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 A004539 Expansion of sqrt(2) in base 2. 21
 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1 (list; constant; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Bailey, Borwein, Crandall, & Pomerance prove a general result that the first n terms contain >> sqrt(x) 1s. Vandehey improves this to sqrt(2*n)(1 + o(1)). - Charles R Greathouse IV, Nov 07 2017 LINKS T. D. Noe, Table of n, a(n) for n = 1..1000 B. Adamczewski and N. Rampersad, On patterns occurring in binary algebraic numbers, Proc. Amer. Math. Soc. 136 (2008), 3105-3109. David H. Bailey, Jonathan M. Borwein, Richard E. Crandall, and Carl Pomerance, On the binary expansions of algebraic numbers, J. Théor. Nombres Bordeaux, 16 (2004), 487-518. R. L. Graham and H. O. Pollak, Note on a nonlinear recurrence related to sqrt(2), Mathematics Magazine, Volume 43, Pages 143-145, 1970. Zbl 201.04705. R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy] Richard Isaac, On the simple normality to base 2 of the square root of s, for s not a perfect square., arXiv:math/0512404 [math.NT], 2005-2006. Mariusz Iwaniuk, Formula for computing sqrt(2) of binary numbers Jason Kimberley, Index of expansions of sqrt(d) in base b Thomas Stoll, On a problem of Erdős and Graham concerning digits, Acta Arithmetica 125(2006), 89-100. Thomas Stoll, A fancy way to obtain the binary digits of 759250125 sqrt{2}, (2009), Amer. Math. Monthly, 117 (2010), 611-617. Joseph Vandehey, On the binary digits of sqrt(2), arXiv:1711.01722 [math.NT], 2017. Eric Weisstein's World of Mathematics, Wolfram's Iteration Eric Weisstein's World of Mathematics, Pythagoras's Constant FORMULA a(n) = 1/2 - (2*arctan(cot(2^(-(3/2)+n)*Pi)))/Pi + arctan(cot(2^(-(1/2)+n)*Pi))/Pi for n > 0 and n in Z. See link for a proof. - Mariusz Iwaniuk, Apr 20 2017 a(n) = floor(2^(-(1/2) + n)) - 2*floor(2^(-(3/2) + n)) for n > 0 and n in Z. See link for a proof. - Mariusz Iwaniuk, Apr 26 2017 EXAMPLE 1.0110101000001001111001... MATHEMATICA N[Sqrt, 200]; RealDigits[%, 2] PROG (bc) obase=2 scale=200 sqrt(2) (Haskell) a004539 n = a004539_list !! (n-1) a004539_list = w 2 0 where    w x r = bit : w (4 * (x - (4 * r + bit) * bit)) (2 * r + bit)      where bit = head (dropWhile (\b -> (4 * r + b) * b < x) [0..]) - 1 -- Reinhard Zumkeller, Dec 16 2013 (PARI) binary(sqrt(2)) \\ Michel Marcus, Nov 06 2017 CROSSREFS Cf. A002193 (decimal version), A233836 (run lengths of 0s and 1s). Sequence in context: A129372 A169591 A189295 * A294878 A023960 A129686 Adjacent sequences:  A004536 A004537 A004538 * A004540 A004541 A004542 KEYWORD nonn,base,cons AUTHOR STATUS approved

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Last modified November 16 20:13 EST 2019. Contains 329206 sequences. (Running on oeis4.)