cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A016655 Decimal expansion of log(32) = 5*log(2).

Original entry on oeis.org

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

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Author

Keywords

Comments

The function exp(x) has its maximum curvature where x = -(1/10)*5*log(2) = -log(2)/2 = 0.34657... - Dimitri Papadopoulos, Oct 27 2022
This maximum curvature occurs at the point with coordinates (x_M = -log(2)/2 = -(this constant)/10; y_M = sqrt(2)/2 = A010503) and is equal to 2*sqrt(3)/9 = A212886. - Bernard Schott, Dec 23 2022

Examples

			3.465735902799726547086160607290882840377500671801276270603400047466968...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 2.

Crossrefs

Cf. A000045, A000032, A060851, A195909, A195913, A195697, A016460 (continued fraction).

Programs

Formula

log(2)/2 = (1 - 1/2 - 1/4) + (1/3 - 1/6 - 1/8) + (1/5 - 1/10 - 1/12) + ... [Jolley, Summation of Series, Dover (1961) eq (73)]
Equals 10*log(2)/2 = 5*log(2) = 5*A002162, so 10*(1/2 - 1/4 + 1/6 - 1/8 + 1/10 - ... + (-1)^(k+1)/(2*k) + ...) = log(32). - Eric Desbiaux, Nov 26 2008
-log(2)/2 = lim_{n->oo} ((Sum_{k=2..n} arctanh(1/k)) - log(n)). - Jean-François Alcover, Aug 07 2014, after Steven Finch
Equals log(sqrt(2)) with offset 0. - Michel Marcus, Feb 19 2017
Equals (5/4)*Sum_{k=1..4} (-1)^(k+1) gamma(0, k/4) where gamma(n,x) denotes the generalized Stieltjes constants. - Peter Luschny, May 16 2018
From Amiram Eldar, Jun 29 2020: (Start)
log(2)/2 = arctanh(1/3) = arcsinh(1/sqrt(8)).
log(2)/2 = Integral_{x=0..Pi/4} tan(x) dx.
log(2)/2 = Sum_{k>=0} (-1)^k/(2*k+2).
log(2)/2 = Sum_{k>=1} 1/A060851(k). (End)
log(2)/2 = Sum_{k>=1} (-1)^(k+1) * arctanh(Lucas(2*k+3)/Fibonacci(2*k+3)^2) (Melham and Shannon, 1995). - Amiram Eldar, Jan 15 2022
Equals 10 * Integral_{1..oo} dx/(x*(1+x^2)). [Nahin] - R. J. Mathar, May 22 2024
Equals -10*Integral_{q=0..1} q*log(sin(Pi*q))dq. [Espinosa] - R. J. Mathar, Aug 13 2024
log(2)/2 = Sum_{k>=2} (-1)^(k) * arccoth(k). - Antonio Graciá Llorente, Sep 16 2024
-0.34657359... = Sum_{k>=0} zeta(2k)/((2k+1)*2^(2k)), [Srivastava (2.20)] - R. J. Mathar, Feb 12 2020
Equals 10*Integral_{x=0..1} Ei((1 + sqrt(2))*log(x)) - li(x) dx, where Ei is the exponential integral and li is the logarithmic integral. - Kritsada Moomuang, Jun 06 2025

A155988 a(n) = (2*n + 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

Views

Author

Jaume Oliver Lafont, Feb 01 2009

Keywords

Crossrefs

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

Programs

  • Magma
    [(2*n+1)*9^n: n in [0..20]]; // Vincenzo Librandi, Jun 08 2011
    
  • Maxima
    makelist((2*n+1)*9^n, n, 0, 20); /* Martin Ettl, Nov 11 2012 */
  • PARI
    a(n)=(2*n+1)*9^n;
    

Formula

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

A069284 Decimal expansion of li(2) = gamma + log(log(2)) + Sum_{k>=1} log(2)^k / ( k*k! ).

Original entry on oeis.org

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

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Author

Frank Ellermann, Mar 13 2002

Keywords

Comments

From Mats Granvik, Jun 14 2013: (Start)
The logarithmic integral li(x) = exponential integral Ei(log(x)).
The generating function for tau A000005, the number of divisors of n is: Sum_{n >= 1} a(n) x^n = Sum_{k > 0} x^k/(1 - x^k). Another way to write the generating function for tau A000005 is Sum_{n>=1} A000005(n) x^n = Sum_{a=1..Infinity} Sum_{b>=1} x^(a*b).
If we instead think of the integral with the same form, evaluate at x = exp(1) = 2.7182818284... = A001113 and set the integration limits to zero and sqrt(log(n)), we get for n >= 0:
Logarithmic integral li(n) = Integral_{a = 0..sqrt(log(n))} Integral_{b=0..sqrt(log(n))} exp(1)^(a*b) + EulerGamma + log(log(n)). (End)
li(2)-1 is the minimum [known to date, for n>1] of |li(n) - PrimePi(n)|. - Jean-François Alcover, Jul 10 2013
The modern logarithmic integral function li(x) = Integral_{t=0..x} (1/log(t)) replaced the Li(x) = Integral_{t=2..x} (1/log(t)) which was sometimes used because it avoids the singularity at x=1. This constant is the offset between the two functions: li(2) = li(x) - Li(x) = Integral_{t=0..2} (1/log(t)). - Stanislav Sykora, May 09 2015

Examples

			1.0451637801174927848445888891946131365226155781512015758329...
		

References

  • S. R. Finch, Mathematical Constants, Cambridge, 2003, p. 425.

Crossrefs

Cf. A069285 (continued fraction), A057754, A057794, A060851.
Euler's constant gamma: A001620, log(2): A002162, k*k!: A001563.

Programs

Extensions

Replaced several occurrences of "Li" with "li" in order to enforce current conventions. - Stanislav Sykora, May 09 2015
Showing 1-3 of 3 results.