A051009 Reduced denominators of Newton's iteration for sqrt(2).
1, 2, 12, 408, 470832, 627013566048, 1111984844349868137938112, 3497379255757941172020851852070562919437964212608
Offset: 1
Examples
G.f. = x + 2*x^2 + 12*x^3 + 408*x^4 + 470832*x^5 + ...
Links
- J. Conrad, Table of n, a(n) for n = 1..12
- Neil J. Calkin, Eunice Y. S. Chan and Robert M. Corless, Computational Discovery with Newton Fractals, Bohemian Matrices, & Mandelbrot Polynomials, arXiv:2109.03765 [math.HO], 2021.
- Eric Weisstein's World of Mathematics, Newton's Iteration
- Eric Weisstein's World of Mathematics, Square Root
- Eric Weisstein's World of Mathematics, Pythagoras's Constant
Programs
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Mathematica
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 *) Do[Print[Fibonacci[2^n,2]],{n,0,10}] (* Sergio Falcon, Dec 04 2008 *)
Formula
a(n) = A000129(2^n).
a(n) = 2*a(n-1)*A001601(n-1). - Joe Keane (jgk(AT)jgk.org), May 31 2002
sqrt(2) = 1 + 1/2 - Sum_{n>=3} (1/a(n)). - Donald S. McDonald, Jan 21 2003
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
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
a(n+1) = (1/(2*sqrt(2)))*((1 + sqrt(2))^(2^n) - (1 - sqrt(2))^(2^n)). - Artur Jasinski, Oct 10 2008
For n>0, a(n) = sqrt((A001601(n)^2-1)/2). - Jose Hortal, Apr 14 2012
a(1)=1, a(2)=2, a(n) = 2 * a(n-1) * cos(2^(n-3) * arccos(3)). - Daniel Suteu, Dec 01 2016
0 = a(n)^2*(2*a(n+1) + a(n+2)) - a(n+1)^3 if n>0. - Michael Somos, Dec 01 2016
a(n) = A001542(2^(n-2)). - A.H.M. Smeets, May 28 2017
Comments