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%I #17 Apr 07 2021 02:49:57
%S 1,4,28,1552,4817152,46409906716672,4307758882900393634270543872,
%T 37113573186414494550922197215584520229965687291643953152
%N a(n) = Sum_{r=0..2^(n-1)} Binomial(2^n,2r)*3^r.
%C a(n)/A084595(n) converges to sqrt(3). Related to Newton's iteration.
%H A. V. Aho and N. J. A. Sloane, <a href="https://www.fq.math.ca/Scanned/11-4/aho-a.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437, <a href="http://neilsloane.com/doc/doubly.html">alternative link</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/NewtonsIteration.html">Newton's Iteration</a>.
%F a(n) = ( (1+sqrt(3))^(2^n) + (1-sqrt(3))^(2^n) )/2.
%F a(n) = A026150(2^n).
%F a(n) = 2*a(n-1)^2 - A001146(n-1), n>1.
%F a(n) = a(n-1)^2 + 3*A084595(n-1)^2.
%t Table[Sum[Binomial[2^n, 2 r]3^r, {r, 0, 2^(n - 1)}], {n, 0, 8}]
%t Table[Simplify[Expand[(1/2) ((1 + Sqrt[3])^(2^n) + (1 - Sqrt[3])^(2^n))]], {n, 0, 7}] (* _Artur Jasinski_, Oct 11 2008 *)
%Y Cf. A001146, A026150, A084595,
%K easy,nonn
%O 0,2
%A Mario Catalani (mario.catalani(AT)unito.it), May 31 2003