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.

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, 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, 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

Offset

0,1

Comments

This constant is transcendental.

The rational approximation z ~ 34988721583274868054335940408411507283/85070591730234615865843651857942052860 is accurate to many thousands of digits.

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.

The complement to this constant is given by A265274.

Links

Table of n, a(n) for n=0..199.

Eric Weisstein's World of Mathematics, Devil's Staircase.

Index entries for transcendental numbers

Formula

The constant z satisfies:

(1) 2*z - 1/2 = Sum_{n>=1} [n*z] / 2^n,

(2) 2*z - 1/2 = Sum_{n>=1} 1 / 2^[n/z],

(3) 3/2 - 2*z = Sum_{n>=1} 1 / 2^[n/(1-z)],

(4) 3/2 - 2*z = Sum_{n>=1} [n*(1-z)] / 2^n,

(5) 1/2 = Sum_{n>=1} {n*(1-z)} / 2^n,

where [x] denotes the integer floor function of x.

Example

z = 0.411290445636334505725904280885932052093903124948409515104407844796...
where z satisfies
(0) 1/2 = {z}/2 + {2*z}/2^2 + {3*z}/2^3 + {4*z}/2^4 + {5*z}/2^5 +...
(1) 2*z - 1/2 = [z]/2 + [2*z]/2^2 + [3*z]/2^3 + [4*z]/2^4 + [5*z]/2^5 +...
(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] +...
The continued fraction of the constant z begins:
[0; 2, 2, 3, 7, 528, 2, 1, 1, 1, 20282564347337181724466999721987, 2, 1, 2, ...]
(the next partial quotient has too many digits to show).
The convergents of the continued fraction of z begin:
[0/1, 1/2, 2/5, 7/17, 51/124, 26935/65489, 53921/131102, 80856/196591, 134777/327693, 215633/524284, 4373590197909358506791992551051357548/10633823966279326983230456482242560001, ...]
The partial quotients of the continued fraction of 2*z - 1/2 are as follows:
[0; 3, 10, 4228, 162260514778697453795735997775904, ..., Q_n, ...]
where
Q_1 = 2^0*(2^(2*1) - 1)/(2^1 - 1) = 3 ;
Q_2 = 2^1*(2^(2*2) - 1)/(2^2 - 1) = 10 ;
Q_3 = 2^2*(2^(3*5) - 1)/(2^5 - 1) = 4228 ;
Q_4 = 2^5*(2^(7*17) - 1)/(2^17 - 1) = 162260514778697453795735997775904 ;
Q_5 = 2^17*(2^(528*124) - 1)/(2^124 - 1) ;
Q_6 = 2^124*(2^(2*65489) - 1)/(2^65489 - 1) ;
Q_7 = 2^65489*(2^(1*131102) - 1)/(2^131102 - 1) ;
Q_8 = 2^131102*(2^(1*196591) - 1)/(2^196591 - 1) ;
Q_9 = 2^196591*(2^(1*327693) - 1)/(2^327693 - 1) ;
Q_10 = 2^327693*(2^(20282564347337181724466999721987*524284) - 1)/(2^524284 - 1) ; ...
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.

Crossrefs

Cf. A265271, A265272, A265274, A265275, A265276.

Sequence in context: A016527 A010325 A355694 * A341932 A293770 A111311

Adjacent sequences: A265270 A265271 A265272 * A265274 A265275 A265276

Keyword

nonn,cons

Author

Paul D. Hanna, Dec 09 2015

Status

approved