

A260332


Labelings of n diamondshaped posets with 4 vertices per diamond where the labels follow the poset relations whose associated reading permutation avoids 231 in the classical sense.


19




OFFSET

0,2


COMMENTS

According to YangJiang (2021) these are the 5Schroeder numbers. If confirmed, this will prove Michael Weiner's conjectures and enable us to extend the sequence. Yang & Jiang (2021) give an explicit formula for the mSchroeder numbers in Theorem 2.4.  N. J. A. Sloane, Mar 28 2021
By diamondshaped poset with 4 vertices, we mean a poset on four elements with e_1 < e_2, e_1 < e_3, e_2 < e_4, e_3 < e_4, and no order relations between e_2 and e_3. In the Hasse diagram for such a poset, we have a least element, two elements in the level above, and one element in the top level, so the diagram resembles a diamond. The associated permutation is the permutation obtained by reading the labels of each poset by levels left to right, starting with the least element.
Also the number of labelings of n diamondshaped posets with 4 vertices per diamond where the labels follow the poset relations whose associated reading permutation avoids 312 in the classical sense via reverse complement Wilf equivalence.
Conjecture: Also the number of lattice paths (Schroeder paths) from (0,0) to (n,4n) with unit steps N=(0,1), E=(1,0) and D=(1,1) staying weakly above the line y = 4x.  Michael D. Weiner, Jul 24 2019


REFERENCES

ShengLiang Yang and Meiyang Jiang, The mSchröder paths and mSchröder numbers, Disc. Math. (2021) Vol. 344, Issue 2, 112209. doi:10.1016/j.disc.2020.112209. See Table 1.


LINKS



FORMULA

There is a complicated recursive formula available in Paukner et al.
Yang & Jiang (2021) give an explicit formula for the 5Schroeder numbers in Theorem 2.4.  N. J. A. Sloane, Mar 28 2021
Conjecture: a(n) = Sum_{k=1..n} binomial(n,k)*binomial(4*n,k1)*2^k/n for n > 0.  Michael D. Weiner, Jul 23 2019
Conjectures: 1) the g.f. A(x) = 1 + 2*x + 18*x^2 + 226*x^3 + ... satisfies A(x)^4 = (1/x) * the series reversion of ((1  x)/(1 + x))^4.
2) Define b(n) = (1/4) * [x^n] ((1 + x)/(1  x))^(4*n). Then A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ).
3) a(n) = 2 * hypergeom([1  n, 4*n], [2], 2) for n >= 1 (equivalent to Weiner's conjecture above).
4) [x^n] A(x)^n = (2*n) * hypergeom([1  n, 1  5*n], [2], 2) for n >= 1. (End)


EXAMPLE

For a single diamond (n=1) poset with 4 vertices, we must label the least element 1 and the greatest element 4, and the two central elements can be labeled either 2, 3 or 3, 2 respectively. Thus the associated permutations are 1234 and 1324.


CROSSREFS



KEYWORD

nonn,more


AUTHOR



STATUS

approved



