A084595 -id:A084595 - cp's OEIS frontend

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A084594 a(n) = Sum_{r=0..2^(n-1)} Binomial(2^n,2r)*3^r.

Original entry on oeis.org

1, 4, 28, 1552, 4817152, 46409906716672, 4307758882900393634270543872, 37113573186414494550922197215584520229965687291643953152

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), May 31 2003

a(n)/A084595(n) converges to sqrt(3). Related to Newton's iteration.

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Eric Weisstein's World of Mathematics, Newton's Iteration.

Cf. A001146, A026150, A084595,

Table[Sum[Binomial[2^n, 2 r]3^r, {r, 0, 2^(n - 1)}], {n, 0, 8}]
Table[Simplify[Expand[(1/2) ((1 + Sqrt[3])^(2^n) + (1 - Sqrt[3])^(2^n))]], {n, 0, 7}] (* Artur Jasinski, Oct 11 2008 *)

a(n) = ( (1+sqrt(3))^(2^n) + (1-sqrt(3))^(2^n) )/2.
a(n) = A026150(2^n).
a(n) = 2*a(n-1)^2 - A001146(n-1), n>1.
a(n) = a(n-1)^2 + 3*A084595(n-1)^2.

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