OFFSET
0,3
COMMENTS
Number of ways to use the elements of {1,..,n} once each to form a sequence of lists, each having length at least 2. - Bob Proctor, Apr 19 2005
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..415
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 493
Robert A. Proctor, Let's Expand Rota's Twelvefold Way For Counting Partitions!, arXiv:math/0606404 [math.CO], 2006-2007.
FORMULA
a(n) = n*a(n-1) + n*(n-1)*a(n-2), with a(0)=1, a(1)=0.
a(n) = Sum(1/5*(-1+3*_alpha)*_alpha^(-1-n), _alpha=RootOf(-1+_Z+_Z^2))*n!.
a(n) = n!*Fibonacci(n-1) for n >= 1. - Bob Proctor, Apr 19 2005
a(0) = 1; a(n) = Sum_{k=2..n} binomial(n,k) * k! * a(n-k). - Ilya Gutkovskiy, Feb 09 2020
a(n)=n!*A212804(n). - R. J. Mathar, Jun 03 2022
MAPLE
spec := [S, {S=Sequence(Prod(Z, Z, Sequence(Z)))}, labeled]: seq(combstruct[count](spec, size=n), n=0..20);
MATHEMATICA
With[{m=20}, CoefficientList[Series[(1-x)/(1-x-x^2), {x, 0, m}], x]* Range[0, m]!] (* G. C. Greubel, May 07 2019 *)
PROG
(PARI) my(x='x+O('x^20)); Vec(serlaplace( (1-x)/(1-x-x^2) )) \\ G. C. Greubel, May 07 2019
(Magma) [1] cat [Fibonacci(n-1)*Factorial(n): n in [1..20]] // G. C. Greubel, May 07 2019
(Sage) [1]+[fibonacci(n-1)*factorial(n) for n in (1..20)] # G. C. Greubel, May 07 2019
(GAP) a:=[0, 2];; for n in [3..20] do a[n]:=n*a[n-1]+n*(n-1)*a[n-2]; od; Concatenation([1], a); # G. C. Greubel, May 07 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
encyclopedia(AT)pommard.inria.fr, Jan 25 2000
STATUS
approved