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A004539
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Expansion of sqrt(2) in base 2.
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21
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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;
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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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
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FORMULA
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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
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EXAMPLE
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1.0110101000001001111001...
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MATHEMATICA
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N[Sqrt[2], 200]; RealDigits[%, 2]
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PROG
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(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
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CROSSREFS
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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
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KEYWORD
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nonn,base,cons
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AUTHOR
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N. J. A. Sloane
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STATUS
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approved
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