

A265275


Least real z > 3/5 such that 1/2 = Sum_{n>=1} {n*z} / 2^n, where {x} denotes the fractional part of x.


5



6, 0, 4, 8, 3, 8, 7, 0, 9, 6, 7, 7, 4, 7, 8, 0, 3, 1, 9, 1, 6, 2, 4, 3, 4, 2, 1, 3, 9, 5, 8, 3, 8, 1, 6, 8, 2, 9, 4, 6, 5, 2, 7, 4, 5, 2, 3, 8, 0, 1, 6, 8, 5, 6, 4, 8, 5, 8, 5, 2, 5, 0, 7, 5, 8, 9, 0, 1, 2, 9, 9, 4, 4, 4, 7, 8, 7, 1, 9, 0, 4, 1, 9, 5, 7, 5, 9, 0, 0, 1, 6, 3, 9, 1, 0, 1, 8, 6, 6, 0, 9, 9, 4, 7, 1, 2, 4, 2, 9, 3, 1, 9, 9, 5, 0, 9, 9, 6, 6, 2, 9, 2, 1, 3, 1, 6, 9, 3, 2, 8, 5, 4, 5, 2, 1, 0, 9, 2, 7, 2, 0, 4, 4, 8, 8, 2, 9, 4, 9, 0, 4, 9, 5, 4, 5, 8, 8, 1, 6, 5, 2, 4, 0, 2, 7, 2, 7, 9, 7, 4, 3, 3, 9, 0, 6, 0, 1, 9, 0, 6, 1, 6, 4, 7, 3, 2, 6, 4, 6, 6, 5, 3, 5, 8, 3, 9, 3, 9, 3, 1
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OFFSET

0,1


COMMENTS

This constant is transcendental.
The rational approximation z ~ 51453986933614638386757392337514900843/85070591730234615865843651857942052860 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 crossreferences for other solutions.
The complement to this constant is given by A265272.


LINKS

Table of n, a(n) for n=0..199.
Eric Weisstein's World of Mathematics, Devil's Staircase.


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/(1z)],
(4) 3/2  2*z = Sum_{n>=1} [n*(1z)] / 2^n,
(5) 1/2 = Sum_{n>=1} {n*(1z)} / 2^n,
where [x] denotes the integer floor function of x.


EXAMPLE

z = 0.6048387096774780319162434213958381682946527452380168564858525075890...
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; 1, 1, 1, 1, 7, 1, 1, 1, 1108378656, 2, 1, 1, 1, 3, 2, 1, 1, 1, 34359738367, 2, 1, 1, 1, 1099511627775, 2, 1, 2, ...]
(the next partial quotient has too many digits to show).
The convergents of the continued fraction of z begin:
[0/1, 1/1, 1/2, 2/3, 3/5, 23/38, 26/43, 49/81, 75/124, 83128399249/137438953425, 166256798573/274877906974, 249385197822/412316860399, ...]
The partial quotients of the continued fraction of 2*z  1/2 are as follows:
[0; 1, 2, 2, 4, 8867029256, 32, 274877906944, 8796093022208, ..., Q_n, ...]
where
Q_1 : 2^0*(2^(1*1)  1)/(2^1  1) = 1;
Q_2 : 2^1*(2^(1*1)  1)/(2^1  1) = 2;
Q_3 : 2^1*(2^(1*2)  1)/(2^2  1) = 2;
Q_4 : 2^2*(2^(1*3)  1)/(2^3  1) = 4;
Q_5 : 2^3*(2^(7*5)  1)/(2^5  1) = 8867029256;
Q_6 : 2^5*(2^(1*38)  1)/(2^38  1) = 32;
Q_7 : 2^38*(2^(1*43)  1)/(2^43  1) = 274877906944;
Q_8 : 2^43*(2^(1*81)  1)/(2^81  1) = 8796093022208;
Q_9 : 2^81*(2^(1108378656*124)  1)/(2^124  1) ;
Q_10 : 2^124*(2^(2*137438953425)  1)/(2^137438953425  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, A265273, A265274, A265276.
Sequence in context: A132709 A197148 A196623 * A113024 A112280 A204850
Adjacent sequences: A265272 A265273 A265274 * A265276 A265277 A265278


KEYWORD

nonn,cons


AUTHOR

Paul D. Hanna, Dec 12 2015


STATUS

approved



