 Mar 30 2012 18:51:39

%S 1,8,32,6272,7250432,1038154236987392,3383826162019367796397224108032,

%T 674838593766753484487654913831820720085359667709963001167872

%N a(n) = 4 * a(n-1) * (3^(2^(n-1))-a(n-1)) with a(0)=1.

%C a(n) is the numerator of b(n)=a(n)/3^(2^n)=a(n)/A011764(n) which is a logistic chaotic sequence of reals in (0,1) with b(n)=4*b(n-1)*(1-b(n-1)) starting at b(0)=1/3; the truncated values of b(n) start: 0.333..., 0.888..., 0.395..., 0.955..., 0.168..., 0.560..., 0.985..., 0.057..., 0.215..., etc.

%K frac,nonn

%O 0,2

%A _Henry Bottomley_, Feb 06 2003