A139262 Total number of two-element anti-chains over all ordered trees on n edges.

0, 0, 1, 8, 47, 244, 1186, 5536, 25147, 112028, 491870, 2135440, 9188406, 39249768, 166656772, 704069248, 2961699667, 12412521388, 51854046982, 216013684528, 897632738722, 3721813363288, 15401045060572, 63616796642368, 262357557683422, 1080387930269464, 4443100381114156

Offset: 0

Author

Lifoma Salaam, Apr 12 2008

Keywords

• nonn

Comments

From Miklos Bona, Mar 04 2009: (Start)

This is the same as the total number of inversions in all 132-avoiding permutations of length n by the well-known bijection between ordered trees on n edges and such permutations.

For example, there are five permutations of length three that avoid 132, namely, 123, 213, 231, 312, and 321. Their numbers of inversions are, respectively, 0,1,2,2, and 3, for a total of eight inversions.

(End)

Appears to be a shifted version of A029760. - R. J. Mathar, Mar 30 2014

a(n) is the number of total East steps below y = x-1 of all North-East paths from (0,0) to (n,n). Details can be found in Section 3.1 and Section 5 in Pan and Remmel's link. - Ran Pan, Feb 01 2016

Examples

			a(3) = 8 because there are 5 ordered trees on 3 edges and two of the trees have 2 two-element anti-chain each, one of the trees has three two element anti-chains, one of the trees has one two element anti-chain and the last tree does not have any two-element anti-chains. Hence in ordered trees on 3 edges there are a total of (2)(2)+1(3)+1(1) = 8 two element anti-chains.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..1650
• Miklós Bóna, Surprising Symmetries in Objects Counted by Catalan Numbers, Electronic J. Combin., 19 (2012), #P62, eq. (2).
• Ran Pan, Jeffrey B. Remmel, Paired patterns in lattice paths, arXiv:1601.07988 [math.CO], 2016.
• Lifoma Salaam, Combinatorial statistics on phylogenetic trees, Ph.D. Dissertation, Howard University, Washington D.C., 2008.
• Sittipong Thamrongpairoj, Dowling Set Partitions, and Positional Marked Patterns, Ph. D. Dissertation, University of California-San Diego (2019).

Crossrefs

Cf. A129182, A139263.

Programs

Maple

0, seq((n+1)*(2*n-1)!/(n!*(n-1)!) - 2^(2*n-1), n=1..30); # Robert Israel, Feb 02 2016

Mathematica

a[0] = 0; a[n_] := (n+1)(2n-1)!/(n! (n-1)!) - 2^(2n-1);
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Aug 19 2018, from Maple *)

Formula

G.f.: (up to offset): A = x^2*(B^3)*(C^2), where B is the generating function for the central binomial coefficients and C is the generating function for the Catalan numbers. Thus A = x^2*(1/sqrt(1-4*x))^3*((1-sqrt(1-4*x))/2*x)^2.

2*a(n) = (n+1)*A000984(n) - 4^n. - J. M. Bergot, Feb 02 2013

Conjecture: n*(n-2)*a(n) +2*(-4*n^2+9*n-3)*a(n-1) +8*(n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Feb 03 2013

The above conjecture follows easily from the formula by J. M. Bergot. - Robert Israel, Feb 02 2016

a(n) = Sum_{k=0..n^2} (n^2-k)/2 * A129182(n,k). - Alois P. Heinz, Mar 31 2018

Extensions

Terms beyond a(9) added by Joerg Arndt, Dec 30 2012