%I M3221 #71 Jun 04 2023 01:39:29
%S 1,4,4,2,6,9,5,0,4,0,8,8,8,9,6,3,4,0,7,3,5,9,9,2,4,6,8,1,0,0,1,8,9,2,
%T 1,3,7,4,2,6,6,4,5,9,5,4,1,5,2,9,8,5,9,3,4,1,3,5,4,4,9,4,0,6,9,3,1,1,
%U 0,9,2,1,9,1,8,1,1,8,5,0,7,9,8,8,5,5,2,6,6,2,2,8,9,3,5,0,6,3,4,4,4,9,6,9,9
%N Decimal expansion of log_2 e.
%C Around 1670, James Gregory discovered by inversion of 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... = log(2) that 1 + 1/2 - 1/12 + 1/24 - 19/720 + (27/1440 = 3/160) - 863/60480 + ... = 1/log(2). See formula with A002206 and A002207. See also A141417 signed /A091137; case i = 0 in A165313. First row in array p. 36 of the reference. - _Paul Curtz_, Sep 12 2011
%C This constant 1/log(2) is also related to the asymptotic evaluation of the maximum number of subtraction steps required to compute gcd(m, n) by the binary Euclidean algorithm, m and n being odd and chosen at random. - _Jean-François Alcover_, Jun 23 2014, after _Steven Finch_
%D Paul Curtz, Intégration numérique des systèmes différentiels .. , note n° 12, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.18 Porter-Hensley constants, p. 159.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Vincenzo Librandi, <a href="/A007525/b007525.txt">Table of n, a(n) for n = 1..5000</a>
%H Simon Plouffe, <a href="http://www.worldwideschool.org/library/books/sci/math/MiscellaneousMathematicalConstants/chap3.html">1/log(2) the inverse of the natural logarithm of 2</a>.
%H Srinivasa Ramanujan, <a href="http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/question/q769.htm">Question 769</a>, J. Ind. Math. Soc.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>.
%F Equals lim_{n->infinity} A000670(n)/A052882(n). - _Mats Granvik_, Aug 10 2009
%F Equals Sum_{k>=-1} A002206(k)/A002207(k). - _Paul Curtz_, Sep 12 2011
%F Also equals integral_{x>=2} 1/(x*log(x)^2). - _Jean-François Alcover_, May 24 2013
%F 1/log(2) = Sum_{n = -infinity..infinity} (2^n / (1 + 2^2^n)). - _Nicolas Nagel_, Mar 16 2018
%F More generally: 1/log(2) = Sum_{n = -infinity..infinity} (2^(n+x) / (1 + 2^2^(n+x))) for all real x. - _Nicolas Nagel_, Jul 02 2019
%F From _Amiram Eldar_, Jun 04 2023: (Start)
%F Equals 1 + Sum_{k>=1} 1/(2^k * (1 + 2^(1/2^k))).
%F Equals Product_{k>=1} ((1 + 2^(1/2^k))/2). (End)
%e 1.442695040888963407359924681...
%t RealDigits[N[1/Log[2], 105]][[1]] (* _Jean-François Alcover_, Oct 30 2012 *)
%o (PARI) 1/log(2) \\ _Charles R Greathouse IV_, Jan 04 2016
%Y Cf. A000670, A002206, A002207, A052882, A091137, A141417, A165313.
%K nonn,cons,easy
%O 1,2
%A _N. J. A. Sloane_