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 A265273 Least positive real z > 2/5 such that 1/2 = Sum_{n>=1} {n*z} / 2^n, where {x} denotes the fractional part of x. 5

%I

%S 4,1,1,2,9,0,4,4,5,6,3,6,3,3,4,5,0,5,7,2,5,9,0,4,2,8,0,8,8,5,9,3,2,0,

%T 5,2,0,9,3,9,0,3,1,2,4,9,4,8,4,0,9,5,1,5,1,0,4,4,0,7,8,4,4,7,9,6,0,9,

%U 7,1,9,5,8,3,3,4,5,2,4,2,2,8,9,4,9,1,4,1,2,6,7,4,1,6,9,4,6,3,7,0,7,0,8,4,5,8,4,6,8,5,5,5,4,9,0,2,1,1,4,6,0,3,1,1,3,9,5,5,5,5,1,9,4,2,5,5,3,5,1,2,1,6,7,3,8,6,3,8,6,5,5,1,8,3,3,5,9,9,8,6,0,7,2,2,0,2,7,6,9,3,6,1,5,3,7,8,9,9,9,2,4,9,7,9,1,7,6,8,9,2,0,7,9,3,7,3,2,9,4,3

%N Least positive real z > 2/5 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 ~ 34988721583274868054335940408411507283/85070591730234615865843651857942052860 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 A265274.

%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 = 0.411290445636334505725904280885932052093903124948409515104407844796...

%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; 2, 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, 1/2, 2/5, 7/17, 51/124, 26935/65489, 53921/131102, 80856/196591, 134777/327693, 215633/524284, 4373590197909358506791992551051357548/10633823966279326983230456482242560001, ...]

%e The partial quotients of the continued fraction of 2*z - 1/2 are as follows:

%e [0; 3, 10, 4228, 162260514778697453795735997775904, ..., Q_n, ...]

%e where

%e Q_1 = 2^0*(2^(2*1) - 1)/(2^1 - 1) = 3 ;

%e Q_2 = 2^1*(2^(2*2) - 1)/(2^2 - 1) = 10 ;

%e Q_3 = 2^2*(2^(3*5) - 1)/(2^5 - 1) = 4228 ;

%e Q_4 = 2^5*(2^(7*17) - 1)/(2^17 - 1) = 162260514778697453795735997775904 ;

%e Q_5 = 2^17*(2^(528*124) - 1)/(2^124 - 1) ;

%e Q_6 = 2^124*(2^(2*65489) - 1)/(2^65489 - 1) ;

%e Q_7 = 2^65489*(2^(1*131102) - 1)/(2^131102 - 1) ;

%e Q_8 = 2^131102*(2^(1*196591) - 1)/(2^196591 - 1) ;

%e Q_9 = 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, A265274, A265275, A265276.

%K nonn,cons

%O 0,1

%A _Paul D. Hanna_, Dec 09 2015

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Last modified December 13 01:23 EST 2019. Contains 329963 sequences. (Running on oeis4.)