OFFSET
1,2
COMMENTS
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
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
REFERENCES
Paul Curtz, Intégration numérique des systèmes différentiels .. , note n° 12, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.18 Porter-Hensley constants, p. 159.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..5000
Simon Plouffe, 1/log(2) the inverse of the natural logarithm of 2.
Srinivasa Ramanujan, Question 769, J. Ind. Math. Soc.
FORMULA
Also equals integral_{x>=2} 1/(x*log(x)^2). - Jean-François Alcover, May 24 2013
1/log(2) = Sum_{n = -infinity..infinity} (2^n / (1 + 2^2^n)). - Nicolas Nagel, Mar 16 2018
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
From Amiram Eldar, Jun 04 2023: (Start)
Equals 1 + Sum_{k>=1} 1/(2^k * (1 + 2^(1/2^k))).
Equals Product_{k>=1} ((1 + 2^(1/2^k))/2). (End)
EXAMPLE
1.442695040888963407359924681...
MATHEMATICA
RealDigits[N[1/Log[2], 105]][[1]] (* Jean-François Alcover, Oct 30 2012 *)
PROG
(PARI) 1/log(2) \\ Charles R Greathouse IV, Jan 04 2016
CROSSREFS
KEYWORD
AUTHOR
STATUS
approved