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

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

Offset

0,2

Comments

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

Links

Table of n, a(n) for n=0..7.

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.

Formula

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.

Mathematica

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 *)

Crossrefs

Cf. A001146, A026150, A084595,

Sequence in context: A081792 A336622 A354842 * A334598 A000838 A218174

Adjacent sequences: A084591 A084592 A084593 * A084595 A084596 A084597

Keyword

easy,nonn,changed

Author

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

Status

approved