OFFSET
0,2
COMMENTS
According to Yang-Jiang (2021) these are the 5-Schroeder 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 m-Schroeder numbers in Theorem 2.4. - N. J. A. Sloane, Mar 28 2021
By diamond-shaped 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 diamond-shaped 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
Sheng-Liang Yang and Mei-yang Jiang, The m-Schröder paths and m-Schröder numbers, Disc. Math. (2021) Vol. 344, Issue 2, 112209. doi:10.1016/j.disc.2020.112209. See Table 1.
LINKS
M. Paukner, L. Pepin, M. Riehl, and J. Wieser, Pattern Avoidance in Task-Precedence Posets, arXiv:1511.00080 [math.CO], 2015.
Lin Yang, Yu-Yuan Zhang, and Sheng-Liang Yang, The halves of Delannoy matrix and Chung-Feller properties of the m-Schröder paths, Linear Alg. Appl. (2024).
Sheng-liang Yang and Mei-yang Jiang, Pattern avoiding problems on the hybrid d-trees, J. Lanzhou Univ. Tech., (China, 2023) Vol. 49, No. 2, 144-150. (in Mandarin)
FORMULA
There is a complicated recursive formula available in Paukner et al.
Yang & Jiang (2021) give an explicit formula for the 5-Schroeder 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,k-1)*2^k/n for n > 0. - Michael D. Weiner, Jul 23 2019
From Peter Bala, Jun 16 2023: (Start)
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
Manda Riehl, Jul 29 2015
STATUS
approved