This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A084595 #29 Feb 16 2025 08:32:49
%S A084595 1,2,16,896,2781184,26794772135936,2487085750646543836443049984,
%T A084595 21427531469765285263614058238314319540132878612321796096
%N A084595 For n > 0: a(n) = Sum_{r=0..2^(n-1)-1} binomial(2^n, 2r+1)*3^r.
%C A084595 A084594(n)/a(n) converges to sqrt(3). Related to Newton's iteration.
%C A084595 a(n) is divisible by 2^n.
%H A084595 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 A084595 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 A084595 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/NewtonsIteration.html">Newton's Iteration</a>.
%F A084595 a(n) = ((1+sqrt(3))^(2^n) - (1-sqrt(3))^(2^n))/(2*sqrt(3)).
%F A084595 For n > 1:
%F A084595 a(n) = 2*a(n-1)*sqrt(3*a(n-1)^2 + A001146(n-1)).
%F A084595 a(n) = 2*a(n-1)*A084594(n-1).
%F A084595 a(n) = A002605(2^n).
%t A084595 For n>0: Table[Sum[Binomial[2^n, 2 r + 1]3^r, {r, 0, 2^(n - 1) - 1}], {n, 1, 8}]
%o A084595 (PARI) a(n) = if (n==0, 1, sum(r=0, 2^(n-1)-1, binomial(2^n, 2*r+1)*3^r)); \\ _Michel Marcus_, Sep 09 2019; corrected Jun 13 2022
%Y A084595 Cf. A001146, A002605, A084594.
%K A084595 easy,nonn
%O A084595 0,2
%A A084595 Mario Catalani (mario.catalani(AT)unito.it), May 31 2003