OFFSET
0,4
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
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).
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
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Lifoma Salaam, Apr 12 2008
EXTENSIONS
Terms beyond a(9) added by Joerg Arndt, Dec 30 2012
STATUS
approved