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%I #17 May 24 2015 22:05:54
%S 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,
%T 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,
%U 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
%N Constant x that satisfies: x = Sum_{n>=1} frac(n*(1-x)) / 2^n.
%C A good approximation to this constant is 680/1533, which is correct to 39 digits.
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/DevilsStaircase.html">Devil's Staircase</a> from MathWorld.
%F This constant satisfies:
%F (1) 1 = x + Sum_{n>=1} {n*x} / 2^n, where {z} denotes the fractional part of z.
%F (2) 1 = 3*x - Sum_{n>=1} [n*x] / 2^n, where [z] denotes the integer floor of z.
%F (3) 1 = 3*x - Sum_{n>=1} 1 / 2^[n/x], a "devil's staircase" sum.
%F (4) 2 = 3*x + Sum_{n>=1} 1 / 2^[n/(1-x)], a "devil's staircase" sum.
%e x = 0.4435746901500326157860404435746901500325851143026...
%e where x = Sum_{n>=1} {n*(1-x)} / 2^n such that x < 1/2 and x > 0.
%e Other series involving x begin:
%e (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 +...
%e (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 +...
%e (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] +...
%e (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)] +...
%e note that (c) and (d) involve Beatty sequences as exponents of 1/2.
%e The complement to this constant is A258072:
%e 1-x = 0.5564253098499673842139595564253098499674148856973...
%e and possesses very similar properties.
%e The CONTINUED FRACTION of 3*x has large partial quotients:
%e 3*x = [1, 3, 42, 4, 41619663273108911871743469597819008, 10889035741470030830827987437816582767104, ...];
%e the number of digits of the partial quotients begin:
%e [1, 1, 2, 1, 35, 41, 115, 270, ...].
%e The initial 1050 digits are:
%e x = 0.44357469015003261578604044357469015003258511430265\
%e 30388125774628431253164859002234302537588527824859\
%e 91958242927156950433109781361196018310495546232260\
%e 21308756488922508493146400309145266931418310463207\
%e 21682655581705538047690626270645407928462460161012\
%e 59348074253641658104334540972276437282369203087870\
%e 91688278965124501135824721922196014571344903077739\
%e 18500944407821228291964676297532896809047816970961\
%e 57232314116611612205997186868899794801384103713379\
%e 27093301379877141640033100637461650643685872664216\
%e 71210915983869597104515647764570759665860825582616\
%e 02202386683742093208255873853725828394394746230649\
%e 17646005634235224736778439225789268521850715113549\
%e 81826246824173592162652534369620590399778082914397\
%e 59525545391741573753969070145840793634947962855349\
%e 40161700909555769785319066569343646229986269080724\
%e 03331326445769921119003906381638002817964787631913\
%e 86332217342163925996090891887058572086607050246768\
%e 90125594710389427906358536708700202340858220467421\
%e 83637059019834780162093855868315000246318027150733\
%e 19542694230313602926455823852338591379348725144367...
%o (PARI) {x=.4; for(i=1, 100, x = (x + sum(n=1, 4000, frac(n*(1-x))/2^n*1.))/2); x}
%o (PARI) /* Series 2-3*x = Sum_{n>=1} 1 / 2^[n/(1-x)] gives faster convergence: */
%o {x=0.4; for(i=1, 10, x = (2 - sum(n=1, 2000, 1./2^floor(n/(1-x))))/3 ); x}
%Y Cf. A258072.
%K nonn,cons
%O 1,2
%A _Paul D. Hanna_, May 21 2015
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