A265222 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.

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, 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, 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

Offset

1,2

Comments

Compare to the trivial sum: 1/2 = Sum_{n>=1} {z^n} / 2^n when z = 2/3.

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?

Links

Table of n, a(n) for n=1..217.

Formula

Constant z satisfies: z/(2-z) = 1/2 + Sum_{n>=1} floor(z^n) / 2^n.

Example

z = 1.35549305184346396043079825451360172153679862662304\
01705970399825183464595965679911752584158592424160\
27810777220810376275540904724462618375927731917056\
54550430511551525325357646884805527825882548524686\
80736700258913704221335857525909384098830553851116\
05126668310187108162998498933972248896649516339678\
49366846073141291529984290772350906229012817370186\
55169666380457011989017081864719838727980967224971\
98474075748848718792660615793372688507739995250655\
89552283138883099708481229931272775563504550828971\
70024113427009808993763831069451402518903559858745\
59505239672622181099687153202444348446965100196443\
80668334503687174824625643625205785168626890858603\
93558591010025659573835216359698255844450783545599\
17427663656403996437836675351715525954403386934996\
84666455482949770524207287282642519434321755773398\
62694330881627744201473062989839201202400657816150\
84391198373759362968161495215799374878373246239017\
13574353581079553116458694020174184103284836868862\
62896573800985278249325536552468829105057478796554...
GENERATING METHOD.
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.
RELATED CONSTANTS.
(1) Least real z > sqrt(2) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.53249422831624633589209045977055204808626394342423\
14743943812327455836621055716194661407233767648443884690...
(2) Least real z > 13^(1/6) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.53687788811637150697634883884345836398432123973966\
82723170954185424082387135095296328278152477537914182407...
(3) Least real z > 113^(1/11) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.53695074329851152802228619086712434244494261099208\
23135334501884406111892113321815249303292032820021290733...
(4) Least real z > 4^(1/3) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.59632077942330242620034231916724745224383614542308\
85710643389591659503459032792156507140054529416666027413...
(5) Least real z > sqrt(3) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.75118759092320930011149976687513099251284547995833\
95627867298408820204655832368182553275152837639807086641...
(6) Least real z > 73562^(1/20) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.75118762663466598663070625956863995162241902713017\
79945479735132207866837514950592202363248785148839017212...
(7) Least real z > 691806^(1/24) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.75118762870531088915040587334001786652479560768473\
66794998600900660167466413594621357623811851518779554331...
(8) Least real z > 13698^(1/17) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.75118802293006688271680253510886080565943343511207\
61212797559379720869345357325961787461767015435821839729...
(9) Least real z > 475^(1/11) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.75121702526493397702106373116694543383377610186430\
88066729332431554693339104712708810382658857189983724634...
(10) Least real z > 832^(1/12) such that 1/2 = Sum_{n>=1} {z^n}/2^n is
z = 1.75123699808590302107034621374375230412694312101003\
33815109181242003520899218074396156255274527703355463573...
It not known where the threshold values lie for all z in the interval (1,2).

Prog

(PARI) N=100 \\ Calculate and Print N digits of the constant
\p500 \\ set precision
{z=1.0; for(i=1, 5*N, z = z + 1/6 - 1/3*suminf(n=1, frac(z^n)/2^n ) ); z}
{a(n) = floor((10^n*z))%10}
{m=0; for(n=0, N, print1( floor((10^n*z))%10, ", "); if(m==50, m=0; print("")); m=m+1)}

Crossrefs

Sequence in context: A329509 A119280 A307634 * A160585 A307447 A016658

Adjacent sequences: A265219 A265220 A265221 * A265223 A265224 A265225

Keyword

nonn,cons

Author

Paul D. Hanna, Dec 05 2015

Status

approved