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 1-5 of 5 results.

A240242 Decimal expansion of Integral_(x=1..c) (log(x)/(1+x))^2 dx, where c = A141251 = e^(LambertW(1/e)+1) corresponds to the maximum of the function.

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Apr 03 2014

Keywords

Examples

			0.13847532425848...

Links

Crossrefs

Cf. A013661, A141251, A195055, A240243.

Programs

  • Mathematica
    w = ProductLog[1/E]; Pi^2/6 - w - w^2 - 2*Log[1+w]*(1+w) + 2*PolyLog[2, -w] // RealDigits[#, 10, 100]& // First
  • PARI
    (w -> Pi^2/6 - w - w^2 - 2*(1+w)*log(1+w) + 2*polylog(2, -w))(lambertw(exp(-1))) \\ Charles R Greathouse IV, Aug 27 2014

Formula

A195055 = A013661 + A240242 + A240243.
(log(c)/(1+c))^2 = (LambertW(1/e))^2 = 0.0775425...

A240243 Decimal expansion of Integral_(x=c..infinity) (log(x)/(1+x))^2 dx, where c = A141251 = e^(LambertW(1/e)+1) corresponds to the maximum of the function.

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Apr 03 2014

Keywords

Examples

			1.50645874258974594605808179809250890162965990098722060615212114365006356...

Links

Crossrefs

Cf. A013661, A141251, A195055, A240242.

Programs

  • Mathematica
    w = ProductLog[1/E]; w + w^2 + 2 *Log[1+w]*(1+w) - 2*PolyLog[2, -w] // RealDigits[#, 10, 100]& // First
  • PARI
    (w -> w + w^2 + 2*(1+w)*log(1+w) - 2*polylog(2, -w))(lambertw(exp(-1)))

Formula

A195055 = A013661 + A240242 + A240243.

A141252 Decimal expansion of number c satisfying c*log(c)=1/2+c.

Original entry on oeis.org

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

Views

Author

David Applegate and N. J. A. Sloane, Sep 04 2008

Keywords

Examples

			3.1809660866389432106835864035475750051608787493965754320539648...

References

  • Michel Goemans and Jon Kleinberg, An improved approximation ratio for the minimum latency problem, Proc. 7th ACM-SIAM Symposium on Discrete Algorithms, 1996, pp. 152-158.

Crossrefs

Cf. A141251.

Programs

  • Mathematica
    RealDigits[ Exp[1 + ProductLog[1/(2*E)]], 10, 99] // First (* Jean-François Alcover, Feb 19 2013 *)

Formula

Equals exp(LambertW(1/(2e))+1).

A143934 Continued fraction expansion of exp(LambertW(1/e)+1).

Original entry on oeis.org

3, 1, 1, 2, 4, 9, 1, 1, 292, 20, 1, 1, 2, 1, 8, 1, 13, 2, 2, 1, 3, 2, 16, 11, 5, 1, 1, 1, 1, 1, 3, 8, 1, 5, 5, 1, 7, 1, 41, 2, 1, 1, 2, 1, 13, 1, 6, 3, 33, 1, 245, 2, 1, 22, 1, 2, 5, 1, 1, 1, 10, 1, 1, 2, 1, 4, 1, 2, 1, 19, 17, 1, 3, 2, 1, 15, 2, 2, 2, 1, 1, 1, 4, 17, 9, 3, 2, 1, 1, 1, 21, 1, 2, 1, 1, 5, 1
Offset: 0

Views

Author

Ross La Haye, Sep 05 2008

Keywords

Links

Crossrefs

Cf. A141251 (decimal expansion).

Programs

  • Mathematica
    ContinuedFraction[E^(ProductLog[1/E] + 1), 111]
  • PARI
    contfrac(exp(lambertw(1/exp(1))+1)) \\ Michel Marcus, Nov 13 2017

Extensions

Offset changed by Andrew Howroyd, Aug 09 2024

A234604 Floor of the solutions to c = exp(1 + n/c) for n >= 0, using recursion.

Original entry on oeis.org

2, 3, 4, 4, 5, 6, 6, 7, 17, 35, 62, 103, 164, 256, 391, 589, 880, 1303, 1919, 2814, 4112, 5993, 8716, 12655, 18353, 26591, 38499, 55710, 80583, 116523, 168453, 243485, 351889, 508506, 734776, 1061672, 1533938, 2216216
Offset: 0

Views

Author

Richard R. Forberg, Dec 28 2013

Keywords

Comments

For n = 1 to 7 recursion produces convergence to single valued solutions.
For n >= 8 a dual-valued oscillating recursion persists between two stable values. The floor of the upper value for each n is included here. (The lower values of c are under 6 and approach exp(1) = 2.71828 for large n.)
At large n, the ratio of a(n)/a(n-1) approaches exp(1/exp(1)) = 1.444667861009 with more digits given by A073229.
At n = 0, c = exp(1).
At n = 1, c = 3.5911214766686 = A141251.
At n = 2, c = 4.3191365662914
At n = 3, c = 4.9706257595442
At n = 4, c = 5.5723925978776
At n = 5, c = 6.1383336446072
At n = 6, c = 6.6767832796664
At n = 7, c = 7.1932188286406
The convergence becomes "dual-valued" at n > exp(2) = 7.3890560989 = A072334.
At values of n = 7 and 8 the convergence is noticeably slower than at either larger or smaller values of n.
The recursion at n = exp(2) is only "quasi-stable" where c reluctantly approaches exp(2) = exp(1 + exp(2)/exp(2)) from any starting value, but never reaches it, and is not quite able to hold it if given the solution, due to machine rounding errors.

Crossrefs

Cf. A073229, A141251.

Formula

a(n) = floor(c) for the solutions to c = exp(1 + n/c) at n = 0 to 7, and the floor of the stable upper values of c for n >= 8.
Conjecture: a(n) = floor(e^(-e^(t^2/e^t - t)*t^2 + t + 1)) for all n > 13. - Jon E. Schoenfield, Jan 11 2014

Extensions

Corrected and edited by Jon E. Schoenfield, Jan 11 2014
Showing 1-5 of 5 results.

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