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 A051009 #68 Feb 16 2025 08:32:41
%S A051009 1,2,12,408,470832,627013566048,1111984844349868137938112,
%T A051009 3497379255757941172020851852070562919437964212608
%N A051009 Reduced denominators of Newton's iteration for sqrt(2).
%C A051009 (2^n)-th Pell numbers. - _Sergio Falcon_, Dec 04 2008
%C A051009 For n>1, Egyptian fraction expansion of 2-sqrt(2), i.e., 2-sqrt(2) = 1/2 + 1/12 + 1/408 + 1/470832 + ... - _Simon Plouffe_, Feb 22 2011
%H A051009 J. Conrad, <a href="/A051009/b051009.txt">Table of n, a(n) for n = 1..12</a>
%H A051009 Neil J. Calkin, Eunice Y. S. Chan and Robert M. Corless, <a href="https://arxiv.org/abs/2109.03765">Computational Discovery with Newton Fractals, Bohemian Matrices, & Mandelbrot Polynomials</a>, arXiv:2109.03765 [math.HO], 2021.
%H A051009 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/NewtonsIteration.html">Newton's Iteration</a>
%H A051009 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SquareRoot.html">Square Root</a>
%H A051009 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PythagorassConstant.html">Pythagoras's Constant</a>
%F A051009 a(n) = A000129(2^n).
%F A051009 a(n) = 2*a(n-1)*A001601(n-1). - Joe Keane (jgk(AT)jgk.org), May 31 2002
%F A051009 sqrt(2) = 1 + 1/2 - Sum_{n>=3} (1/a(n)). - _Donald S. McDonald_, Jan 21 2003
%F A051009 For n>1, a(n) = 2*a(n-1)*sqrt(2*a(n-1)^2+1). - Mario Catalani (mario.catalani(AT)unito.it), May 27 2003
%F A051009 For n>0: a(n) = Sum_{r=0..2^(n-1)-1} binomial(2^n, 2*r+1)*2^r. - Mario Catalani (mario.catalani(AT)unito.it), May 30 2003
%F A051009 For n>=4, a(n) = A098890(n-2) - A098890(n-3). - _Kieren MacMillan_, Dec 19 2007
%F A051009 a(n+1) = (1/(2*sqrt(2)))*((1 + sqrt(2))^(2^n) - (1 - sqrt(2))^(2^n)). - _Artur Jasinski_, Oct 10 2008
%F A051009 For n>0, a(n) = sqrt((A001601(n)^2-1)/2). - _Jose Hortal_, Apr 14 2012
%F A051009 a(1)=1, a(2)=2, a(n) = 2 * a(n-1) * cos(2^(n-3) * arccos(3)). - _Daniel Suteu_, Dec 01 2016
%F A051009 0 = a(n)^2*(2*a(n+1) + a(n+2)) - a(n+1)^3 if n>0. - _Michael Somos_, Dec 01 2016
%F A051009 a(n) = A001542(2^(n-2)). - _A.H.M. Smeets_, May 28 2017
%e A051009 G.f. = x + 2*x^2 + 12*x^3 + 408*x^4 + 470832*x^5 + ...
%t A051009 Table[Simplify[Expand[(1/(2 Sqrt[2])) ((1 + Sqrt[2])^(2^n) - (1 - Sqrt[2])^(2^n))]], {n, 0, 7}] (* _Artur Jasinski_, Oct 10 2008 *)
%t A051009 Do[Print[Fibonacci[2^n,2]],{n,0,10}] (* _Sergio Falcon_, Dec 04 2008 *)
%Y A051009 Cf. A000129, A001542, A001601, A051009, A098890.
%K A051009 nonn,frac
%O A051009 1,2
%A A051009 _Eric W. Weisstein_