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%I #17 May 14 2019 21:21:17
%S 3,9,5,1,6,1,2,9,0,3,2,2,5,2,1,9,6,8,0,8,3,7,5,6,5,7,8,6,0,4,1,6,1,8,
%T 3,1,7,0,5,3,4,7,2,5,4,7,6,1,9,8,3,1,4,3,5,1,4,1,4,7,4,9,2,4,1,0,9,8,
%U 7,0,0,5,5,5,2,1,2,8,0,9,5,8,0,4,2,4,0,9,9,8,3,6,0,8,9,8,1,3,3,9,0,0,5,2,8,7,5,7,0,6,8,0,0,4,9,0,0,3,3,7,0,7,8,6,8,3,0,6,7,1,4,5,4,7,8,9,0,7,2,7,9,5,5,1,1,7,0,5,0,9,5,0,4,5,4,1,1,8,3,4,7,5,9,7,2,7,2,0,2,5,6,6,0,9,3,9,8,0,9,3,8,3,5,2,6,7,3,5,3,3,4,6,4,1,6,0,6,0,6,8
%N Least real z > 1/3 such that 1/2 = Sum_{n>=1} {n*z} / 2^n, where {x} denotes the fractional part of x.
%C This constant is transcendental.
%C The rational approximation z ~ 33616604796619977479086259520427152017/85070591730234615865843651857942052860 is accurate to many thousands of digits.
%C This constant is one of 6 solutions to the equation 1/2 = Sum_{n>=1} {n*z}/2^n, where z is in the interval (0,1) - see cross-references for other solutions.
%C The complement to this constant is given by A265275.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DevilsStaircase.html">Devil's Staircase</a>.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F The constant z satisfies:
%F (1) 2*z - 1/2 = Sum_{n>=1} [n*z] / 2^n,
%F (2) 2*z - 1/2 = Sum_{n>=1} 1 / 2^[n/z],
%F (3) 3/2 - 2*z = Sum_{n>=1} 1 / 2^[n/(1-z)],
%F (4) 3/2 - 2*z = Sum_{n>=1} [n*(1-z)] / 2^n,
%F (5) 1/2 = Sum_{n>=1} {n*(1-z)} / 2^n,
%F where [x] denotes the integer floor function of x.
%e z = 0.395161290322521968083756578604161831705347254761983143514147492410987...
%e where z satisfies
%e (0) 1/2 = {z}/2 + {2*z}/2^2 + {3*z}/2^3 + {4*z}/2^4 + {5*z}/2^5 +...
%e (1) 2*z - 1/2 = [z]/2 + [2*z]/2^2 + [3*z]/2^3 + [4*z]/2^4 + [5*z]/2^5 +...
%e (2) 2*z - 1/2 = 1/2^[1/z] + 1/2^[2/z] + 1/2^[3/z] + 1/2^[4/z] + 1/2^[5/z] +...
%e The continued fraction of the constant z begins:
%e [0; 2, 1, 1, 7, 1, 1, 1, 1108378656, 2, 1, 1, 1, 3, 2, 1, 1, 1, 34359738367, 2, 1, 1, 1, 1099511627775, 2, 1, 2, ...]
%e (the next partial quotient has too many digits to show).
%e The convergents of the continued fraction of z begin:
%e [0/1, 1/2, 1/3, 2/5, 15/38, 17/43, 32/81, 49/124, 54310554176/137438953425, 108621108401/274877906974, 162931662577/412316860399, ...].
%e The partial quotients of the continued fraction of 2*z - 1/2 are as follows:
%e [0; 3, 2, 4, 8867029256, 32, 274877906944, 8796093022208, ..., Q_n, ...]
%e where
%e Q_1 = 2^0*(2^(2*1) - 1)/(2^1 - 1) = 3 ;
%e Q_2 = 2^1*(2^(1*2) - 1)/(2^2 - 1) = 2 ;
%e Q_3 = 2^2*(2^(1*3) - 1)/(2^3 - 1) = 4 ;
%e Q_4 = 2^3*(2^(7*5) - 1)/(2^5 - 1) = 8867029256 ;
%e Q_5 = 2^5*(2^(1*38) - 1)/(2^38 - 1) = 32 ;
%e Q_6 = 2^38*(2^(1*43) - 1)/(2^43 - 1) = 274877906944 ;
%e Q_7 = 2^43*(2^(1*81) - 1)/(2^81 - 1) = 8796093022208 ;
%e Q_8 = 2^81*(2^(1108378656*124) - 1)/(2^124 - 1) ;
%e Q_9 = 2^124*(2^(2*137438953425) - 1)/(2^137438953425 - 1) ;
%e Q_10 = 2^137438953425*(2^(1*274877906974) - 1)/(2^274877906974 - 1) ;...
%e These partial quotients can be calculated from the simple continued fraction of z and the denominators in the convergents of the continued fraction of z; see the Mathworld link entitled "Devil's Staircase" for more details.
%Y Cf. A265271, A265273, A265274, A265275, A265276.
%K nonn,cons
%O 0,1
%A _Paul D. Hanna_, Dec 09 2015
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