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%I #12 May 14 2019 21:25:05
%S 6,0,4,8,3,8,7,0,9,6,7,7,4,7,8,0,3,1,9,1,6,2,4,3,4,2,1,3,9,5,8,3,8,1,
%T 6,8,2,9,4,6,5,2,7,4,5,2,3,8,0,1,6,8,5,6,4,8,5,8,5,2,5,0,7,5,8,9,0,1,
%U 2,9,9,4,4,4,7,8,7,1,9,0,4,1,9,5,7,5,9,0,0,1,6,3,9,1,0,1,8,6,6,0,9,9,4,7,1,2,4,2,9,3,1,9,9,5,0,9,9,6,6,2,9,2,1,3,1,6,9,3,2,8,5,4,5,2,1,0,9,2,7,2,0,4,4,8,8,2,9,4,9,0,4,9,5,4,5,8,8,1,6,5,2,4,0,2,7,2,7,9,7,4,3,3,9,0,6,0,1,9,0,6,1,6,4,7,3,2,6,4,6,6,5,3,5,8,3,9,3,9,3,1
%N Least real z > 3/5 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 ~ 51453986933614638386757392337514900843/85070591730234615865843651857942052860 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 A265272.
%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 = 0.6048387096774780319162434213958381682946527452380168564858525075890...
%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; 1, 1, 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, 1/1, 1/2, 2/3, 3/5, 23/38, 26/43, 49/81, 75/124, 83128399249/137438953425, 166256798573/274877906974, 249385197822/412316860399, ...]
%e The partial quotients of the continued fraction of 2*z - 1/2 are as follows:
%e [0; 1, 2, 2, 4, 8867029256, 32, 274877906944, 8796093022208, ..., Q_n, ...]
%e where
%e Q_1 : 2^0*(2^(1*1) - 1)/(2^1 - 1) = 1;
%e Q_2 : 2^1*(2^(1*1) - 1)/(2^1 - 1) = 2;
%e Q_3 : 2^1*(2^(1*2) - 1)/(2^2 - 1) = 2;
%e Q_4 : 2^2*(2^(1*3) - 1)/(2^3 - 1) = 4;
%e Q_5 : 2^3*(2^(7*5) - 1)/(2^5 - 1) = 8867029256;
%e Q_6 : 2^5*(2^(1*38) - 1)/(2^38 - 1) = 32;
%e Q_7 : 2^38*(2^(1*43) - 1)/(2^43 - 1) = 274877906944;
%e Q_8 : 2^43*(2^(1*81) - 1)/(2^81 - 1) = 8796093022208;
%e Q_9 : 2^81*(2^(1108378656*124) - 1)/(2^124 - 1) ;
%e Q_10 : 2^124*(2^(2*137438953425) - 1)/(2^137438953425 - 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, A265272, A265273, A265274, A265276.
%K nonn,cons
%O 0,1
%A _Paul D. Hanna_, Dec 12 2015
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