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.

A202543 Decimal expansion of the number x satisfying e^(x/2) - e^(-x/2) = 1.

Original entry on oeis.org

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

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Author

Clark Kimberling, Dec 21 2011

Keywords

Comments

See A202537 for a guide to related sequences. The Mathematica program includes a graph.
W. Gawronski et al. in their paper - see ref. below - obtained the asymptotics for the Chebyshev-Stirling numbers. In the algebraic description of the respective "asymptotic coefficients" the number x = 2*log phi, where phi is the golden section, play the central role. - Roman Witula, Feb 02 2015
Also two times the Lévy measure for the continued fraction of the golden section, i.e., A202543/log(2) is the mean number of bits gained from the next convergent of the continued fraction representation. (See also Dan Lascu in links.) - A.H.M. Smeets, Jun 06 2018

Examples

			0.9624236501192068949955178268487368462703686...
		

Crossrefs

Programs

  • Mathematica
    u = 1/2; v = 1/2;
    f[x_] := E^(u*x) - E^(-v*x); g[x_] := 1
    Plot[{f[x], g[x]}, {x, 0, 2}, {AxesOrigin -> {0, 0}}]
    r = x /. FindRoot[f[x] == g[x], {x, .9, 1}, WorkingPrecision -> 110]
    RealDigits[r]    (* A202543 *)
    RealDigits[ Log[ (3+Sqrt[5])/2], 10, 99] // First (* Jean-François Alcover, Feb 27 2013 *)
    RealDigits[ FindRoot[ Exp[x/2] == 1 +  Exp[-x/2] , {x, 0}, WorkingPrecision -> 128][[1, 2]]][[1]] (* Robert G. Wilson v, Jun 13 2018 *)
  • PARI
    2*asinh(1/2) \\ Michel Marcus, Jun 24 2018, after A002390

Formula

Equals 2*A002390. - A.H.M. Smeets, Jun 06 2018
From Amiram Eldar, Aug 21 2020: (Start)
Equals log(A104457) = log(1 + A001622).
Equals 2*arcsinh(1/2). [corrected by Georg Fischer, Jul 12 2021]
Equals Sum_{k>=0} (-1)^k*binomial(2*k,k)/((2*k+1)*16^k). (End)
Equals Pi*i + Sum_{k>=0} arctanh(phi^(2^k))/2^k, with phi = A001622 and i = sqrt(-1). - Antonio Graciá Llorente, Feb 13 2025

Extensions

Typo in name fixed by Jean-François Alcover, Feb 27 2013