A073000 -id:A073000 - 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

A058962 a(n) = 2^(2n)(2*n+1).

Original entry on oeis.org

1, 12, 80, 448, 2304, 11264, 53248, 245760, 1114112, 4980736, 22020096, 96468992, 419430400, 1811939328, 7784628224, 33285996544, 141733920768, 601295421440, 2542620639232, 10720238370816, 45079976738816, 189115999977472, 791648371998720

Offset: 0

N. J. A. Sloane, Jan 13 2001

Denominators in expansion of -1/2*i*Pi+i*arcsin((1+1/4*x^2)/(1-1/4*x^2)), where i=sqrt(-1); numerators are all 1.
Bisection of A001787. That is, a(n) = A001787(2n+1). - Graeme McRae, Jul 12 2006
Denominators of odd terms in expansion of 2*arctanh(s/2); numerators are all 1. - Gerry Martens, Jul 26 2015
Reciprocals of coefficients of Taylor series expansion of sinh(x/2) / (x/2). - Tom Copeland, Feb 03 2016

Harry J. Smith, Table of n, a(n) for n = 0..200
G. A. Campbell, Physical theory of the electric wave-filter, Bell Syst. Tech. J., 1 (1922), 1-32, see Eq. (15d). Also reprinted in M. E. Van Valkebburg, ed., Circuit Theory, Dowden, Hutchinson and Ross, 1974.
Index entries for linear recurrences with constant coefficients, signature (8,-16).

Cf. A001787, A002391, A073000, A118413, A118415.
Cf. A154920. - Jaume Oliver Lafont, Jan 29 2009
Factor of the LS1[-2,n] matrix coefficients in A160487. - Johannes W. Meijer, May 24 2009

[2^(2*n)*(2*n+1) : n in [0..30]]; // Wesley Ivan Hurt, Aug 07 2015

a[n_] := 1/SeriesCoefficient[2 ArcTanh[s/2],{s,0,n}]
Table[a[n], {n, 1, 40, 2}] (* Gerry Martens, Jul 26 2015 *)
Table[2^(2 n) (2 n + 1), {n, 0, 40}] (* Vincenzo Librandi, Aug 08 2015 *)
a[ n_] := With[{m = 2 n + 2}, If[ n < 0, -a[-1 - n] 4^(m - 1), m! SeriesCoefficient[ x^2 D[x Sinc[I x]^2, x]/2, {x, 0, m}]]]; (* Michael Somos, Jun 18 2017 *)

A058962(n)=2^(2*n)*(2*n+1) \\ M. F. Hasler, Aug 11 2015

{a(n) = my(m = 2*n + 2); if( n<0, -a(-1 - n) * 4^(m - 1), m! * polcoeff( x^2 * deriv(x * sinc(I*x + x * O(x^m))^2, x) / 2, m))}; /* Michael Somos, Jun 18 2017 */

Central terms of the triangle in A118416: a(n) = A118416(2*n+1, n+1) - Reinhard Zumkeller, Apr 27 2006
Sum_{n>=0} 1/a(n) = log(3). - Jaume Oliver Lafont, May 22 2007; corrected by Jaume Oliver Lafont, Jan 26 2009
a(n) = 4((2n+1)/(2n-1))*a(n-1) = 4*a(n-1)+2^(2n+1) = 8*a(n-1)-16*a(n-2). - Jaume Oliver Lafont, Dec 09 2008
G.f.: (1+4*x)/(1-4*x)^2. - Jaume Oliver Lafont, Jan 29 2009
E.g.f.: exp(4*x)*(1+8*x). - Robert Israel, Aug 10 2015
a(n) = -a(-1-n) * 4^(2*n+1) for all n in Z. - Michael Somos, Jun 18 2017
a(n) = Sum_{k = 0..n} (2*k + 1)^2*binomial(2*n + 1, n - k). - Peter Bala, Feb 25 2019
Sum_{n>=0} (-1)^n/a(n) = 2 * arctan(1/2) = 2 * A073000. - Amiram Eldar, Jul 03 2020

A105199 Decimal expansion of arctan(2).

Original entry on oeis.org

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

Offset: 1

Bryan Jacobs (bryanjj(AT)gmail.com), Apr 12 2005

arctan(2) + A073000 = Pi/2.
arctan(2) is the (minimal) central angle of a regular icosahedron, which is the platonic solid having 20 faces and 12 vertices. The (minimal) central angle is AOB, where A and B are any neighboring pair of vertices and O is the center. To evaluate AOB, it is helpful to start with 12 vertices: (0,c*t,d), (d,0,c*t), (c*t,d,0) where c=(1 or -1) and d=(1 or -1) and t is the golden ratio, (1+sqrt(5))/2. For neighboring vertices, one can select (t,1,0) and (0,t,1). - Clark Kimberling, Feb 10 2009
Lesser interior angle (in radians) of a golden rhombus; i.e., either of the angles bisected by the longer diagonal. A137218 is the greater interior angle. - Rick L. Shepherd, Apr 10 2017
The apex angle in the isosceles triangle that is the triangle with angles A, B and C in which the maximum values of sin(A) + sin(B)*sin(C) is attained. The maximum value is phi (A001622) (Rabinowitz, 2007). - Amiram Eldar, Aug 04 2022
Also <5_1> in Conway et al. (1999). - Eric W. Weisstein, Nov 06 2024

			1.107148717794090503017065460...

G. C. Greubel, Table of n, a(n) for n = 1..5000
G. Boros and V. Moll, Sums of arctangents and some formulas of Ramanujan, Scientia Ser. A Math. Sci 11 (2005) 13-24 [MR2196063] eq. (2.11). [From _R. J. Mathar_, Apr 12 2010]
J. H. Conway, C. Radin, L. and Sadun, On Angles Whose Squared Trigonometric Functions Are Rational. Discr. Computat. Geom., Vol. 22, (1999), pp. 321-332.
Stanley Rabinowitz, Problem 3297, Crux Mathematicorum, Vol. 33, No. 8 (2007), pp. 486 and 488; Solution to Problem 3297 by D. J. Smeenk, ibid., Vol. 34, No. 8 (2008), pp. 499-500.
Eric Weisstein's World of Mathematics, Dehn Invariant.
Eric Weisstein's World of Mathematics, Golden Rhombus.
Index entries for transcendental numbers

Cf. A001622, A195693, A197292, A197376.

Cf. A137218 (larger interior angle of the golden rhombus).

RealDigits[ArcTan[2], 10, 105][[1]] (* Indranil Ghosh, Apr 10 2017 *)

default(realprecision, 120);
atan(2) \\ Rick L. Shepherd, Apr 10 2017

Equals Sum_{k>=1} arctan(8k/(4k^4+5)). [Boros and Moll, from R. J. Mathar, Apr 12 2010]
Equals 2*A195693. - Rick L. Shepherd, Apr 10 2017
Equals arcsin(2/sqrt(5)) = arccos(1/sqrt(5)). - Amiram Eldar, Aug 04 2022
Equals 2 - log(5) + (Integral_{x=0..2} log(1 + x^2) dx)/2. - Vaclav Kotesovec, Oct 06 2023
Equals 3*A197292 = A197376/2. - Hugo Pfoertner, Nov 06 2024

Offset corrected by R. J. Mathar, Apr 12 2010

A034910 One quarter of octo-factorial numbers.

Original entry on oeis.org

1, 12, 240, 6720, 241920, 10644480, 553512960, 33210777600, 2258332876800, 171633298636800, 14417197085491200, 1326382131865190400, 132638213186519040000, 14324927024144056320000, 1661691534800710533120000, 206049750315288106106880000

Offset: 1

Wolfdieter Lang

A034910 occurs in connection with the Vandermonde permanent of (1,3,5,7,9,...); see the Mathematica section of A203516. - Clark Kimberling, Jan 03 2012

			G.f. = x + 12*x^2 + 240*x^3 + 6720*x^4 + 241920*x^5 + 10644480*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 1..325
Index entries for sequences related to factorial numbers.

Cf. A001147, A045755, A034908, A034909, A034911, A034912, A203516.

[n le 2 select 12^(n-1) else (7*n-3)*Self(n-1) +4*(n-1)*(2*n-3)*Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 20 2022

[seq((2*n)!/(n)!*2^(n-2), n=1..14)]; # Zerinvary Lajos, Sep 25 2006

s=1;lst={s};Do[s+=n*s;AppendTo[lst, s], {n, 11, 5!, 8}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 08 2008 *)
a[ n_] := Pochhammer[ 1/2, n] 8^n / 4; (* Michael Somos, Feb 04 2015 *)

{a(n) = if( n==1, 1, n>1, a(n-1) * (8*n - 4), a(n+1) / (8*n + 4))}; /* Michael Somos, Feb 04 2015 */

[2^(3*n-2)*rising_factorial(1/2, n) for n in range(1,40)] # G. C. Greubel, Oct 20 2022

4*a(n) = (8*n-4)(!^8) = Product_{j=1..n} (8*j-4) = 4^n*A001147(n) = 2^n*(2*n)!/n!, A001147(n) = (2*n-1)!!.
E.g.f. (-1+(1-8*x)^(-1/2))/4.
a(n) = A090802(2n-1, n). - Ross La Haye, Oct 18 2005
a(n) = ((2*n)!/n!)*2^(n-2). - Zerinvary Lajos, Sep 25 2006
G.f.: x/(1-12*x/(1-8*x/(1-20*x/(1-16*x/(1-28*x/(1-24*x/(1-36*x/(1-32*x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2011
From Peter Bala, Feb 01 2015: (Start)
Recurrence equation: a(n) = (7*n - 3)*a(n-1) + 4*(n - 1)*(2*n - 3)*a(n-2).
The sequence b(n) := a(n)* Sum_{k = 0..n-1} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) beginning [1, 11, 222, 6210, 223584, ...] satisfies the same recurrence. This leads to the finite continued fraction expansion b(n)/a(n) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/(7*n - 3) )))) for n >= 3.
Letting n tend to infinity gives the continued fraction expansion Sum_{k>=0} (-1)^k/( 2^k*(2*k + 1)*binomial(2*k,k) ) = (4/3)*log(2) = 1/(1 + 1/(11 + 24/(18 + 60/(25 + ... + 4*(n - 1)*(2*n - 3)/((7*n - 3) + ... ))))). (End)
From Peter Bala, Feb 03 2015: (Start)
This sequence satisfies several other second order recurrence equations leading to some continued fraction expansions.
1) a(n) = (9*n + 4)*a(n-1) - 4*n*(2*n - 1)*a(n-2).
This recurrence is also satisfied by the (integer) sequence c(n) := a(n)*Sum_{k = 0..n} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ). From this we can obtain the continued fraction expansion Sum_{k >= 0} 1/( 2^k*(2*k + 1)*binomial(2*k,k) ) = (8/sqrt(7))*arctan(sqrt(7)/7) = (8/sqrt(7))*A195699 = 1 + 1/(12 - 24/(22 - 60/(31 - ... - 4*n*(2*n - 1)/((9*n + 4) - ... )))).
2) a(n) = (12*n + 2)*a(n-1) - 8*(2*n - 1)^2*a(n-2).
This recurrence is also satisfied by the (integer) sequence d(n) := a(n)*Sum_{k = 0..n} 1/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum_{k >= 0} 1/( (2*k + 1)*2^k ) = (1/sqrt(2))*log(3 + 2*sqrt(2)) = 1 + 2/(12 - 8*3^2/(26 - 8*5^2/(38 - ... - 8*(2*n - 1)^2/((12*n + 2) - ... )))). Cf. A002391.
3) a(n) = (4*n + 6)*a(n-1) + 8*(2*n - 1)^2*a(n-2).
This recurrence is also satisfied by the (integer) sequence e(n) := a(n)*Sum_{k = 0..n} (-1)^k/( (2*k + 1)*2^k ). From this we can obtain the continued fraction expansion Sum_{k >= 0} (-1)^k/( (2*k + 1)*2^k ) = (1/sqrt(2))*arctan(sqrt(2)/2) = 1 - 2/(12 + 8*3^2/(14 + 8*5^2/(18 + ... + 8*(2*n - 1)^2/((4*n + 6) + ... )))). Cf. A073000. (End)
a(n) = (-1)^n / (16*a(-n)) for all n in Z. - Michael Somos, Feb 04 2015
From Amiram Eldar, Jan 08 2022: (Start)
Sum_{n>=1} 1/a(n) = e^(1/8)*sqrt(2*Pi)*erf(1/(2*sqrt(2))), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = e^(-1/8)*sqrt(2*Pi)*erfi(1/(2*sqrt(2))), where erfi is the imaginary error function. (End)

A105531 Decimal expansion of arctan(1/3).

Original entry on oeis.org

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

Offset: 0

Bryan Jacobs (bryanjj(AT)gmail.com), Apr 12 2005

arctan(1/3) + A073000 = 2*arctan(1/3) + A105533 = Pi/4.

			0.3217505543966421934014046143...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 242.

Peter Bala, New series for old functions
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
Eric Weisstein's World of Mathematics, Machin-Like Formulas
Index entries for transcendental numbers

Cf. A003881 (Pi/4), A073000 (arctan(1/2)), A105533 (arctan(1/7)).

RealDigits[ArcTan[1/3],10,120][[1]] (* Harvey P. Dale, Oct 28 2011 *)

PARI
```
atan(1/3) \\ Michel Marcus, Mar 29 2016
```

From Peter Bala, Feb 04 2015: (Start)
Equals (1/3)*Sum_{k >= 0} (-1)^k/((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)^k/((2*k + 1)*9^k). Both sequences satisfy the same recurrence equation u(n) = (32*n + 20)*u(n-1) + 36*(2*n - 1)^2*u(n-2). From this observation we find the continued fraction expansion arctan(1/3) = (1/3)*(1 - 2/(54 + 36*3^2/(84 + 36*5^2/(116 + ... + 36*(2*n - 1)^2/((32*n + 20) + ...))))).
Equals (3/10) * Sum_{k >= 0} (2/5)^k/( (2*k + 1)*binomial(2*k,k) ).
Define a pair of integer sequences C(n) = 10^n*(2*n + 1)!/n! and D(n) = C(n)*Sum_{k = 0..n} (2/5)^k/( (2*k + 1)*binomial(2*k,k) ). Both sequences satisfy the same recurrence equation u(n) = (44*n + 20)*u(n-1) - 80*n*(2*n - 1)*u(n-2). From this observation we obtain the continued fraction expansion arctan(1/3) = (3/10)*( 1 + 4/(60 - 480/(108 - 1200/(152 - ... - 80*n*(2*n - 1)/((44*n + 20) - ...))))). (End)
Equals Sum_{k>=1} arctan(L(4*k+2)/F(4*k+2)^2) where L=A000032 and F=A000045. See also A033890 and A246453. - Michel Marcus, Mar 29 2016 [corrected by Jason Yuen, Jan 18 2025]
From Amiram Eldar, Aug 09 2020: (Start)
Equals Sum_{k>=2} arctan(1/(2*k^2)) = Sum_{k>=2} (-1)^k arctan(2/k^2).
Equals Integral_{x=1..2} 1/(x^2 + 1) dx. (End)
Equals Sum_{n>=0} arctan(1/F(2*n+5)) = Sum_{n>=0} (-1)^n arctan(F(2*n+1)) where F=A000045. - Gleb Koloskov, Oct 01 2021

A098443 Expansion of 1/sqrt(1-8x-4x^2).

Original entry on oeis.org

1, 4, 26, 184, 1366, 10424, 80996, 637424, 5064166, 40528984, 326251276, 2638751504, 21426682876, 174563719984, 1426219233416, 11681133293024, 95877105146246, 788433553532824, 6494463369141116, 53576199709855184

Offset: 0

Paul Barry, Sep 07 2004

Binomial transform of A098444. Second binomial transform of A084770. Third binomial transform of A098264.

			G.f. = 1 + 4*x + 26*x^2 + 184*x^3 + 1366*x^4 + 10424*x^5 + 80996*x^6 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
Hacène Belbachir, Abdelghani Mehdaoui, László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Column k=2 of A386621.

Cf. A006139, A126869.

CoefficientList[Series[1/Sqrt[1 - 8*x - 4*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 15 2012, updated Mar 21 2024 *)

x='x+O('x^66); Vec(1/sqrt(1-8*x-4*x^2)) \\ Joerg Arndt, May 11 2013

E.g.f.: exp(4*x) * BesselI(0, 2*sqrt(5)*x).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k) * binomial(2(n-k), n) * 2^(n-2k).
D-finite with recurrence: n*a(n) = 4*(2*n-1)*a(n-1) + 4*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 15 2012
a(n) ~ sqrt(50+20*sqrt(5))*(4+2*sqrt(5))^n/(10*sqrt(Pi*n)). Equivalently, a(n) ~ 2^(n-1/2) * phi^(3*n + 3/2) / (5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Oct 15 2012, updated Mar 21 2024
G.f.: 1/(1 - 2*x*(2+x)*Q(0)), where Q(k)= 1 + (4*k+1)*x*(2+x)/(k+1 - x*(2+x)*(2*k+2)*(4*k+3)/(2*x*(2+x)*(4*k+3) + (2*k+3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 15 2013
G.f.: Q(0), where Q(k) = 1 + 2*x*(x+2)*(4*k+1)/( 2*k+1 - x*(x+2)*(2*k+1)*(4*k+3)/(x*(x+2)*(4*k+3) + (k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 16 2013
From Peter Bala, Mar 16 2024: (Start)
a(n) = (-2*i)^n * P(n, 2*i), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial.
Sum_{n >= 1} (-1)^(n+1)*4^n/(n*a(n-1)*a(n)) = 2*arctan(1/2) = 2*A073000. (End)
From Seiichi Manyama, Aug 29 2025: (Start)
a(n) = Sum_{k=0..n} (2-i)^k * (2+i)^(n-k) * binomial(n,k)^2, where i is the imaginary unit.
a(n) = Sum_{k=0..floor(n/2)} 5^k * 4^(n-2*k) * binomial(n,2*k) * binomial(2*k,k).
a(n) = [x^n] (1+4*x+5*x^2)^n. (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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A058962 a(n) = 2^(2*n)*(2*n+1).

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A105199 Decimal expansion of arctan(2).

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A034910 One quarter of octo-factorial numbers.

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A105531 Decimal expansion of arctan(1/3).

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A058962 a(n) = 2^(2n)(2*n+1).

A098443 Expansion of 1/sqrt(1-8x-4x^2).

A254619 a(n) = 4^n(2n + 1)!/n!.