A154920 -id:A154920 - cp's OEIS frontend

A002391 Decimal expansion of natural logarithm of 3.

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

A165998 Denominators of Taylor series expansion of 1/(3x)log((1+x)/(1-x)^2).

Original entry on oeis.org

1, 6, 3, 12, 5, 18, 7, 24, 9, 30, 11, 36, 13, 42, 15, 48, 17, 54, 19, 60, 21, 66, 23, 72, 25, 78, 27, 84, 29, 90, 31, 96, 33, 102, 35, 108, 37, 114, 39, 120, 41, 126, 43, 132, 45, 138, 47, 144, 49, 150, 51, 156, 53, 162, 55, 168, 57, 174, 59, 180, 61, 186, 63, 192, 65, 198

Offset: 0

Jaume Oliver Lafont, Oct 03 2009

Numerators are all 1.
Setting x=1/3 into 1/(3*x)*log((1+x)/(1-x)^2) = Sum_{k>=0} x^k/((2-(-1)^k)*(k+1)),
log(3) = Sum_{k>=0} 1/((2-(-1)^k)*(k+1)*3^k) = Sum_{k>=0} (9/(2k+1)+1/(2k+2))/9^(k+1) is obtained.
It appears that this is also the first differences of the generalized decagonal numbers A074377. - Omar E. Pol, Sep 10 2011
It appears that this is also A005408 and positive terms of A008588 interleaved. - Omar E. Pol, May 28 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1004
Tian-Xiao He, Peter J.-S. Shiue, Zihan Nie, and Minghao Chen, Recursive sequences and Girard-Waring identities with applications in sequence transformation, Electronic Research Archive (2020) Vol. 28, No. 2, 1049-1062.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A005408, A008588, A074377, A154920.

[(2-(-1)^n)*(n+1): n in [0..350]]; // Vincenzo Librandi, Apr 04 2011

LinearRecurrence[{0,2,0,-1}, {1, 6, 3, 12}, 50] (* Vincenzo Librandi, Feb 22 2012 *)

PARI
```
a(n)=(2-(-1)^n)*(n+1)
```

G.f.: (1+6*x+x^2)/(1-x^2)^2.
a(n) = (2-(-1)^n)*(n+1) (see PARI's code by Jaume Oliver Lafont).
a(2n)= 2n+1. a(2n+1) = 6*(n+1). - R. J. Mathar, Apr 02 2011
With offset 1 this sequence is multiplicative (in fact, a generalized totient function): a(p^e) = p^e for any odd prime p and a(2^e) = 3*2^e for e >= 1. - Charles R Greathouse IV, Mar 09 2015
With offset 1, Dirichlet g.f.: zeta(s-1) * (1 + 2^(2-s)). - Amiram Eldar, Oct 25 2023

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

A155988 a(n) = (2n + 1)9^n.

Original entry on oeis.org

1, 27, 405, 5103, 59049, 649539, 6908733, 71744535, 731794257, 7360989291, 73222472421, 721764371007, 7060738412025, 68630377364883, 663426981193869, 6382625094934119, 61149666232110753, 583701359488329915, 5553501505988967477, 52683216989246691471, 498464283821334080841

Offset: 0

Jaume Oliver Lafont, Feb 01 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200
David H. Bailey, A Compendium of BBP-Type Formulas for Mathematical Constants, 2017, page 14.
Xavier Gourdon and Pascal Sebah, Collection of formulas for log 2.
Index entries for linear recurrences with constant coefficients, signature (18,-81).

Cf. A058962 for the similar (2n+1)4^n.

Cf. A001019, A005408, A060851, A096949, A096950, A154920, A164985, A165132.

[(2*n+1)*9^n: n in [0..20]]; // Vincenzo Librandi, Jun 08 2011

makelist((2*n+1)*9^n, n, 0, 20); /* Martin Ettl, Nov 11 2012 */

PARI
```
a(n)=(2*n+1)*9^n;
```

G.f.: (1 + 9*x)/(1 - 9*x)^2.
a(n) = 18*a(n-1) - 81*a(n-2) for n>=2.
Sum_{n>=0} 1/a(n) = (3/2)*log(2).
a(n) = A005408(n) * A001019(n).
a(n) = (2*n - 1)*3^(2*n-1)/3 = A060851(n)/3.
Sum_{n>=0} (-1)^n/a(n) = 3*arctan(1/3). - Amiram Eldar, Feb 26 2022
E.g.f.: exp(9*x)*(1 + 18*x). - Stefano Spezia, May 07 2023

cp's OEIS Frontend

A002391 Decimal expansion of natural logarithm of 3.

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A165998 Denominators of Taylor series expansion of 1/(3*x)*log((1+x)/(1-x)^2).

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

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A155988 a(n) = (2*n + 1)*9^n.

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A164985 Denominators of ternary BBP-type series for log(5).

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A157024 a(0)=1, a(n) = (3n-1)*3n*(3n+1)/2 for n>0.

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A157718 Greedy Egyptian fraction expansion of log(3).

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A165998 Denominators of Taylor series expansion of 1/(3x)log((1+x)/(1-x)^2).

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

A155988 a(n) = (2n + 1)9^n.

A157024 a(0)=1, a(n) = (3n-1)3n(3n+1)/2 for n>0.