0,1

ln(2) = 1/4*(3 - sum(1/(n*(n+1)*(2*n+1)),n=1...infinity)) - from and by Alexander R. Povolotsky [From Alexander R. Povolotsky (pevnev(AT)juno.com), Dec 16 2008]

ln(2) = 105*(sum(1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7)),n=1...infinity) - 319/44100) ln(2) = (319/420 - 3/2*sum(1/(6*n^2+39*n+63),n=1...infinity)) [From Alexander R. Povolotsky (pevnev(AT)juno.com), Dec 16 2008]

i^2*ln(1/2) [From Eric Desbiaux (moongerms(AT)wanadoo.fr), Jun 27 2009]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.3.3.

D. W. Sweeney, On the computation of Euler's constant, Math. Comp., 17 (1963), 170-178.

Uhler, Horace S.; Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17. Proc. Nat. Acad. Sci. U. S. A. 26, (1940). 205-212.

Harry J. Smith, Table of n, a(n) for n=0,...,20000

D. H. Bailey and J. M. Borwein, Experimental Mathematics: Examples, Methods and Implications

Paul Cooijmans, Odds.

X. Gourdon and P. Sebah, The logarithm constant:log(2)

S. Plouffe, log(2), natural logarithm of 2 to 2000 places

S. Ramanujan, Question 260, J. Ind. Math. Soc.

Eric Weisstein's World of Mathematics, Natural Logarithm of 2

Eric Weisstein's World of Mathematics, Masser-Gramain Constant

Eric Weisstein's World of Mathematics, Logarithmic Integral

log(2) = Sum_{ k >= 1 } 1/(k*2^k) = Sum_{j >= 1} (-1)^(j+1)/j.

log(2) = Integral_{t = 0..1 } dt/(1+t).

log(2) = 2/3 * (1 + Sum{k=1..inf, 2/[(4k)^3-4k]}) (Ramanujan).

log(2)=4*sum_{k=0..inf} [3-2*sqrt(2)]^(2k+1)/(2k+1) (Y. Luke) - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 13 2006

log(2) = 1-(1/2)Sum_{ k >= 1 } 1/(k*(2k+1)) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Jan 06 2009, Jan 08 2009]

log(2) = 4*Sum_{ k >= 0 } 1/((4k+1)(4k+2)(4k+3)) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Jan 08 2009]

Equals 7/12+24*sum_{k=1..infinity} 1/(A052787(k+4)*A000079(k)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jan 23 2009]

Beginning of the contribution from Alexander R. Povolotsky In addition to formula posted by me in Program/Maple section on Dec 16, 2008 ln(2) = 1/4*(3 - sum(1/(n*(n+1)*(2*n+1)), n=1...infinity)) Here is the BBP style formula I came up with today ln(2)=(230166911/9240-Sum((1/2)^k* (11/k+10/(k+1)+9/(k+2)+8/(k+3)+7/(k+4)+6/(k+5)-6/(k+7)-7/(k+8)-8/(k+9) -9/(k+10)-10/(k+11)), k = 1 .. infinity))/35917 End of the contribution from Alexander R. Povolotsky [From Alexander R. Povolotsky (pevnev(AT)juno.com), Jul 04 2009]

log(2) = A052882/A000670. [From Mats Granvik (mats.granvik(AT)abo.fi), Aug 10 2009]

From log(1-x-x^2) at x=1/2, log(2)=(1/2)*Sum_{k>=1}L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Oct 24 2009]

.6931471805599453...

0.693147180559945309417232121458176568075500134360255254120680009493393... [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), May 16 2009]

ln(2) = 1/4*(3 - sum(1/(n*(n+1)*(2*n+1)), n=1...infinity)) [From Alexander R. Povolotsky (pevnev(AT)juno.com), Dec 16 2008]

(PARI) { default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Apr 21 2009]

Cf. A016730 Continued fraction. [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), May 16 2009]

Sequence in context: A129938 A022698 A013707 this_sequence A072365 A085138 A153872

Adjacent sequences: A002159 A002160 A002161 this_sequence A002163 A002164 A002165

cons,nonn

N. J. A. Sloane (njas(AT)research.att.com).

Fixed my PARI program, had -n Harry J. Smith (hjsmithh(AT)sbcglobal.net), May 19 2009