A265271 - OEIS

%I #25 May 14 2019 21:21:24

%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 cross-references 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/(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 = 0.321428569299834107234456044563859867169931036121886358119124...

%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