%I M4595 N1960 #105 Jul 01 2024 13:16:30

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

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

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

%N Decimal expansion of natural logarithm of 3.

%D Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 221.

%D W. E. Mansell, Tables of Natural and Common Logarithms. Royal Society Mathematical Tables, Vol. 8, Cambridge Univ. Press, 1964, p. 2.

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

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

%H Harry J. Smith, <a href="/A002391/b002391.txt">Table of n, a(n) for n = 1..20000</a>

%H D. H. Bailey, <a href="http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-formulas.pdf">Compendium to BBP formulas</a>. [Broken link]

%H Peter Bala, <a href="/A002117/a002117.pdf">New series for old functions</a>.

%H G. Huvent, <a href="https://web.archive.org/web/20150911213120/http://gery.huvent.pagesperso-orange.fr/articlespdf/bbp-base3.pdf">Formules BBP en base 3</a>. - _Jaume Oliver Lafont_, Oct 12 2009

%H Melissa Larson, <a href="https://www.d.umn.edu/~jgreene/masters_reports/BBP%20Paper%20final.pdf">Verifying and discovering BBP-type formulas</a>, 2008.

%H Simon Plouffe, Plouffe's Inverter, <a href="http://www.plouffe.fr/simon/constants/log3.txt">The natural logarithm of 3 to 10000 digits</a>.

%H Simon Plouffe, <a href="https://web.archive.org/web/20150911213121/http://www.worldwideschool.org/library/books/sci/math/MiscellaneousMathematicalConstants/chap61.html">log(3), natural logarithm of 3 to 2000 places</a>.

%H S. Ramanujan, <a href="http://www.imsc.res.in/~rao/ramanujan/NoteBooks/NoteBook2/chapterII/page2.htm">Notebook entry</a>.

%H Horace S. Uhler, <a href="https://doi.org/10.1073/pnas.26.3.205">Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17</a>, Proc. Nat. Acad. Sci. U. S. A. 26, (1940). 205-212.

%H Eric Weisstein's World of Mathematics, <a href="http://www.mathworld.wolfram.com/BBP-TypeFormula.html">BBP-Type Formula</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula">Bailey-Borwein-Plouffe formula</a>.

%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>

%F log(3) = Sum_{n>=1} (9*n-4)/((3*n-2)*(3*n-1)*3*n). [Jolley, Summation of Series, Dover (1961) eq 74]

%F log(3) = (1/4)*(1 + Sum_{m>=0} (1/9)^(k+1)*(27/(2*k+1) + 4/(2*k+2) + 1/(2*k+3))) (a BBP-type formula). - _Alexander R. Povolotsky_, Dec 01 2008

%F log(3) = 4/5 + (1/5)*Sum_{n>=0} (1/4)^n*(1/(2*n+1) + 1/(2*n+3)). - _Alexander R. Povolotsky_, Dec 18 2008

%F log(3) = Sum_{k>=0} (1/9)^(k+1)*(9/(2k+1) + 1/(2k+2)). - _Jaume Oliver Lafont_, Dec 22 2008

%F Sum_{i>=1} 1/(9^i*i) + Sum_{i>=0} 1/(9^i*(i+1/2)) = 2*log(3) (Huvent 2001). - _Jaume Oliver Lafont_, Oct 12 2009

%F Conjecture: log(3) = Sum_{k>=1} A191907(3,k)/k. - _Mats Granvik_, Jun 19 2011

%F log(3) = lim_{n->oo} Sum_{k=3^n..3^(n+1)-1} 1/k. Also see A002162. By analogy to the integral of 1/x, log(m) = lim_{n->oo} Sum_{k=m^n..m^(n+1)-1} 1/k, for any value of m > 1. - _Richard R. Forberg_, Aug 16 2014

%F From _Peter Bala_, Feb 04 2015: (Start)

%F log(3) = Sum {k >= 0} 1/((2*k + 1)*4^k).

%F Define a pair of integer sequences A(n) = 4^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*4^k). Both sequences satisfy the same second-order recurrence equation u(n) = (20*n + 6)*u(n-1) - 16*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(3) = 1 + 2/(24 - 16*3^2/(46 - 16*5^2/(66 - ... - 16*(2*n - 1)^2/((20*n + 6) - ... )))). Cf. A002162, A073000 and A105531 for similar expansions.

%F log(3) = 2 * Sum_{k >= 1} (-1)^(k+1)*(4/3)^k/(k*binomial(2*k,k)).

%F log(3) = (1/4) * Sum_{k >= 1} (-1)^(k+1) (55*k - 23)*(8/9)^k/( 2*k*(2*k - 1)*binomial(3*k,k) ).

%F log(3) = (1/4) * Sum_{k >= 1} (7*k + 1)*(8/3)^k/( 2*k*(2*k - 1)*binomial(3*k,k) ). (End)

%F log(3) = -lim_{n->oo} (n+1)th derivative of zeta(n) / n-th derivative of zeta(n). By n = 1000 there is convergence to 25 digits. A related expression: lim_{n->oo} n-th derivative of zeta(n-1) / n-th derivative of zeta(n) = 3. Also see A002581. - _Richard R. Forberg_, Feb 24 2015

%F From _Peter Bala_, Nov 02 2019: (Start)

%F log(3) = 2*Integral_{x = 0..1} (1 - x^2)/(1 + x^2 + x^4) dx = 2*( 1 - (2/3) + 1/5 + 1/7 - (2/9) + 1/11 + 1/13 - (2/15) + ... ).

%F log(3) = 16*Sum_{n >= 0} 1/( (6*n + 1)*(6*n + 3)*(6*n + 5) ).

%F log(3) = 4/5 + 64*Sum_{n >= 0} (18*n + 1)/((6*n - 5)*(6*n - 3)*(6*n - 1)*(6*n + 1)*(6*n + 7)). (End)

%F From _Amiram Eldar_, Jul 05 2020: (Start)

%F Equals 2*arctanh(1/2).

%F Equals Sum_{k>=1} (2/3)^k/k.

%F Equals Integral_{x=0..Pi} sin(x)dx/(2 + cos(x)). (End)

%F log(3) = Integral_{x = 0..1} (x^2 - 1)/log(x) dx. - _Peter Bala_, Nov 14 2020

%F From _Peter Bala_, Oct 28 2023: (Start)

%F The series representation log(3) = 16*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*(6*n + 5)) given above appears to be the case k = 0 of the following infinite family of series representations for log(3):

%F log(3) = c(k) + (-1)^k*d(k)*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*...*(6*n + 12*k + 5)), where c(k) is a rational approximation to log(3) and d(k) = 2^(6*k+3)/27^k * (6*k + 2)!.

%F The first few values of c(k) for k >= 0 are [0, 2996/2673, 89195548/81236115, 23239436137364/21153065697225, 3345533089100222564/3045237239236561677, ...]. Cf A304656. (End)

%F log(3) = 1 + 2*Sum_{k>=1} 1/((3*k)^3 - 3*k) [Ramanujan]. - _Stefano Spezia_, Jul 01 2024

%e 1.098612288668109691395245236922525704647490557822749451734694333637494...

%t RealDigits[Log[3],10,120][[1]] (* _Harvey P. Dale_, Apr 23 2011 *)

%o (PARI) log(3) \\ _Charles R Greathouse IV_, Jan 24 2012

%o (Python) # Use some guard digits when computing.

%o # BBP formula P(1, 4, 2, (1, 0)).

%o from decimal import Decimal as dec, getcontext

%o def BBPlog3(n: int) -> dec:

%o getcontext().prec = n

%o s = dec(0); f = dec(1); g = dec(4)

%o for k in range(2 * n):

%o s += f / dec(2 * k + 1)

%o f /= g

%o return s

%o print(BBPlog3(200)) # _Peter Luschny_, Nov 03 2023

%Y Cf. A058962, A154920, A002162, A016731 (continued fraction), A073000, A105531, A254619.

%K nonn,cons

%O 1,3

%A _N. J. A. Sloane_

%E Editing and more terms from _Charles R Greathouse IV_, Apr 20 2010