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%I #25 May 24 2015 22:07:55
%S 0,5,5,6,4,2,5,3,0,9,8,4,9,9,6,7,3,8,4,2,1,3,9,5,9,5,5,6,4,2,5,3,0,9,
%T 8,4,9,9,6,7,4,1,4,8,8,5,6,9,7,3,4,6,9,6,1,1,8,7,4,2,2,5,3,7,1,5,6,8,
%U 7,4,6,8,3,5,1,4,0,9,9,7,7,6,5,6,9,7,4,6,2,4,1,1,4,7,2,1,7,5,1,4,0,0,8,0,4,1,7,5,7,0,7,2,8,4,3,0,4,9,5,6,6,8,9,0,2,1,8,6,3,8,8,0,3,9,8,1,6,8,9,5,0,4,4,5,3,7,6,7,7,3,9,7
%N Constant x that satisfies: x = Sum_{n>=1} frac(n*(1-x)) / 2^n.
%C A good approximation to this constant is 853/1533, which is correct to 39 digits.
%H Paul D. Hanna, <a href="/A258072/b258072.txt">Table of n, a(n) for n = 1..1051</a>
%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 = 0.5564253098499673842139595564253098499674148856973...
%e where x = Sum_{n>=1} {n*(1-x)} / 2^n such that x > 1/2 and x < 1.
%e Other series involving x begin:
%e (a) 3*x-1 = 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*x]/2^n +...
%e (b) 2-3*x = 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*(1-x)]/2^n +...
%e (c) 3*x-1 = 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/x] +...
%e (d) 2-3*x = 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/(1-x)] +...
%e note that (c) and (d) involve Beatty sequences as exponents of 1/2.
%e The complement to this constant is A258075:
%e 1-x = 0.4435746901500326157860404435746901500325851143026...
%e and possesses very similar properties.
%e The CONTINUED FRACTION of this constant has large partial quotients:
%e x = [0, 1, 1, 3, 1, 13, 2, 2, 2, 13873221091036303957247823199273002, 2, 1, 3629678580490010276942662479272194255700, 1, 2, 1641750258183103300511626670839317241878322469602726944497845558510238407145724198024905280171526972004597661433855, 1, 2];
%e the next partial quotient has 269 digits.
%e The initial 1050 digits are:
%e x = 0.5564253098499673842139595564253098499674148856973\
%e 46961187422537156874683514099776569746241147217514\
%e 00804175707284304956689021863880398168950445376773\
%e 97869124351107749150685359969085473306858168953679\
%e 27831734441829446195230937372935459207153753983898\
%e 74065192574635834189566545902772356271763079691212\
%e 90831172103487549886417527807780398542865509692226\
%e 08149905559217877170803532370246710319095218302903\
%e 84276768588338838779400281313110020519861589628662\
%e 07290669862012285835996689936253834935631412733578\
%e 32878908401613040289548435223542924033413917441738\
%e 39779761331625790679174412614627417160560525376935\
%e 08235399436576477526322156077421073147814928488645\
%e 01817375317582640783734746563037940960022191708560\
%e 24047445460825842624603092985415920636505203714465\
%e 05983829909044423021468093343065635377001373091927\
%e 59666867355423007888099609361836199718203521236808\
%e 61366778265783607400390910811294142791339294975323\
%e 10987440528961057209364146329129979765914177953257\
%e 81636294098016521983790614413168499975368197284926\
%e 68045730576968639707354417614766140862065127485563...
%o (PARI) {x=.5; 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=.5; for(i=1,10, x = (2 - sum(n=1,2000, 1./2^floor(n/(1-x))))/3 );x}
%Y Cf. A258075.
%K nonn,cons
%O 1,2
%A _Paul D. Hanna_, May 18 2015
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