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

%I #8 May 17 2015 21:37:43

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

%T 2,1,8,5,5,9,0,2,6,1,2,2,9,4,5,3,6,2,8,1,5,1,3,0,3,6,0,8,4,3,9,9,3,0,

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

%N Constant x that satisfies: x = 1 + Sum_{n>=1} abs(frac(x^n) - 1/2) / 2^n.

%H Paul D. Hanna, <a href="/A258058/b258058.txt">Table of n, a(n) for n = 0..1100</a>

%F Also, x - 1/2 = Sum_{n>=0} |{x^n} - 1/2| / 2^n, where {z} is the fractional part of z.

%e x = 1.2179364054803528100744074581172662185590261229453\

%e 62815130360843993080221280778102481222183901022114\

%e 02274342126910119357753015766132236449015777369699\

%e 71145304161033755031305673096551171712286197875547\

%e 48823302883527141954945458090962854471161815193024\

%e 62246298903853014685004240626261506449260059710890\

%e 76857314350309486666774044734875208618333937982494\

%e 36401457702599188865310365350038283454621194033888\

%e 51068396350020384282795362650901688529469910528781\

%e 80247522363011638169474275487869500491157022021016\

%e 40175370376589773200215823614275967960954030740900\

%e 09084693322032985498621296412158225311744454307662\

%e 18587821830786207975558749016596758060682955860100\

%e 13281692542057245549665593099837640583860718782376\

%e 75098714951532403004124930808921749440024659008321\

%e 65401825881736204856914863100124554256261104009835\

%e 56903447047264149289420594809680408319231055400107\

%e 99303252951033057183591981903287999009697641393092\

%e 06390285519808405620411103645940830148401808385405\

%e 57712356851967778793950578605983493382464529417909\

%e 62311166426346997304070850177183561881410571667958\

%e 5581829539158840572182422517591104015912086698085294002...

%K nonn,cons

%O 0,2

%A _Paul D. Hanna_, May 17 2015