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%I #32 Dec 06 2015 00:30:36
%S 1,3,5,5,4,9,3,0,5,1,8,4,3,4,6,3,9,6,0,4,3,0,7,9,8,2,5,4,5,1,3,6,0,1,
%T 7,2,1,5,3,6,7,9,8,6,2,6,6,2,3,0,4,0,1,7,0,5,9,7,0,3,9,9,8,2,5,1,8,3,
%U 4,6,4,5,9,5,9,6,5,6,7,9,9,1,1,7,5,2,5,8,4,1,5,8,5,9,2,4,2,4,1,6,0,2,7,8,1,0,7,7,7,2,2,0,8,1,0,3,7,6,2,7,5,5,4,0,9,0,4,7,2,4,4,6,2,6,1,8,3,7,5,9,2,7,7,3,1,9,1,7,0,5,6,5,4,5,5,0,4,3,0,5,1,1,5,5,1,5,2,5,3,2,5,3,5,7,6,4,6,8,8,4,8,0,5,5,2,7,8,2,5,8,8,2,5,4,8,5,2,4,6,8,6,8,0,7,3,6,7,0,0,2,5,8,9,1,3,7,0
%N Decimal expansion of the least real z > 1 that satisfies: 1/2 = Sum_{n>=1} {z^n} / 2^n, where {x} denotes the fractional part of x.
%C Compare to the trivial sum: 1/2 = Sum_{n>=1} {z^n} / 2^n when z = 2/3.
%C Are there an infinite number of solutions to z such that 1/2 = Sum_{n>=1} {z^n}/2^n for z in the interval (1,2)? Can it be shown that these solutions are irrational?
%F Constant z satisfies: z/(2-z) = 1/2 + Sum_{n>=1} floor(z^n) / 2^n.
%e z = 1.35549305184346396043079825451360172153679862662304\
%e 01705970399825183464595965679911752584158592424160\
%e 27810777220810376275540904724462618375927731917056\
%e 54550430511551525325357646884805527825882548524686\
%e 80736700258913704221335857525909384098830553851116\
%e 05126668310187108162998498933972248896649516339678\
%e 49366846073141291529984290772350906229012817370186\
%e 55169666380457011989017081864719838727980967224971\
%e 98474075748848718792660615793372688507739995250655\
%e 89552283138883099708481229931272775563504550828971\
%e 70024113427009808993763831069451402518903559858745\
%e 59505239672622181099687153202444348446965100196443\
%e 80668334503687174824625643625205785168626890858603\
%e 93558591010025659573835216359698255844450783545599\
%e 17427663656403996437836675351715525954403386934996\
%e 84666455482949770524207287282642519434321755773398\
%e 62694330881627744201473062989839201202400657816150\
%e 84391198373759362968161495215799374878373246239017\
%e 13574353581079553116458694020174184103284836868862\
%e 62896573800985278249325536552468829105057478796554...
%e GENERATING METHOD.
%e Set z = 1, then 5*N iterations of: z = z + 1/6 - 1/3*suminf(n=1, frac(z^n)/2^n ) yields about N digits.
%e RELATED CONSTANTS.
%e (1) Least real z > sqrt(2) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.53249422831624633589209045977055204808626394342423\
%e 14743943812327455836621055716194661407233767648443884690...
%e (2) Least real z > 13^(1/6) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.53687788811637150697634883884345836398432123973966\
%e 82723170954185424082387135095296328278152477537914182407...
%e (3) Least real z > 113^(1/11) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.53695074329851152802228619086712434244494261099208\
%e 23135334501884406111892113321815249303292032820021290733...
%e (4) Least real z > 4^(1/3) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.59632077942330242620034231916724745224383614542308\
%e 85710643389591659503459032792156507140054529416666027413...
%e (5) Least real z > sqrt(3) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.75118759092320930011149976687513099251284547995833\
%e 95627867298408820204655832368182553275152837639807086641...
%e (6) Least real z > 73562^(1/20) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.75118762663466598663070625956863995162241902713017\
%e 79945479735132207866837514950592202363248785148839017212...
%e (7) Least real z > 691806^(1/24) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.75118762870531088915040587334001786652479560768473\
%e 66794998600900660167466413594621357623811851518779554331...
%e (8) Least real z > 13698^(1/17) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.75118802293006688271680253510886080565943343511207\
%e 61212797559379720869345357325961787461767015435821839729...
%e (9) Least real z > 475^(1/11) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.75121702526493397702106373116694543383377610186430\
%e 88066729332431554693339104712708810382658857189983724634...
%e (10) Least real z > 832^(1/12) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
%e z = 1.75123699808590302107034621374375230412694312101003\
%e 33815109181242003520899218074396156255274527703355463573...
%e It not known where the threshold values lie for all z in the interval (1,2).
%o (PARI) N=100 \\ Calculate and Print N digits of the constant
%o \p500 \\ set precision
%o {z=1.0; for(i=1,5*N, z = z + 1/6 - 1/3*suminf(n=1, frac(z^n)/2^n ) ); z}
%o {a(n) = floor((10^n*z))%10}
%o {m=0;for(n=0,N, print1( floor((10^n*z))%10,","); if(m==50,m=0;print(""));m=m+1)}
%K nonn,cons
%O 1,2
%A _Paul D. Hanna_, Dec 05 2015
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