A260332 Labelings of n diamond-shaped posets with 4 vertices per diamond where the labels follow the poset relations whose associated reading permutation avoids 231 in the classical sense.
1, 2, 18, 226, 3298, 52450, 881970
Offset: 0
Examples
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.
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.
- Manda Riehl, A 231-avoiding diamond whose associated permutation is 1234.
- Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See p. 11.
- 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)
Crossrefs
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)
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