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Least positive real z such that 1/2 = Sum_{n>=1} {n*z} / 2^n, where {x} denotes the fractional part of x.
+10
5
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, 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, 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
COMMENTS
This constant is transcendental.
The rational approximation z ~ 345131297/1073741820 is accurate to over 5 million 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 A265276.
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 = 0.321428569299834107234456044563859867169931036121886358119124...
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; 3, 8, 1, 599185, 2, 1, 1, 3, 1, 2, ...]
(the next partial quotient has over 5 million digits).
The convergents of the continued fraction of z begin:
[0/1, 1/3, 8/25, 9/28, 5392673/16777205, 10785355/33554438, 16178028/50331643, 26963383/83886081, 97068177/301989886, 124031560/385875967, 345131297/1073741820, ...].
The partial quotients of the continued fraction of 2*z - 1/2 are as follows:
[0; 7, 4793490, 8, ..., Q_n, ...]
where
Q_1 = 2^0*(2^(3*1) - 1)/(2^1 - 1) = 7 ;
Q_2 = 2^1*(2^(8*3) - 1)/(2^3 - 1) = 4793490 ;
Q_3 = 2^3*(2^(1*25) - 1)/(2^25 - 1) = 8 ;
Q_4 = 2^25*(2^(599185*28) - 1)/(2^28 - 1) ;
Q_5 = 2^28*(2^(2*16777205) - 1)/(2^16777205 - 1) = 2^28*(2^16777205 + 1) ;
Q_6 = 2^16777205*(2^(1*33554438) - 1)/(2^33554438 - 1) = 2^16777205 ;
Q_7 = 2^33554438*(2^(1*50331643) - 1)/(2^50331643 - 1) = 2^33554438 ;
Q_8 = 2^50331643*(2^(3*83886081) - 1)/(2^83886081 - 1) ;
Q_9 = 2^83886081*(2^(1*301989886) - 1)/(2^301989886 - 1) ;
Q_10 = 2^301989886*(2^(2*385875967) - 1)/(2^385875967 - 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.
Least real z > 1/3 such that 1/2 = Sum_{n>=1} {n*z} / 2^n, where {x} denotes the fractional part of x.
+10
5
3, 9, 5, 1, 6, 1, 2, 9, 0, 3, 2, 2, 5, 2, 1, 9, 6, 8, 0, 8, 3, 7, 5, 6, 5, 7, 8, 6, 0, 4, 1, 6, 1, 8, 3, 1, 7, 0, 5, 3, 4, 7, 2, 5, 4, 7, 6, 1, 9, 8, 3, 1, 4, 3, 5, 1, 4, 1, 4, 7, 4, 9, 2, 4, 1, 0, 9, 8, 7, 0, 0, 5, 5, 5, 2, 1, 2, 8, 0, 9, 5, 8, 0, 4, 2, 4, 0, 9, 9, 8, 3, 6, 0, 8, 9, 8, 1, 3, 3, 9, 0, 0, 5, 2, 8, 7, 5, 7, 0, 6, 8, 0, 0, 4, 9, 0, 0, 3, 3, 7, 0, 7, 8, 6, 8, 3, 0, 6, 7, 1, 4, 5, 4, 7, 8, 9, 0, 7, 2, 7, 9, 5, 5, 1, 1, 7, 0, 5, 0, 9, 5, 0, 4, 5, 4, 1, 1, 8, 3, 4, 7, 5, 9, 7, 2, 7, 2, 0, 2, 5, 6, 6, 0, 9, 3, 9, 8, 0, 9, 3, 8, 3, 5, 2, 6, 7, 3, 5, 3, 3, 4, 6, 4, 1, 6, 0, 6, 0, 6, 8
COMMENTS
This constant is transcendental.
The rational approximation z ~ 33616604796619977479086259520427152017/85070591730234615865843651857942052860 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 A265275.
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 = 0.395161290322521968083756578604161831705347254761983143514147492410987...
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, 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, 1/2, 1/3, 2/5, 15/38, 17/43, 32/81, 49/124, 54310554176/137438953425, 108621108401/274877906974, 162931662577/412316860399, ...].
The partial quotients of the continued fraction of 2*z - 1/2 are as follows:
[0; 3, 2, 4, 8867029256, 32, 274877906944, 8796093022208, ..., Q_n, ...]
where
Q_1 = 2^0*(2^(2*1) - 1)/(2^1 - 1) = 3 ;
Q_2 = 2^1*(2^(1*2) - 1)/(2^2 - 1) = 2 ;
Q_3 = 2^2*(2^(1*3) - 1)/(2^3 - 1) = 4 ;
Q_4 = 2^3*(2^(7*5) - 1)/(2^5 - 1) = 8867029256 ;
Q_5 = 2^5*(2^(1*38) - 1)/(2^38 - 1) = 32 ;
Q_6 = 2^38*(2^(1*43) - 1)/(2^43 - 1) = 274877906944 ;
Q_7 = 2^43*(2^(1*81) - 1)/(2^81 - 1) = 8796093022208 ;
Q_8 = 2^81*(2^(1108378656*124) - 1)/(2^124 - 1) ;
Q_9 = 2^124*(2^(2*137438953425) - 1)/(2^137438953425 - 1) ;
Q_10 = 2^137438953425*(2^(1*274877906974) - 1)/(2^274877906974 - 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.
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.
+10
5
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
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.
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.
Least real z > 3/5 such that 1/2 = Sum_{n>=1} {n*z} / 2^n, where {x} denotes the fractional part of x.
+10
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
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 cross-references for other solutions.
The complement to this constant is given by A265272.
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 = 0.6048387096774780319162434213958381682946527452380168564858525075890...
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.
Least real z > 2/3 such that 1/2 = Sum_{n>=1} {n*z} / 2^n, where {x} denotes the fractional part of x.
+10
5
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, 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, 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
COMMENTS
This constant is transcendental.
The rational approximation z ~ 728610523/1073741820 is accurate to over 5 million 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 A265271.
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 = 0.6785714307001658927655439554361401328300689638781136418808759819003...
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, 2, 8, 1, 599185, 2, 1, 1, 3, 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, 2/3, 17/25, 19/28, 11384532/16777205, 22769083/33554438, 34153615/50331643, 56922698/83886081, 204921709/301989886, 261844407/385875967, ...].
The partial quotients of the continued fraction of 2*z - 1/2 are as follows:
[0; 1, 6, 4793490, 8, ..., Q_n, ...]
where
Q_1 : 2^0*(2^(1*1) - 1)/(2^1 - 1) = 1;
Q_2 : 2^1*(2^(2*1) - 1)/(2^1 - 1) = 6;
Q_3 : 2^1*(2^(8*3) - 1)/(2^3 - 1) = 4793490;
Q_4 : 2^3*(2^(1*25) - 1)/(2^25 - 1) = 8;
Q_5 : 2^25*(2^(599185*28) - 1)/(2^28 - 1) ;
Q_6 : 2^28*(2^(2*16777205) - 1)/(2^16777205 - 1) ;
Q_7 : 2^16777205*(2^(1*33554438) - 1)/(2^33554438 - 1) ;
Q_8 : 2^33554438*(2^(1*50331643) - 1)/(2^50331643 - 1) ;
Q_9 : 2^50331643*(2^(3*83886081) - 1)/(2^83886081 - 1) ;
Q_10 : 2^83886081*(2^(1*301989886) - 1)/(2^301989886 - 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.
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