A191907 -id:A191907 - cp's OEIS frontend

A002162 Decimal expansion of the natural logarithm of 2.

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

Offset: 0

N. J. A. Sloane

Newton calculated the first 16 terms of this sequence.
Area bounded by y = tan x, y = cot x, y = 0. - Clark Kimberling, Jun 26 2020
Choose four values independently and uniformly at random from the unit interval [0,1]. Sort them, and label them a,b,c,d from least to greatest (so that a b^2+c^2. - Akiva Weinberger, Dec 02 2024
Define the trihyperboloid to be the intersection of the three solid hyperboloids x^2+y^2-z^2<1, x^2-y^2+z^2<1, and -x^2+y^2+z^2<1. This fits perfectly within the cube [-1,1]^3. Then this is the ratio of the volume of the trihyperboloid to its bounding cube. - Akiva Weinberger, Dec 02 2024

			0.693147180559945309417232121458176568075500134360255254120680009493393...

G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 227.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 250.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3, 2.21, 6.2, and 7.2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 25 and appendix A, equations 25:14:3 and A:7:3 at pages 232, 670.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 29.

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, Notices of the AMS, May 2005, Volume 52, Issue 5.
Peter Bala, New series for old functions
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Paul Cooijmans, Odds.
X. Gourdon and P. Sebah, The logarithm constant:log(2), 2001.
X. Gourdon and P. Sebah, The logarithm constant:log(2), 2004. (Revised and extended version of above article.)
M. Kontsevich and D. Zagier, Periods, pp. 4-5.
Mathematical Reflections, Solution to Problem U376, Issue 4, 2016, p 17.
Isaac Newton, The method of fluxions and infinite series; with its application to the geometry of curve-lines, 1736; see p. 96.
Michael Penn, an alternating floor sum., YouTube video, 2020.
Michael Penn, A wonderful limit from the 2011 Virginia Tech Regional Math Competition, YouTube video, 2022.
Simon Plouffe, log(2), natural logarithm of 2 to 2000 places.
S. Ramanujan, Question 260, J. Ind. Math. Soc., III, p. 43.
Albert Stadler, Problem 3567, Crux Mathematicorum, Vol. 36 (Oct. 2010), p. 396; Oliver Geupel, Solution, Crux Mathematicorum, Vol. 37 (Oct. 2011), pp. 400-401.
D. W. Sweeney, On the computation of Euler's constant, Math. Comp., 17 (1963), 170-178.
Horace S. Uhler, 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.
Eric Weisstein's World of Mathematics, Natural Logarithm of 2, Masser-Gramain Constant, Logarithmic Integral, Dirichlet Eta Function.
Wikipedia, Natural logarithm of 2.
Wikipedia, Period (algebraic geometry).
Index entries for transcendental numbers.

Cf. A016730 (continued fraction), A002939, A008288, A142979, A142992.

RealDigits[N[Log[2],200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)
RealDigits[Log[2],10,120][[1]] (* Harvey P. Dale, Jan 25 2024 *)

{ default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } \\ Harry J. Smith, Apr 21 2009

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} 2/((4*k)^3-4*k)) (Ramanujan).
log(2) = 4*Sum_{k>=0} (3-2*sqrt(2))^(2*k+1)/(2*k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = 1 - (1/2)*Sum_{k>=1} 1/(k*(2*k+1)). - Jaume Oliver Lafont, Jan 06 2009, Jan 08 2009
log(2) = 4*Sum_{k>=0} 1/((4*k+1)*(4*k+2)*(4*k+3)). - Jaume Oliver Lafont, Jan 08 2009
log(2) = 7/12 + 24*Sum_{k>=1} 1/(A052787(k+4)*A000079(k)). - R. J. Mathar, Jan 23 2009
From Alexander R. Povolotsky, Jul 04 2009: (Start)
log(2) = (1/4)*(3 - Sum_{n>=1} 1/(n*(n+1)*(2*n+1))).
log(2) = (230166911/9240 - Sum_{k>=1} (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)))/35917. (End)
log(2) = A052882/A000670. - Mats Granvik, 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). - Jaume Oliver Lafont, Oct 24 2009
log(2) = Sum_{k>=1} 1/(cos(k*Pi/3)*k*2^k) (cf. A176900). - Jaume Oliver Lafont, Apr 29 2010
log(2) = (Sum_{n>=1} 1/(n^2*(n+1)^2*(2*n+1)) + 11)/16. - Alexander R. Povolotsky, Jan 13 2011
log(2) = ((Sum_{n>=1} (2*n+1)/(Sum_{k=1..n} k^2)^2)+396)/576. - Alexander R. Povolotsky, Jan 14 2011
From Alexander R. Povolotsky, Dec 16 2008: (Start)
log(2) = 105*(319/44100 - Sum_{n>=1} 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7))).
log(2) = 319/420 - (3/2)*Sum_{n>=1} 1/(6*n^2+39*n+63). (End)
log(2) = Sum_{k>=1} A191907(2,k)/k. - Mats Granvik, Jun 19 2011
log(2) = Integral_{x=0..oo} 1/(1 + e^x) dx. - Jean-François Alcover, Mar 21 2013
log(2) = lim_{s->1} zeta(s)*(1-1/2^(s-1)). - Mats Granvik, Jun 18 2013
From Peter Bala, Dec 10 2013: (Start)
log(2) = 2*Sum_{n>=1} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
log(2) = (1/3)*Sum_{n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = (1/12)*Sum_{n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
log(2) = (3/16)*Sum_{n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = (1/12)*Sum_{n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = Sum_{n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
log(2) = lim_{n->oo} Sum_{k=2^n..2^(n+1)-1} 1/k. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 03 2015: (Start)
log(2) = (2/3)*Sum_{k >= 0} 1/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second-order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = (2/3)*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
log(2) = Sum_{n>=1} (Zeta(2*n)-1)/n. - Vaclav Kotesovec, Dec 11 2015
From Peter Bala, Oct 30 2016: (Start)
Asymptotic expansions:
for N even, log(2) - Sum_{k = 1..N/2} (-1)^(k-1)/k ~ (-1)^(N/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1. See Borwein et al., Theorem 1 (b);
for N odd, log(2) - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/k ~ (-1)^((N-1)/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), by Borwein et al., Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then set x = (N - 1)/2, where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. (End)
log(2) = lim_{n->oo} Sum_{k=1..n} sin(1/(n+k)). See Mathematical Reflections link. - Michel Marcus, Jan 07 2017
log(2) = Sum_{n>=1} A006519(n) / ((1 + 2^A006519(n)) * A000265(n) * (1 + A000265(n))). - Nicolas Nagel, Mar 19 2018
From Amiram Eldar, Jul 02 2020: (Start)
Equals Sum_{k>=2} zeta(k)/2^k.
Equals -Sum_{k>=2} log(1 - 1/k^2).
Equals Sum_{k>=1} 1/A002939(k).
Equals Integral_{x=0..Pi/3} tan(x) dx. (End)
log(2) = Integral_{x=0..Pi/2} (sec(x) - tan(x)) dx. - Clark Kimberling, Jul 08 2020
From Peter Bala, Nov 14 2020: (Start)
log(2) = Integral_{x = 0..1} (x - 1)/log(x) dx (Boros and Moll, p. 97).
log(2) = (1/2)*Integral_{x = 0..1} (x + 2)*(x - 1)^2/log(x)^2 dx.
log(2) = (1/4)*Integral_{x = 0..1} (x^2 + 3*x + 4)*(x - 1)^3/log(x)^3 dx. (End)
log(2) = 2*arcsinh(sqrt(2)/4) = 2*sqrt(2)*Sum_{n >= 0} (-1)^n*C(2*n,n)/ ((8*n+4)*32^n) = 3*Sum_{n >= 0} (-1)^n/((8*n+4)*(2^n)*C(2*n,n)). - Peter Bala, Jan 14 2022
log(2) = Integral_{x=0..oo} ( e^(-x) * (1-e^(-2x)) * (1-e^(-4x)) * (1-e^(-6x)) ) / ( x * (1-e^(-14x)) ) dx (see Crux Mathematicorum link). - Bernard Schott, Jul 11 2022
From Peter Bala, Oct 22 2023: (Start)
log(2) = 23/32 + 2!^3/16 * Sum_{n >= 1} (-1)^n * (n + 1)/(n*(n + 1)*(n + 2))^2 = 707/1024 - 4!^3/(16^2 * 2!^2) * Sum_{n >= 1} (-1)^n * (n + 2)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^2 = 42611/61440 + 6!^3/(16^3 * 3!^2) * Sum_{n >= 1} (-1)^n * (n + 3)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6))^2.
More generally, it appears that for k >= 0, log(2) = c(k) + (2*k)!^3/(16^k * k!^2) * Sum_{n >= 1} (-1)^(n+k+1) * (n + k)/(n*(n + 1)*...*(n + 2*k))^2 , where c(k) is a rational approximation to log(2). The first few values of c(k) are [0, 23/32, 707/1024, 42611/61440, 38154331/55050240, 76317139/110100480, 26863086823/38755368960, ...].
Let P(n,k) = n*(n + 1)*...*(n + k).
Conjecture: for k >= 0 and r even with r - 1 <= k, the series Sum_{n >= 1} (-1)^n * (d/dn)^r (P(n,k)) / (P(n,k)^2  = A(r,k)*log(2) + B(r,k), where A(r,k) and B(r,k) are both rational numbers. (End)
From Peter Bala, Nov 13 2023: (Start)
log(2) = 5/8 + (1/8)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)^2 / ( k*(k + 1) )^4
  = 257/384 + (3!^5/2^9)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)^2*(2*k + 5) / ( k*(k + 1)*(k + 2)*(k + 3) )^4
  = 267515/393216 + (5!^5/2^19)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)*(2*k + 5)^2*(2*k + 7)*(2*k + 9) / ( k*(k + 1)*(k + 2)*(k + 3)*(k + 4)*(k + 5) )^4
log(2) = 3/4 - 1/128 * Sum_{k >= 0} (-1/16)^k * (10*k + 12)*binomial(2*k+2,k+1)/ ((k + 1)*(2*k + 3)). The terms of the series are O(1/(k^(3/2)*4^n)). (End)
log(2) = eta(1) is a period, where eta(x) is the Dirichlet eta function. - Andrea Pinos, Mar 19 2024
log(2) = K_{n>=0} (n^2 + [n=0])/1, where K is the Gauss notation for an infinite continued fraction. In the expanded form, log(2) = 1/(1 + 1/(1 + 4/(1 + 9/1 + 16/(1 + 25/(1 + ... (see Clawson at p. 227). - Stefano Spezia, Jul 01 2024
log(2) = lim_{n->oo} Sum_{k=1..n} 1/(n + k) = lim_{x->0} (2^x - 1)/x = lim_{x->0} (2^x - 2^(-x))/(2*x) (see Finch). - Stefano Spezia, Oct 19 2024
From Colin Linzer, Nov 08 2024: (Start)
log(2) = Integral_{t=0...oo} (1 - tanh(t)) dt.
log(2) = Integral_{t=0...1} arctanh(t) dt.
log(2) = (1/2) * Integral_{t=-1...1} |arctanh(t)| dt. (End)
log(2) = 1 + Sum_{n >= 1} (-1)^n/(n*(4*n^2 - 1)) = 1/2 + (1/2)*Sum_{n >= 1} 1/(n*(4*n^2 - 1)). - Peter Bala, Jan 07 2025
log(2) = Integral_{x=0..1} Integral_{y=0..1} 1/((1 - x*y)*(1 + x)*(1 + y)) dy dx. - Kritsada Moomuang, May 22 2025

A002391 Decimal expansion of natural logarithm of 3.

Original entry on oeis.org

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, 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, 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

Offset: 1

N. J. A. Sloane

			1.098612288668109691395245236922525704647490557822749451734694333637494...

Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 221.
W. E. Mansell, Tables of Natural and Common Logarithms. Royal Society Mathematical Tables, Vol. 8, Cambridge Univ. Press, 1964, p. 2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Harry J. Smith, Table of n, a(n) for n = 1..20000
D. H. Bailey, Compendium to BBP formulas. [Broken link]
Peter Bala, New series for old functions.
G. Huvent, Formules BBP en base 3. - _Jaume Oliver Lafont_, Oct 12 2009
Melissa Larson, Verifying and discovering BBP-type formulas, 2008.
Simon Plouffe, Plouffe's Inverter, The natural logarithm of 3 to 10000 digits.
Simon Plouffe, log(3), natural logarithm of 3 to 2000 places.
S. Ramanujan, Notebook entry.
Horace S. Uhler, 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.
Eric Weisstein's World of Mathematics, BBP-Type Formula.
Wikipedia, Bailey-Borwein-Plouffe formula.
Index entries for transcendental numbers

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

RealDigits[Log[3],10,120][[1]]  (* Harvey P. Dale, Apr 23 2011 *)

log(3) \\ Charles R Greathouse IV, Jan 24 2012

# Use some guard digits when computing.
# BBP formula P(1, 4, 2, (1, 0)).
from decimal import Decimal as dec, getcontext
def BBPlog3(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(4)
    for k in range(2 * n):
        s += f / dec(2 * k + 1)
        f /= g
    return s
print(BBPlog3(200))  # Peter Luschny, Nov 03 2023

log(3) = Sum_{n>=1} (9*n-4)/((3*n-2)*(3*n-1)*3*n). [Jolley, Summation of Series, Dover (1961) eq 74]
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
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
log(3) = Sum_{k>=0} (1/9)^(k+1)*(9/(2k+1) + 1/(2k+2)). - Jaume Oliver Lafont, Dec 22 2008
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
Conjecture: log(3) = Sum_{k>=1} A191907(3,k)/k. - Mats Granvik, Jun 19 2011
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
From Peter Bala, Feb 04 2015: (Start)
log(3) = Sum {k >= 0} 1/((2*k + 1)*4^k).
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.
log(3) = 2 * Sum_{k >= 1} (-1)^(k+1)*(4/3)^k/(k*binomial(2*k,k)).
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) ).
log(3) = (1/4) * Sum_{k >= 1} (7*k + 1)*(8/3)^k/( 2*k*(2*k - 1)*binomial(3*k,k) ). (End)
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
From Peter Bala, Nov 02 2019: (Start)
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) + ... ).
log(3) = 16*Sum_{n >= 0} 1/( (6*n + 1)*(6*n + 3)*(6*n + 5) ).
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)
From Amiram Eldar, Jul 05 2020: (Start)
Equals 2*arctanh(1/2).
Equals Sum_{k>=1} (2/3)^k/k.
Equals Integral_{x=0..Pi} sin(x)dx/(2 + cos(x)). (End)
log(3) = Integral_{x = 0..1} (x^2 - 1)/log(x) dx. - Peter Bala, Nov 14 2020
From Peter Bala, Oct 28 2023: (Start)
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):
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)!.
The first few values of c(k) for k >= 0 are [0, 2996/2673, 89195548/81236115, 23239436137364/21153065697225, 3345533089100222564/3045237239236561677, ...]. Cf A304656. (End)
log(3) = 1 + 2*Sum_{k>=1} 1/((3*k)^3 - 3*k) [Ramanujan]. - Stefano Spezia, Jul 01 2024

Editing and more terms from Charles R Greathouse IV, Apr 20 2010

A016627 Decimal expansion of log(4).

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

This constant (negated) is the 1-dimensional analog of Madelung's constant. - Jean-François Alcover, May 20 2014
This constant is the sum over the reciprocals of the hexagonal numbers A000384(n), n >= 1. See the Downey et al. link, and the formula by Robert G. Wilson v below. - Wolfdieter Lang, Sep 12 2016
log(4) - 1 is the mean ratio between the smaller length and the larger length of the two parts of a stick that is being broken at a point that is uniformly chosen at random (Mosteller, 1965). - Amiram Eldar, Jul 25 2020
From Bernard Schott, Sep 11 2020: (Start)
This constant was the subject of the problem B5 during the 42nd Putnam competition in 1981 (see formula Sep 11 2020 and Putnam link).
Jeffrey Shallit generalizes this result obtained for base 2 to any base b (see Amer. Math. Month. link): Sum_{k>=1} digsum(k)_b / (k*(k+1)) = (b/(b-1)) * log(b) where digsum(k)_b is the sum of the digits of k when expressed in base b (for base 10 see A334388). (End)

			1.38629436111989061883446424291635313615100026872051050824136...

Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 2.
Frederick Mosteller, Fifty challenging problems of probability, Dover, New York, 1965. See problem 42, pp. 10 and 63.
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 2, equation 2:13:8 at page 23.

Harry J. Smith, Table of n, a(n) for n = 1..20000
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Lawrence Downey, Boon W. Ong, and James A. Sellers, Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers, Coll. Math. J., 39, no. 5 (2008), 391-394.
I. S. Gradsteyn and I. M. Ryzhik, Table of integrals, series and products (6th ed.), 2000, (eq. 0.236.1 and 0.236.6).
D. H. Lehmer, Interesting series involving the Central Binomial Coefficient, Am. Math. Monthly, Vol. 92, No. 7 (1985) pp. 449-457.
Allon G. Percus, Gabriel Istrate, Bruno Goncalves, Robert Z. Sumi, and Stefan Boettcher, The Peculiar Phase Structure of Random Graph Bisection, arXiv:0808.1549 [cond-mat.stat-mech], 2008.
H.-J. Seiffert, Problem B-771, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 32, No. 4 (1994), p. 374; More Sums, Solution to Problem B-771 by Don Redmond, ibid., Vol. 33, No. 5 (1995), pp. 470-471.
J. O. Shallit, Solutions of Advanced Problems, 6450, The American Mathematical Monthly, Vol. 92, No. 7, Aug.-Sep., 1985, pp. 513-514; DOI: 10.2307/2322523.
42nd Putnam Competition, Problem B5, 1981.
Index entries for sequences related to Olympiads and other Mathematical competitions.
Index entries for transcendental numbers.

Cf. A016732 (continued fraction).
Cf. A002162 (half), A133362 (reciprocal).
Cf. A000384, A001787, A002378, A007531, A066066, A069072, A191907.
Cf. A000045, A000120, A334388.

RealDigits[Log@ 4, 10, 111][[1]] (* Robert G. Wilson v, Aug 31 2014 *)

default(realprecision, 20080); x=log(4); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b016627.txt", n, " ", d)); \\ Harry J. Smith, May 16 2009, corrected May 19 2009

A016627_vec(N)=digits(floor(log(precision(4.,N))*10^(N-1))) \\ Or: default(realprecision,N);digits(log(4)\.1^N) \\ M. F. Hasler, Oct 20 2013

log(4) = Sum_{k >= 1} H(k)/2^k where H(k) is the k-th harmonic number. - Benoit Cloitre, Jun 15 2003
Equals 1 - Sum_{k >= 1} (-1)^k/A002378(k) = 1 + 2*Sum_{k >= 0} 1/A069072(k) = 5/4 - Sum_{k >= 1} (-1)^k/A007531(k+2). - R. J. Mathar, Jan 23 2009
Equals 2*A002162 = Sum_{n >= 1} binomial(2*n, n)/(n*4^n) [D. H. Lehmer, Am. Math. Monthly 92 (1985) 449 and Jolley eq. 262]. - R. J. Mathar, Mar 04 2009
log(4) = Sum_{k >= 1} A191907(4, k)/k, (conjecture). - Mats Granvik, Jun 19 2011
log(4) = lim_{n -> infinity} A066066(n)/n. - M. F. Hasler, Oct 20 2013
Equals Sum_{k >= 1} 1/( 2*k^2 - k ). - Robert G. Wilson v, Aug 31 2014
Equals gamma(0, 1/2) - gamma(0, 1) = -(EulerGamma + polygamma(0, 1/2)), where gamma(n,x) denotes the generalized Stieltjes constants, see A020759. - Peter Luschny, May 16 2018
From Amiram Eldar, Jul 25 2020: (Start)
Equals Sum_{k>=1} (3/4)^k/k.
Equals Sum_{k>=1} 1/(k*2^(k-1)) = Sum_{k>=1} 1/A001787(k).
Equals Integral_{x=0..1} log(1+1/x) dx. (End)
Equals Sum_{k>=1} A000120(k) / (k*(k+1)). - Bernard Schott, Sep 11 2020
Equals 1 + Sum_{k>=1} zeta(2*k+1)/4^k. - Amiram Eldar, May 27 2021
Equals Sum_{k>=1} (2*k+1)*Fibonacci(k)/(k*(k+1)*2^k) (Seiffert, 1994). - Amiram Eldar, Jan 15 2022
Continued fraction: log(4) = 1 + 1/(2 + (1*2)/(2 + (2*3)/(2 + (3*4)/(2 + (4*5)/(2 + ... ))))) due to Euler. - Peter Bala, Mar 05 2024
log(4) = 2*Sum_{k>=1} 1/(k*P(k, 5/3)*P(k-1, 5/3)), where P(k, x) denotes the k-th Legendre polynomial. The first 20 terms of the series gives log(4) correct to 18 decimal places. - Peter Bala, Mar 18 2024
Equals Sum_{k>=1} (2*k - 1)!!/(k*(2*k)!!) [Ross] (see Spanier at p. 23). - Stefano Spezia, Dec 27 2024
Equals 1 + Sum_{k>=1} 1/(k*(4*k^2-1)). - Sean A. Irvine, Apr 05 2025
Equals Sum_{k>=1} (12*k^2-1)/(k*(4*k^2-1)^2). - Sean A. Irvine, Apr 06 2025
Equals Integral_{x=0..1} arctanh(sqrt(x))/sqrt(x) dx. - Kritsada Moomuang, Jun 06 2025
From Kritsada Moomuang, Jun 18 2025: (Start)
Equals Integral_{x=0..1} (x^(n - 1)*(x^(3*n) - 1))/log(x) dx, for n > 0.
Equals Integral_{x=0..Pi} sin(x)/(1 + abs(cos(x))) dx. (End)

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A002162 Decimal expansion of the natural logarithm of 2.

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A002391 Decimal expansion of natural logarithm of 3.

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A016627 Decimal expansion of log(4).

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A191904 Square array read by antidiagonals up: T(n,k) = 1-k if k divides n, else 1.

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A191910 Triangle read by rows: T(n,n)=n; T(n,k) = k-1 if k divides n and k < n, otherwise -1.

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