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%I #13 May 14 2019 21:25:08
%S 5,8,8,7,0,9,5,5,4,3,6,3,6,6,5,4,9,4,2,7,4,0,9,5,7,1,9,1,1,4,0,6,7,9,
%T 4,7,9,0,6,0,9,6,8,7,5,0,5,1,5,9,0,4,8,4,8,9,5,5,9,2,1,5,5,2,0,3,9,0,
%U 2,8,0,4,1,6,6,5,4,7,5,7,7,1,0,5,0,8,5,8,7,3,2,5,8,3,0,5,3,6,2,9,2,9,1,5,4,1,5,3,1,4,4,4,5,0,9,7,8,8,5,3,9,6,8,8,6,0,4,4,4,4,8,0,5,7,4,4,6,4,8,7,8,3,2,6,1,3,6,1,3,4,4,8,1,6,6,4,0,0,1,3,9,2,7,7,9,7,2,3,0,6,3,8,4,6,2,1,0,0,0,7,5,0,2,0,8,2,3,1,0,7,9,2,0,6,2,6,7,0,5,6
%N Least real z > 1/2 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 ~ 50081870146959747811507711449530545577/85070591730234615865843651857942052860 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 A265273.
%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 = 0.58870955436366549427409571911406794790609687505159048489559215520...
%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, 2, 3, 7, 528, 2, 1, 1, 1, 20282564347337181724466999721987, 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, 3/5, 10/17, 73/124, 38554/65489, 77181/131102, 115735/196591, 192916/327693, 308651/524284, 6260233768369968476438463931191202453/10633823966279326983230456482242560001, ...]
%e The partial quotients of the continued fraction of 2*z - 1/2 are as follows:
%e [0; 1, 2, 10, 4228, 162260514778697453795735997775904, ..., 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^(2*2) - 1)/(2^2 - 1) = 10;
%e Q_4 : 2^2*(2^(3*5) - 1)/(2^5 - 1) = 4228;
%e Q_5 : 2^5*(2^(7*17) - 1)/(2^17 - 1) = 162260514778697453795735997775904;
%e Q_6 : 2^17*(2^(528*124) - 1)/(2^124 - 1) ;
%e Q_7 : 2^124*(2^(2*65489) - 1)/(2^65489 - 1) ;
%e Q_8 : 2^65489*(2^(1*131102) - 1)/(2^131102 - 1) ;
%e Q_9 : 2^131102*(2^(1*196591) - 1)/(2^196591 - 1) ;
%e Q_10 : 2^196591*(2^(1*327693) - 1)/(2^327693 - 1) ;
%e Q_10 = 2^327693*(2^(20282564347337181724466999721987*524284) - 1)/(2^524284 - 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, A265275, A265276.
%K nonn,cons
%O 0,1
%A _Paul D. Hanna_, Dec 12 2015
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