A105531 -id:A105531 - 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

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

Original entry on oeis.org

2, 4, 14, 52, 194, 724, 2702, 10084, 37634, 140452, 524174, 1956244, 7300802, 27246964, 101687054, 379501252, 1416317954, 5285770564, 19726764302, 73621286644, 274758382274, 1025412242452, 3826890587534, 14282150107684, 53301709843202, 198924689265124

Offset: 0

N. J. A. Sloane

a(n) gives values of x satisfying x^2 - 3*y^2 = 4; corresponding y values are given by 2*A001353(n).
If M is any given term of the sequence, then the next one is 2*M + sqrt(3*M^2 - 12). - Lekraj Beedassy, Feb 18 2002
For n > 0, the three numbers a(n) - 1, a(n), and a(n) + 1 form a Fleenor-Heronian triangle, i.e., a Heronian triangle with consecutive sides, whose area A(n) may be obtained from the relation [4*A(n)]^2 = 3([a(2n)]^2 - 4); or A(n) = 3*A001353(2*n)/2 and whose semiperimeter is 3*a[n]/2. The sequence is symmetrical about a[0], i.e., a[-n] = a[n].
For n > 0, a(n) + 2 is the number of dimer tilings of a 2*n X 2 Klein bottle (cf. A103999).
Tsumura shows that, for prime p, a(p) is composite (contrary to a conjecture of Juricevic). - Charles R Greathouse IV, Apr 13 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 12 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 192 = 0. - Colin Barker, Feb 16 2014
A268281(n) - 1 is a member of this sequence iff A268281(n) is prime. - Frank M Jackson, Feb 27 2016
a(n) gives values of x satisfying 3*x^2 - 4*y^2 = 12; corresponding y values are given by A005320. - Sture Sjöstedt, Dec 19 2017
Middle side lengths of almost-equilateral Heronian triangles. - Wesley Ivan Hurt, May 20 2020
For all elements k of the sequence, 3*(k-2)*(k+2) is a square. - Davide Rotondo, Oct 25 2020

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 82.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p.91.
Michael P. Cohen, Generating Heronian Triangles With Consecutive Integer Sides. Journal of Recreational Mathematics, vol. 30 no. 2 1999-2000 p. 123.
L. E. Dickson, History of The Theory of Numbers, Vol. 2 pp. 197;198;200;201. Chelsea NY.
Charles R. Fleenor, Heronian Triangles with Consecutive Integer Sides, Journal of Recreational Mathematics, Volume 28, no. 2 (1996-7) 113-115.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
V. D. To, "Finding All Fleenor-Heronian Triangles", Journal of Recreational Mathematics vol. 32 no.4 2003-4 pp. 298-301 Baywood NY.

T. D. Noe, Table of n, a(n) for n=0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
R. A. Beauregard and E. R. Suryanarayan, The Brahmagupta Triangles, The College Mathematics Journal 29(1) 13-7 1998 MAA.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Daniel Birmajer, Juan B. Gil and Michael D. Weiner, Linear recurrence sequences with indices in arithmetic progression and their sums, arXiv preprint arXiv:1505.06339 [math.NT], 2015.
K. S. Brown's Mathpages, Some Properties of the Lucas Sequence(2, 4, 14, 52, 194, ...)
H. W. Gould, A triangle with integral sides and area, Fib. Quart., 11 (1973), 27-39.
Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2, ..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Hideyuki Ohtskua, proposer, Problem B-1351, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 258.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Yu Tsumura, On compositeness of special types of integers, arXiv:1004.1244 [math.NT], 2010.
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001075, A001353, A001835.
Cf. A011945 (areas), A334277 (perimeters).
Cf. this sequence (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).
Cf. A001570, A002530, A005320, A006051, A048788, A174500, A268281.
Cf. A011943, A102344, A019673, A105531.

a003500 n = a003500_list !! n
a003500_list = 2 : 4 : zipWith (-)
   (map (* 4) $ tail a003500_list) a003500_list
-- Reinhard Zumkeller, Dec 17 2011

I:=[2,4]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2018

A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*procname(n-1)-procname(n-2); fi;
end proc;

a[0]=2; a[1]=4; a[n_]:= a[n]= 4a[n-1] -a[n-2]; Table[a[n], {n, 0, 23}]
LinearRecurrence[{4,-1},{2,4},30] (* Harvey P. Dale, Aug 20 2011 *)
Table[Round@LucasL[2n, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

x='x+O('x^99); Vec(-2*(-1+2*x)/(1-4*x+x^2)) \\ Altug Alkan, Apr 04 2016

[lucas_number2(n,4,1) for n in range(0, 24)] # Zerinvary Lajos, May 14 2009

a(n) = ( 2 + sqrt(3) )^n + ( 2 - sqrt(3) )^n.
a(n) = 2*A001075(n).
G.f.: 2*(1 - 2*x)/(1 - 4*x + x^2). Simon Plouffe in his 1992 dissertation.
a(n) = A001835(n) + A001835(n+1).
a(n) = trace of n-th power of the 2 X 2 matrix [1 2 / 1 3]. - Gary W. Adamson, Jun 30 2003 [corrected by Joerg Arndt, Jun 18 2020]
From the addition formula, a(n+m) = a(n)*a(m) - a(m-n), it is easy to derive multiplication formulas, such as: a(2*n) = (a(n))^2 - 2, a(3*n) = (a(n))^3 - 3*(a(n)), a(4*n) = (a(n))^4 - 4*(a(n))^2 + 2, a(5*n) = (a(n))^5 - 5*(a(n))^3 + 5*(a(n)), a(6*n) = (a(n))^6 - 6*(a(n))^4 + 9*(a(n))^2 - 2, etc. The absolute values of the coefficients in the expansions are given by the triangle A034807. - John Blythe Dobson, Nov 04 2007
a(n) = 2*A001353(n+1) - 4*A001353(n). - R. J. Mathar, Nov 16 2007
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n=0..infinity} (1 + x^(4*n + 1))/(1 + x^(4*n + 3)). Let alpha = 2 - sqrt(3). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.24561 99455 06551 88869 ... = 2 + 1/(4 + 1/(14 + 1/(52 + ...))). Cf. A174500.
Also F(-alpha) = 0.74544 81786 39692 68884 ... has the continued fraction representation 1 - 1/(4 - 1/(14 - 1/(52 - ...))) and the simple continued fraction expansion 1/(1 + 1/((4 - 2) + 1/(1 + 1/((14 - 2) + 1/(1 + 1/((52 - 2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((4^2 - 4) + 1/(1 + 1/((14^2 - 4) + 1/(1 + 1/((52^2 - 4) + 1/(1 + ...))))))).
(End)
a(2^n) = A003010(n). - John Blythe Dobson, Mar 10 2014
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 12*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
E.g.f.: 2*exp(2*x)*cosh(sqrt(3)*x). - Ilya Gutkovskiy, Apr 27 2016
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*n*(n - k - 1)!/(k!*(n - 2*k)!)*4^(n - 2*k) for n >= 1. - Peter Luschny, May 10 2016
From Peter Bala, Oct 15 2019: (Start)
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; -1, 4].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
2*Sum_{n >= 1} 1/( a(n) - 6/a(n) ) = 1.
6*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 2/a(n) ) = 1.
8*Sum_{n >= 1} 1/( a(n) + 24/(a(n) - 12/(a(n))) ) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 8/(a(n) + 4/(a(n))) ) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/2 - 6*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 6)),
Sum_{n >= 1} 1/a(n) = 1/8 + 24*Sum_{n >= 1} 1/(a(n)*(a(n)^2 + 12)),
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/6 + 2*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 2)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 12)).
Sum_{n >= 1} 1/a(n) = ( theta_3(2-sqrt(3))^2 - 1 )/4 = 0.34770 07561 66992 06261 .... See Borwein and Borwein, Proposition 3.5 (i), p.91.
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - theta_3(sqrt(3)-2)^2 )/4. Cf. A003499 and A153415. (End)
a(n) = tan(Pi/12)^n + tan(5*Pi/12)^n. - Greg Dresden, Oct 01 2020
From Wolfdieter Lang, Sep 06 2021: (Start)
a(n) = S(n, 4) - S(n-2, 4) = 2*T(n, 2), for n >= 0, with S and T Chebyshev polynomials, with S(-1, x) = 0 and S(-2, x) = -1. S(n, 4) = A001353(n+1), for n >= -1, and T(n, 2) = A001075(n).
a(2*k) = A067902(k), a(2*k+1) = 4*A001570(k+1), for k >= 0. (End)
a(n) = sqrt(2 + 2*A011943(n+1)) = sqrt(2 + 2*A102344(n+1)), n>0. - Ralf Steiner, Sep 23 2021
Sum_{n>=1} arctan(3/a(n)^2) = Pi/6 - arctan(1/3) = A019673 - A105531 (Ohtskua, 2024). - Amiram Eldar, Aug 29 2024

More terms from James Sellers, May 03 2000

Additional comments from Lekraj Beedassy, Feb 14 2002

A073000 Decimal expansion of arctangent of 1/2.

Original entry on oeis.org

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

Offset: 0

Robert G. Wilson v, Aug 03 2002

The angle at which you must shoot a cue ball on a standard pool table so that it will strike all four sides and return to its origin. [Barrow] - Robert G. Wilson v, Nov 29 2015

			Arctan(1/2)
=0.463647609000806116214256231461214402028537054286120263810933088720197864165... radians
=26°.56505117707798935157219372045329467120421429964522102798601631528806582148474...
=26°33'.9030706246793610943316232271976802722528579787132616791609789172839492890...
=26°33'54".184237480761665659897393631860816335171478722795700749658735037036957...
complement = 63°.43494882292201064842780627954670532879578570035477897201398368471...
supplement = 153°.4349488229220106484278062795467053287957857003547789720139836847...

John D. Barrow, One Hundred Essential Things You Didn't Know You Didn't Know, W. W. Norton & Co., NY & London, 2008.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 242.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Peter Bala, New series for old functions.
R. J. Mathar, Hierarchical Subdivision of the Simple Cubic Lattice, arXiv preprint arXiv:1309.3705 [math.MG], 2013.
Simon Plouffe, arctan(1/2).
Index entries for transcendental numbers

Cf. A001622, A002162, A002391, A105531, A254619.

evalf(arctan(0.5)) ; # R. J. Mathar, Aug 22 2013

Mathematica
```
RealDigits[ ArcTan[1/2], 10, 110] [[1]]
```

default(realprecision,2000); atan(1/2) \\ Anders Hellström, Nov 30 2015

Equals Pi/2 - A105199 = A019669-A105199. - R. J. Mathar, Aug 21 2013
From Peter Bala, Feb 04 2015: (Start)
Arctan(1/2) = 1/2*Sum_{k >= 0} (-1)^k/((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)^k/((2*k + 1)*4^k). Both sequences satisfy the same second order recurrence equation u(n) = (12*n + 10)*u(n-1) + 16*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion 2*arctan(1/2) = 1 - 2/(24 + 16*3^2/(34 + 16*5^2/(46 + ... + 16*(2*n - 1)^2/((12*n + 10) + ...)))). See A002391, A105531 and A002162 for similar expansions.
Arctan(1/2) = 2/5 * Sum_{k >= 0} (4/5)^k/((2*k + 1)*binomial(2*k,k)).
Define a pair of integer sequences C(n) = 5^n*(2*n + 1)!/n! and D(n) = C(n)*Sum_{k = 0..n} (4/5)^k/((2*k + 1)*binomial(2*k,k)). Both sequences satisfy the same second order recurrence equation u(n) = (24*n + 10)*u(n-1) - 40*n*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion 5/2*arctan(1/2) = 1 + 4/(30 - 240/(58 - 600/(82 - ... - 40*n*(2*n - 1)/((24*n + 10) - ... )))).
Arctan(1/2) = 2/25 * Sum_{k >= 0} (24*k + 17)*(4/5)^(2*k)/( (4*k + 1)*(4*k + 3)*binomial(4*k,2*k) ).
Arctan(1/2) = 2/125 * Sum_{k >= 0} (1116*k^2 + 1446*k + 433)*(4/5)^(3*k)/( (6*k + 1)*(6*k + 3)*(6*k + 5)*binomial(6*k,3*k) ). (End)
Equals Integral_{x = 0..oo} exp(-2*x)*sin(x)/x dx. - Peter Bala, Nov 05 2019
Equals 2 * arccot(phi^3), where phi is the golden ratio (A001622). - Amiram Eldar, Jul 06 2023
Equals Sum_{n >= 1} i/(n*P(n, 2*i)*P(n-1, 2*i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A098443(n)*A098443(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(2))^n ). - Peter Bala, Mar 16 2024

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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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A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

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A073000 Decimal expansion of arctangent of 1/2.

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A105533 Decimal expansion of arctan(1/7).

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A195729 Decimal expansion of arctan(3).

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A254620 a(n) = 9^n(2n + 1)!/n!.