OFFSET
1,1
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).
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..200
J. M. Gandhi, On logarithmic numbers, Math. Student, 31 (1963), 73-83. [Annotated scanned copy]
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
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 *)
PROG
(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
CROSSREFS
KEYWORD
sign
AUTHOR
EXTENSIONS
More terms from Jeffrey Shallit
More terms from Sean A. Irvine, Aug 11 2014
STATUS
approved