A206948 Number of nonisomorphic graded posets with 0 and non-uniform Hasse graph of rank n, with exactly 2 elements of each rank level above 0.

0, 0, 0, 2, 19, 131, 791, 4446, 23913, 124892, 638878, 3218559, 16027375, 79093773, 387540260, 1887974063, 9154751912, 44221373872, 212931964415, 1022594028515, 4900116587043, 23437066655010, 111923110602497

Offset: 0

Author

David Nacin, Feb 13 2012

Keywords

• nonn
• easy

Comments

We do not assume all maximal elements have maximal rank and thus use graded poset to mean: For every element x, all maximal chains among those with x as greatest element have the same finite length. Here, the term uniform used in the sense of Retakh, Serconek and Wilson.

References

• R. Stanley, Enumerative combinatorics. Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• V. Retakh, S. Serconek, and R. Wilson, Hilbert Series of Algebras Associated to Directed Graphs and Order Homology, arXiv:1010.6295 [math.RA], 2010-2011.
• Wikipedia, Graded poset
• Index entries for linear recurrences with constant coefficients, signature (11, -40, 55, -30, 6).

Crossrefs

a(n) = A086405(n) - A012781(n+1).

Cf. A206947 (unique maximal element added).

Cf. A206949, A206950 (allowing one or two elements in each rank level above 0 with and without maximal element).

Programs

Mathematica

LinearRecurrence[{11, -40, 55, -30, 6}, {0, 0, 0, 2, 19, 131}, 23] (* David Nacin, Feb 29 2012; a(0) added by Georg Fischer, Apr 03 2019 *)

Python

def a(n, adict={0:0, 1:0, 2:0, 3:2, 4:19, 5:131}):
  if n in adict:
    return adict[n]
  adict[n]=11*a(n-1)-40*a(n-2)+55*a(n-3)-30*a(n-4)+6*a(n-5)
  return adict[n]
for n in range(0,40):
  print(a(n))

Formula

a(n) = 11*a(n-1) - 40*a(n-2) + 55*a(n-3) - 30*a(n-4) + 6*a(n-5), a(0)=0, a(1)=0, a(2)=0, a(3)=2, a(4)=19, a(5)=131.

G.f.: (x^3*(2 - 3*x + 2*x^2))/((1 - 6*x + 6*x^2)*(1 - 5*x + 4*x^2 - x^3)).