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.

A002742 Logarithmic numbers.

Original entry on oeis.org

2, -6, 24, -80, 450, -2142, 17696, -112464, 1232370, -9761510, 132951192, -1258797696, 20476388114, -225380451870, 4261074439680, -53438049741152, 1151146814425506, -16199301256675974, 391615698778725080, -6109914386833902960
Offset: 1

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Comments

From Peter Bala, Sep 06 2022: (Start)
Conjectures: Let k be a positive integer.
1) for n >= 1, a(n+2*k) - a(n) is divisible by 2*k; if true, then the reduction of the sequence modulo 2*k gives a periodic sequence with period dividing 2*k.
2) for n >= 1, a(n+2*k+1) + a(n) is divisible by 2*k+1; if true, then the reduction of the sequence modulo 2*k+1 gives a periodic sequence with period dividing 4*k + 2. (End)

References

  • J. M. Gandhi, On logarithmic numbers, Math. Student, 31 (1963), 73-83.
  • 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).

Crossrefs

Cf. A002741.

Programs

  • Mathematica
    Table[(-1)^(n-1)Sum[Binomial[n+1,2k+1](n-2k)/(k+1)(2k+1)!,{k,0,n}],{n,0,100}] (* Emanuele Munarini, Dec 16 2017 *)
  • Maxima
    makelist((-1)^(n-1)*sum(binomial(n+1,2*k+1)*(n-2*k)/(k+1)*(2*k+1)!,k,0,n),n,0,12); /* Emanuele Munarini, Dec 16 2017 */
    
  • PARI
    first(n) = x='x+O('x^(n+1)); Vec(serlaplace((2*x/(1-x^2)+log(1-x^2))*exp(-x))) \\ Iain Fox, Dec 16 2017

Formula

E.g.f.: (2*x/(1-x^2)+log(1-x^2))*exp(-x). - Sean A. Irvine, Aug 11 2014
a(n) = 2*A002747(n) - a(n-1). - R. J. Mathar, Jul 24 2015
From Emanuele Munarini, Dec 16 2017: (Start)
a(n) = (-1)^(n-1)*Sum_{k=0..n} binomial(n+1,2*k+1)*((n-2*k)/(k+1))*(2*k+1)!.
a(n+3)+a(n+2)-(n+2)*(n+3)*a(n+1)-(n+2)*(n+3)*a(n) = 2*(-1)^n*(n+3).
(n+3)*a(n+4)+(2*n+7)*a(n+3)-(n+2)*(n+4)^2*a(n+2)-(n+3)*(n+4)*(2*n+5)*a(n+1)-(n+2)*(n+3)*(n+4)*a(n) = 0.
E.g.f.: A(x) = - D(exp(-x)*log(1-x^2)), where D is the derivative with respect to x. (End)
a(n) ~ n! * (exp(-1) - (-1)^n * exp(1)). - Vaclav Kotesovec, Dec 16 2017

Extensions

More terms from Jeffrey Shallit
More terms from Sean A. Irvine, Aug 11 2014