A029767 a(n) = (n-1)!*(2^n-1) for n>=1, a(0)=0.
0, 1, 3, 14, 90, 744, 7560, 91440, 1285200, 20603520, 371226240, 7428153600, 163459296000, 3923502105600, 102017281766400, 2856571067750400, 85698439706880000, 2742370993410048000, 93240969463369728000, 3356681303055015936000, 127554011161191014400000
Offset: 0
References
- François Bergeron, Gilbert Labelle, and Pierre Leroux, Combinatorial Species and Tree-Like Structures, Cambridge University Press, 1998, pp. 12, 55, 409.
- Richard P. Stanley, Enumerative Combinatorics, Cambridge University Press, Vol. 2, 1999; see Example 5.1.5.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..400
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 498.
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 777.
Crossrefs
Cf. A001865.
Programs
-
GAP
Concatenation([0],List([1..20],n->Factorial(n-1)*(2^n-1))); # Muniru A Asiru, Aug 09 2018
-
Magma
[0] cat [Factorial(n-1)*(2^n-1): n in [1..20]]; // Vincenzo Librandi, Apr 18 2015
-
Maple
with(combinat): seq(stirling1(j,1)*stirling2(j+1,2)*(-1)^(j+1), j=0..16); # Zerinvary Lajos, Mar 30 2007
-
Mathematica
a=x/(1-x); Range[0,20]! CoefficientList[Series[Log[1/(1-a)], {x,0,20}], x] (* Geoffrey Critzer, Dec 07 2011 *) Join[{0}, Table[(n - 1)! (2^n - 1), {n, 20}]] (* Vincenzo Librandi, Apr 18 2015 *)
-
PARI
concat([0], for(n=1,25, print1((n-1)!*(2^n -1), ", "))) \\ G. C. Greubel, Jan 19 2017
Formula
E.g.f.: log(1-x)-log(1-2*x).
From Karol A. Penson, Oct 15 2002: (Start)
Representation as an infinite sum: a(n) = (1/2)*Sum_{k>=0} (n+k)!/((k+1)!*2^k), n >= 1.
Representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*(1/2)*exp(-x)/x*(2*exp(1/2*x)-2), n >= 1. (End)
D-finite with recurrence: a(n) +3*(-n+1)*a(n-1) +2*(n-1)*(n-2)*a(n-2) = 0. - R. J. Mathar, Jan 08 2013
a(n) = n!*Sum_{k=0..n-1} binomial(n-1,k)/(k+1). - J. M. Bergot, Jul 30 2015
a(n) = (1/zeta(n)) * Integral_{x=0..1} (log(1/x))^(n-1) / (sqrt(x) * (1-x)) dx. - Amrik Singh Nimbran, May 06 2018
a(n) = polygamma(n-1, 1/2) / ((-1)^n*zeta(n)) for n >= 2. - Amiram Eldar, Aug 08 2025
Comments