%I
%S 3,2,1,4,2,8,5,6,9,2,9,9,8,3,4,1,0,7,2,3,4,4,5,6,0,4,4,5,6,3,8,5,9,8,
%T 6,7,1,6,9,9,3,1,0,3,6,1,2,1,8,8,6,3,5,8,1,1,9,1,2,4,0,1,8,0,9,9,6,2,
%U 1,0,0,5,7,2,7,4,2,8,9,6,4,2,5,5,1,1,3,0,2,1,4,8,9,6,5,3,8,1,6,4,0,8,1,1,9,4,1,1,7,9,6,7,7,6,2,4,9,2,4,7,7,0,0,9,0,4,4,8,7,4,4,9,3,1,9,9,8,6,4,3,7,7,0,8,0,8,8,8,9,6,0,8,1,1,8,2,7,1,8,5,7,9,4,0,6,7,3,2,9,8,9,1,2,7,6,8,4,3,4,4,0,8,1,8,9,4,8,4,5,0,7,5,5,1,3,5,9,0,4,0
%N Least positive real z 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 ~ 345131297/1073741820 is accurate to over 5 million 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 crossreferences for other solutions.
%C The complement to this constant is given by A265276.
%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/(1z)],
%F (4) 3/2  2*z = Sum_{n>=1} [n*(1z)] / 2^n,
%F (5) 1/2 = Sum_{n>=1} {n*(1z)} / 2^n,
%F where [x] denotes the integer floor function of x.
%e z = 0.321428569299834107234456044563859867169931036121886358119124...
%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; 3, 8, 1, 599185, 2, 1, 1, 3, 1, 2, ...]
%e (the next partial quotient has over 5 million digits).
%e The convergents of the continued fraction of z begin:
%e [0/1, 1/3, 8/25, 9/28, 5392673/16777205, 10785355/33554438, 16178028/50331643, 26963383/83886081, 97068177/301989886, 124031560/385875967, 345131297/1073741820, ...].
%e The partial quotients of the continued fraction of 2*z  1/2 are as follows:
%e [0; 7, 4793490, 8, ..., Q_n, ...]
%e where
%e Q_1 = 2^0*(2^(3*1)  1)/(2^1  1) = 7 ;
%e Q_2 = 2^1*(2^(8*3)  1)/(2^3  1) = 4793490 ;
%e Q_3 = 2^3*(2^(1*25)  1)/(2^25  1) = 8 ;
%e Q_4 = 2^25*(2^(599185*28)  1)/(2^28  1) ;
%e Q_5 = 2^28*(2^(2*16777205)  1)/(2^16777205  1) = 2^28*(2^16777205 + 1) ;
%e Q_6 = 2^16777205*(2^(1*33554438)  1)/(2^33554438  1) = 2^16777205 ;
%e Q_7 = 2^33554438*(2^(1*50331643)  1)/(2^50331643  1) = 2^33554438 ;
%e Q_8 = 2^50331643*(2^(3*83886081)  1)/(2^83886081  1) ;
%e Q_9 = 2^83886081*(2^(1*301989886)  1)/(2^301989886  1) ;
%e Q_10 = 2^301989886*(2^(2*385875967)  1)/(2^385875967  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. A265272, A265273, A265274, A265275, A265276.
%K nonn,cons
%O 0,1
%A _Paul D. Hanna_, Dec 08 2015
