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A004539
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Expansion of sqrt(2) in base 2.
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24
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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;
constant;
graph;
refs;
listen;
history;
text;
internal format)
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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(n) 1's. 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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FORMULA
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a(k) = floor(Sum_{n>=1} A005875(n)/exp(Pi*n/(2^((2/3)*k+(1/3))))) mod 2. Will give the k-th binary digit of sqrt(2). A005875 : number of ways to write n as sum of 3 squares. - Simon Plouffe, Dec 30 2023
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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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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