A016730 -id:A016730 - 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

A067882 Factorial expansion of log(2) = Sum_{n>=1} a(n)/n!.

Original entry on oeis.org

0, 1, 1, 0, 3, 1, 0, 3, 6, 2, 5, 4, 6, 11, 4, 11, 5, 12, 3, 5, 13, 2, 22, 6, 22, 13, 20, 7, 1, 0, 1, 20, 2, 6, 4, 1, 18, 14, 35, 2, 11, 31, 16, 19, 42, 36, 41, 0, 14, 31, 25, 43, 4, 13, 34, 53, 50, 57, 2, 30, 12, 25, 45, 24, 2, 39, 57, 51, 30, 41, 65, 15, 9, 55, 23, 4, 35, 18, 77, 43

Offset: 1

Benoit Cloitre, Mar 10 2002

			log(2) = 0 + 1/2! + 1/3! + 0/4! + 3/5! + 1/6! + 0/7! + 3/8! + 6/9! + ...

Cf. A002162 (decimal expansion), A016730 (continued fraction).

Cf. A322334 (log(3)), A322333 (log(5)), A068460 (log(7)), A068461 (log(11)).

SetDefaultRealField(RealField(250)); [Floor(Log(2))] cat [Floor(Factorial(n)*Log(2)) - n*Floor(Factorial((n-1))*Log(2)) : n in [2..80]]; // G. C. Greubel, Nov 26 2018

With[{b = Log[2]}, Table[If[n == 1, Floor[b], Floor[n!*b] - n*Floor[(n - 1)!*b]], {n, 1, 100}]] (* G. C. Greubel, Nov 26 2018 *)

default(realprecision, 250); b = log(2); for(n=1, 80, print1(if(n==1, floor(b), floor(n!*b) - n*floor((n-1)!*b)), ", ")) \\ G. C. Greubel, Nov 26 2018

def A067882(n):
    if (n==1): return floor(log(2))
    else: return expand(floor(factorial(n)*log(2)) - n*floor(factorial(n-1)*log(2)))
[A067882(n) for n in (1..80)] # G. C. Greubel, Nov 26 2018

a(n) = floor(n!*log(2)) - n*floor((n-1)!*log(2)).

A129935 Numbers n such that ceiling( 2/(2^(1/n)-1) ) is not equal to floor( 2n/log(2) ).

Original entry on oeis.org

777451915729368, 140894092055857794, 1526223088619171207, 3052446177238342414, 54545811706258836911039145, 624965662836733496131286135873807507, 1667672249427111806462471627630318921648499, 36465374036664559522628534720215805439659141

Offset: 1

Richard Stanley, Apr 30 2007 (who sent a(1))

nonn

If n belongs to this sequence and m = ceiling(2/(2^(1/n)-1)), then 0 < m/(2n) - 1/log(2) < (log(2)/3) * (1/(2n)^2) implying that m/(2n) is a convergent of 1/log(2) (note that m and 2n are not necessarily coprime). - Max Alekseyev, Jun 06 2007
From David Applegate, Jun 07 2007: (Start)
"Some background to Max Alekseyev's comments: The key point is that the Laurent series for 2/(2^(1/n)-1) about n=infinity is 2/log(2)*n - 1 + (1/6)*log(2)/n + O(1/n^3).
"Also, since 2/log(2) is irrational, 2n/log(2) is never integral, so floor(2n/log(2)) = ceiling(2n/log(2)-1).
"So the question becomes: when is 2n/log(2)-1 so close to an integer that 2/(2^(1/n)-1) is on the other side of the integer? That is why the continued fraction expansion of 2/log(2) is relevant." (End)
The appropriate generalization of ceiling(2/(2^(1/n)-1)) = ? floor(2n/log(2)) is floor(a/(b^(1/n)-1)+a/2) = ceiling(an/log(b)). When a=2, the a/2 can be hidden in floor() + 1 = ceiling(). - David Applegate, Jun 08 2007 [edited Jun 11 2007]

S. W. Golomb and A. W. Hales, "Hypercube Tic-Tac-Toe", in "More Games of No Chance", ed. R. J. Nowakowski, MSRI Publications 42, Cambridge University Press, 2002, pp. 167-182. Here it is stated that the first counterexample is at n=6847196937, an error due to faulty multiprecision arithmetic. The correct value was found by J. Buhler in 2004 and is reported in S. Golomb, "Martin Gardner and Tictacktoe," in Demaine, Demaine, and Rodgers, eds., A Lifetime of Puzzles, A K Peters, 2008, pp. 293-301.
Dean Hickerson, Email to Jon Perry and N. J. A. Sloane, Dec 16 2002. Gives first three terms: 777451915729368, 140894092055857794, 1526223088619171207, as well as five later terms. - N. J. A. Sloane, Apr 30 2014

Max Alekseyev and Robert Gerbicz, Table of n, a(n) for n = 1..100
K. O'Bryant, The sequence of fractional parts of roots, arXiv preprint arXiv:1410.2927 [math.NT], 2014-2015.
Max Alekseyev and others, Integer Parts [in Russian]
Art of Problem Solving, Logarithmic identity
S. W. Golomb and A. W. Hales, Hypercube Tic-Tac-Toe, More Games of No Chance, MSRI Publications, Vol. 42, 2002.
N. J. A. Sloane, Two Sequences that Agree for a Long Time (Vugraph from a talk about the OEIS)

Cf. A078608 for the sequence ceiling( 2/(2^(1/n)-1) ).

Cf. A016730, A120754, A120755.

(* Mma 9.0.1 code from Bill Gosper, Mar 15 2013. He comments: "This reproduces the hundred values in the b-file, and probably works up to around half a billion digits. When Mathematica gets fixed, change 999999999 to infinity." *)
$MaxExtraPrecision = 999999999; For[{lo = {0, 1}, hi = {1, 0}, nu = {0, 0}, n = 0}, nu[[2]] < 10^386, nu = lo + hi; For[{k = nu[[2]]}, Floor[k*2/Log[2]] != Ceiling[2/(2^(1/k) - 1)], k += nu[[2]], Print[{++n, k}]];
  If[nu[[1]]*Log[2] > 2*nu[[2]], hi = nu, lo = nu]]

prec=1500;default(realprecision,prec);c=contfrac(log(2)/2);default(realprecision,prec*2+50); i=0;for(n=2,#c-1, cand=contfracpnqn(vecextract(c,2^n-1))[1,1];forstep(m=cand,c[n+1]*cand,cand, if(ceil(2/(2^(1/m)-1)) != floor(2*m/log(2)), i++;print(i" "m), break))) /* Phil Carmody, Mar 20 2013 */

More terms from Max Alekseyev, Jun 06 2007

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 08 2007

A317557 Number of binary digits to which the n-th convergent of the continued fraction expansion of log(2) matches the correct value.

Original entry on oeis.org

0, -1, 3, 6, 9, 13, 14, 17, 19, 20, 23, 20, 25, 20, 33, 37, 35, 38, 41, 43, 45, 43, 47, 48, 52, 54, 58, 61, 68, 70, 74, 77, 78, 81, 86, 89, 92, 93, 92, 99, 105, 109, 113, 116, 118, 121, 127, 133, 136, 135, 139, 141, 145, 149, 154, 159, 161, 165, 171, 173, 172, 180

Offset: 1

A.H.M. Smeets, Jul 31 2018

Binary expansion of log(2) in A068426.
For number of correct decimal digits see A317558.
For the similar case of number of correct binary digits of Pi see A305879.
The denominator of the k-th convergent obtained from a continued fraction satisfying the Gauss-Kuzmin distribution will tend to exp(k*A100199), A100199 being the inverse of Lévy's constant; the error between the k-th convergent and the constant itself tends to exp(-2*k*A100199), or in binary digits 2*k*A100199/log(2) bits after the binary point.
The sequence for quaternary digits is obtained by floor(a(n)/2), the sequence for octal digits is obtained by floor(a(n)/3), and the sequence for hexadecimal digits is obtained by floor(a(n)/4).

			   n   convergent         binary expansion         a(n)
  ==  ============  =============================  ====
   1     0 / 1      0.0                              0
   2     1 / 1      1.0                             -1
   3     2 / 3      0.1010...                        3
   4     7 / 10     0.1011001...                     6
   5     9 / 13     0.1011000100...                  9
   6    61 / 88     0.10110001011101...             13
   7   192 / 277    0.101100010111000...            14
   8   253 / 365    0.101100010111001001...         17
   9   445 / 642    0.10110001011100100000...       19
  10  1143 / 1649   0.101100010111001000011...      20
  oo  lim = log(2)  0.101100010111001000010111...   --

A.H.M. Smeets, Table of n, a(n) for n = 1..20000

Cf. A016730, A068426, A086702, A100199, A305607, A317558.

a[n_] := Block[{k = 1, a = RealDigits[ Log@2, 2, 4 + 10][[1]], b = RealDigits[ FromContinuedFraction@ ContinuedFraction[Log@2, n + 1], 2, 4n + 10][[1]]}, While[ a[[k]] == b[[k]], k++]; k - 1]; a[1] = 0; a[2] = -1; Array[a, 61] (* Robert G. Wilson v, Aug 09 2018 *)

Lim_{n -> oo} a(n)/n = 2*log(A086702)/log(2) = 2*A100199/log(2) = 2*A305607.

a(40) onward from Robert G. Wilson v, Aug 09 2018

a(n) are also the incrementally largest terms in the continued fraction of log(2) = [a_0; a_1, a_2, ...] = [0; 1, 2, 3, 1, 6, 3, 1, 1, 2] (excluding the a_0 term). - Eric W. Weisstein, Aug 20 2013

a(n) are also the positions of incrementally highest terms in the continued fraction of log(2) = [a_0; a_1, a_2, ...] = [0; 1, 2, 3, 1, 6, 3, 1, 1, 2] (excluding the term a_0) - Eric W. Weisstein, Aug 20 2013

cp's OEIS Frontend

A120754 Records in continued fraction expansion of 1/log(2) (cf. A016730).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Extensions

A120755 Where records occur in continued fraction expansion of 1/log(2) (cf. A016730).

Views

Author

Keywords

Comments

Links

Crossrefs

Extensions

A002162 Decimal expansion of the natural logarithm of 2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A067882 Factorial expansion of log(2) = Sum_{n>=1} a(n)/n!.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

A129935 Numbers n such that ceiling( 2/(2^(1/n)-1) ) is not equal to floor( 2n/log(2) ).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A317557 Number of binary digits to which the n-th convergent of the continued fraction expansion of log(2) matches the correct value.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A317558 Number of decimal digits to which the n-th convergent of the continued fraction expansion of log(2) matches the correct value.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula