A084594 - OEIS

%I #21 Feb 16 2025 08:32:49

%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://web.archive.org/web/2024*/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.

%H A. V. Aho and N. J. A. Sloane, <a href="/A000058/a000058.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)

%H Eric Weisstein's World of Mathematics, <a href="https://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,changed

%O 0,2

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