A006531 Semiorders on n elements.

1, 1, 3, 19, 183, 2371, 38703, 763099, 17648823, 468603091, 14050842303, 469643495179, 17315795469063, 698171064855811, 30561156525545103, 1443380517590979259, 73161586346500098903, 3961555049961803092531, 228225249142441259147103, 13938493569348563803135339

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• nice
• easy

Comments

Labeled semiorders on n elements: (1+3) and (2+2)-free posets. - Detlef Pauly (dettodet(AT)yahoo.de), Dec 27 2002

Labeled incomplete binary trees (every vertex has a left child, a right child, neither, or both) in which every vertex with a right child but no left child has a label greater than the label of its right child. - Ira M. Gessel, Nov 09 2013

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
• R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.30.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• J. L. Chandon, J. LeMaire and J. Pouget, Dénombrement des quasi-ordres sur un ensemble fini, Math. Sci. Humaines, No. 62 (1978), 61-80.
• J. L. Chandon, J. LeMaire and J. Pouget, Enumeration of semiorders on a finite set, Preprint (English) of "Dénombrement des quasi-ordres sur un ensemble fini".
• J. L. Chandon, Letter to N. J. A. Sloane, May 1981
• Julie Christophe, Jean-Paul Doignon and Samuel Fiorini, Counting Biorders, J. Integer Seqs., Vol. 6, 2003.
• Bérénice Delcroix-Oger and Clément Dupont, Lie-operads and operadic modules from poset cohomology, arXiv:2505.06094 [math.CO], 2025. See p. 28.
• Yan X. Zhang, Four Variations on Graded Posets, arXiv preprint arXiv:1508.00318 [math.CO], 2015.

Crossrefs

Cf. A000108 (unlabeled semiorders: Catalan numbers), A052895.

Programs

Maple

A006531 := n->add(stirling2(n,k)*k!*A001006(k-1),k=1..n);

Mathematica

m[n_] := m[n] = m[n-1] + Sum[ m[k]*m[n-k-2], {k, 0, n-2}]; m[0] = a[0] = 1; a[n_] := Sum[ StirlingS2[n, k]*k!*m[k-1], {k, 1, n}]; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Jul 24 2012, after Maple *)

PARI

{a(n)=polcoeff(sum(m=0, n, (2*m)!/(m+1)!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)} /* Paul D. Hanna, Jul 20 2011 */

Formula

E.g.f.: C(1-exp(-x)), where C(x) = (1 - sqrt(1 - 4*x)) / (2*x) is the ordinary g.f. for the Catalan numbers A000108. [corrected by Joel B. Lewis, Mar 29 2011]

a(n) = Sum_{k=1..n} S(n, k) * k! * M(k-1), S(n, k): Stirling number of the second kind, M(n): Motzkin number, A001006. - Detlef Pauly, Jun 06 2002

O.g.f.: Sum_{n>=1} (2*n)!/(n+1)! * x^n / Product_{k=0..n} (1+k*x). - Paul D. Hanna, Jul 20 2011

a(n) ~ n! * sqrt(3)*(log(4/3))^(1/2-n)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 13 2013

E.g.f.: 1/(1 + (exp(-x) - 1)/(1 + (exp(-x) - 1)/(1 + (exp(-x) - 1)/(1 + (exp(-x) - 1)/(1 + ...))))), a continued fraction. - Ilya Gutkovskiy, Nov 18 2017

From Peter Bala, Jan 15 2018: (Start)

a(n) = Sum_{k = 0..n} (-1)^(n+k)*Catalan(k)*k!*Stirling2(n,k). Cf. A052895.

Conjecture: for fixed k = 1,2,..., the sequence a(n) (mod k) is eventually periodic with the exact period dividing phi(k), where phi(k) is Euler's totient function A000010. For example, modulo 10 the sequence becomes (1, 1, 3, 9, 3, 1, 3, 9, 3, ...) with an apparent period 1, 3, 9, 3 of length 4 = phi(10) beginning at a(1). (End)

Consider the transformation of a sequence u given by T(u)(m) = (-1)^m*Sum_{n=0..m} (u(n)/(n+1))*(Sum_{k=0..n}(-1)^k*binomial(n,k)*k^m). If u(n) = 1 then T(u)(n) = Bernoulli(n) (with Bernoulli(1) = 1/2), if u(n) = binomial(2*n,n) then T(u)(n) = a(n). - Peter Luschny, Jul 09 2020