%I #9 May 16 2019 21:10:30
%S 6,7,8,5,7,1,4,3,0,7,0,0,1,6,5,8,9,2,7,6,5,5,4,3,9,5,5,4,3,6,1,4,0,1,
%T 3,2,8,3,0,0,6,8,9,6,3,8,7,8,1,1,3,6,4,1,8,8,0,8,7,5,9,8,1,9,0,0,3,7,
%U 8,9,9,4,2,7,2,5,7,1,0,3,5,7,4,4,8,8,6,9,7,8,5,1,0,3,4,6,1,8,3,5,9,1,8,8,0,5,8,8,2,0,3,2,2,3,7,5,0,7,5,2,2,9,9,0,9,5,5,1,2,5,5,0,6,8,0,0,1,3,5,6,2,2,9,1,9,1,1,1,0,3,9,1,8,8,1,7,2,8,1,4,2,0,5,9,3,2,6,7,0,1,0,8,7,2,3,1,5,6,5,5,9,1,8,1,0,5,1,5,4,9,2,4,4,8,6,4,0,9,5,9
%N Least real z > 2/3 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 ~ 728610523/1073741820 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 cross-references for other solutions.
%C The complement to this constant is given by A265271.
%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 = 0.6785714307001658927655439554361401328300689638781136418808759819003...
%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, 2, 8, 1, 599185, 2, 1, 1, 3, 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, 2/3, 17/25, 19/28, 11384532/16777205, 22769083/33554438, 34153615/50331643, 56922698/83886081, 204921709/301989886, 261844407/385875967, ...].
%e The partial quotients of the continued fraction of 2*z - 1/2 are as follows:
%e [0; 1, 6, 4793490, 8, ..., Q_n, ...]
%e where
%e Q_1 : 2^0*(2^(1*1) - 1)/(2^1 - 1) = 1;
%e Q_2 : 2^1*(2^(2*1) - 1)/(2^1 - 1) = 6;
%e Q_3 : 2^1*(2^(8*3) - 1)/(2^3 - 1) = 4793490;
%e Q_4 : 2^3*(2^(1*25) - 1)/(2^25 - 1) = 8;
%e Q_5 : 2^25*(2^(599185*28) - 1)/(2^28 - 1) ;
%e Q_6 : 2^28*(2^(2*16777205) - 1)/(2^16777205 - 1) ;
%e Q_7 : 2^16777205*(2^(1*33554438) - 1)/(2^33554438 - 1) ;
%e Q_8 : 2^33554438*(2^(1*50331643) - 1)/(2^50331643 - 1) ;
%e Q_9 : 2^50331643*(2^(3*83886081) - 1)/(2^83886081 - 1) ;
%e Q_10 : 2^83886081*(2^(1*301989886) - 1)/(2^301989886 - 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, A265275.
%K nonn,cons
%O 0,1
%A _Paul D. Hanna_, Dec 12 2015
|