A258075 Constant x that satisfies: x = Sum_{n>=1} frac(n*(1-x)) / 2^n.

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

Offset

1,2

Comments

A good approximation to this constant is 680/1533, which is correct to 39 digits.

Links

Table of n, a(n) for n=1..152.

Eric Weisstein, Devil's Staircase from MathWorld.

Formula

This constant satisfies:

(1) 1 = x + Sum_{n>=1} {n*x} / 2^n, where {z} denotes the fractional part of z.

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

(3) 1 = 3*x - Sum_{n>=1} 1 / 2^[n/x], a "devil's staircase" sum.

(4) 2 = 3*x + Sum_{n>=1} 1 / 2^[n/(1-x)], a "devil's staircase" sum.

Example

x = 0.4435746901500326157860404435746901500325851143026...
where x = Sum_{n>=1} {n*(1-x)} / 2^n such that x < 1/2 and x > 0.
Other series involving x begin:
(a) 3*x-1 = 0/2 + 0/2^2 + 1/2^3 + 1/2^4 + 2/2^5 + 2/2^6 + 3/2^7 + 3/2^8 + 3/2^9 + 4/2^10 + 4/2^11 + 5/2^12 + 5/2^13 + 6/2^14 +...+ [n*x]/2^n +...
(b) 2-3*x = 0/2 + 1/2^2 + 1/2^3 + 2/2^4 + 2/2^5 + 3/2^6 + 3/2^7 + 4/2^8 + 5/2^9 + 5/2^10 + 6/2^11 + 6/2^12 + 7/2^13 + 7/2^14 +...+ [n*(1-x)]/2^n +...
(c) 3*x-1 = 1/2^2 + 1/2^4 + 1/2^6 + 1/2^9 + 1/2^11 + 1/2^13 + 1/2^15 + 1/2^18 + 1/2^20 + 1/2^22 + 1/2^24 + 1/2^27 + 1/2^29 +...+ 1/2^[n/x] +...
(d) 2-3*x = 1/2^1 + 1/2^3 + 1/2^5 + 1/2^7 + 1/2^8 + 1/2^10 + 1/2^12 + 1/2^14 + 1/2^16 + 1/2^17 + 1/2^19 + 1/2^21 + 1/2^23 +...+ 1/2^[n/(1-x)] +...
note that (c) and (d) involve Beatty sequences as exponents of 1/2.
The complement to this constant is A258072:
1-x = 0.5564253098499673842139595564253098499674148856973...
and possesses very similar properties.
The CONTINUED FRACTION of 3*x has large partial quotients:
3*x = [1, 3, 42, 4, 41619663273108911871743469597819008, 10889035741470030830827987437816582767104, ...];
the number of digits of the partial quotients begin:
[1, 1, 2, 1, 35, 41, 115, 270, ...].
The initial 1050 digits are:
x = 0.44357469015003261578604044357469015003258511430265\
30388125774628431253164859002234302537588527824859\
91958242927156950433109781361196018310495546232260\
21308756488922508493146400309145266931418310463207\
21682655581705538047690626270645407928462460161012\
59348074253641658104334540972276437282369203087870\
91688278965124501135824721922196014571344903077739\
18500944407821228291964676297532896809047816970961\
57232314116611612205997186868899794801384103713379\
27093301379877141640033100637461650643685872664216\
71210915983869597104515647764570759665860825582616\
02202386683742093208255873853725828394394746230649\
17646005634235224736778439225789268521850715113549\
81826246824173592162652534369620590399778082914397\
59525545391741573753969070145840793634947962855349\
40161700909555769785319066569343646229986269080724\
03331326445769921119003906381638002817964787631913\
86332217342163925996090891887058572086607050246768\
90125594710389427906358536708700202340858220467421\
83637059019834780162093855868315000246318027150733\
19542694230313602926455823852338591379348725144367...

Prog

(PARI) {x=.4; for(i=1, 100, x = (x + sum(n=1, 4000, frac(n*(1-x))/2^n*1.))/2); x}
(PARI) /* Series 2-3*x = Sum_{n>=1} 1 / 2^[n/(1-x)] gives faster convergence: */
{x=0.4; for(i=1, 10, x = (2 - sum(n=1, 2000, 1./2^floor(n/(1-x))))/3 ); x}

Crossrefs

Cf. A258072.

Sequence in context: A073321 A055620 A072420 * A286296 A023530 A337365

Adjacent sequences: A258072 A258073 A258074 * A258076 A258077 A258078

Keyword

nonn,cons

Author

Paul D. Hanna, May 21 2015

Status

approved