A056940 Number of antichains (or order ideals) in the poset 4*m*n or plane partitions with at most m rows and n columns and entries <= 4.
1, 1, 1, 1, 5, 1, 1, 15, 15, 1, 1, 35, 105, 35, 1, 1, 70, 490, 490, 70, 1, 1, 126, 1764, 4116, 1764, 126, 1, 1, 210, 5292, 24696, 24696, 5292, 210, 1, 1, 330, 13860, 116424, 232848, 116424, 13860, 330, 1, 1, 495, 32670, 457380, 1646568, 1646568, 457380, 32670, 495, 1
Offset: 0
Examples
Triangle begins as: 1. 1, 1. 1, 5, 1. 1, 15, 15, 1. 1, 35, 105, 35, 1. 1, 70, 490, 490, 70, 1. 1, 126, 1764, 4116, 1764, 126, 1. 1, 210, 5292, 24696, 24696, 5292, 210, 1. 1, 330, 13860, 116424, 232848, 116424, 13860, 330, 1. - _Roger L. Bagula_, Mar 08 2010
Links
- G. C. Greubel, Rows n = 0..100 of triangle, flattened
- Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
- J. Berman and P. Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), 103-124. [Annotated scanned copy]
- Johann Cigler, Pascal triangle, Hoggatt matrices, and analogous constructions, arXiv:2103.01652 [math.CO], 2021.
- Johann Cigler, Some observations about Hoggatt triangles, Universität Wien (Austria, 2021).
- P. A. MacMahon, Combinatory analysis, sect. 495, 1916.
- R. P. Stanley, Theory and application of plane partitions, II. Studies in Appl. Math. 50 (1971), p. 259-279. DOI:10.1002/sapm1971503259. Thm. 18.1.
- Index entries for sequences related to posets
Crossrefs
Programs
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Mathematica
c[n_, q_] = Product[i + j, {j, 0, q}, {i, 1, n}]; T[n_, m_, q_] = c[n, q]/(c[m, q]*c[n - m, q]); Table[T[n, k, 3], {n, 0, 10}, {k, 0, n}]//Flatten (* Roger L. Bagula, Mar 08 2010 *)(* modified by G. C. Greubel, Apr 13 2019 *)
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PARI
A056940(n,m)=prod(k=0,3,binomial(n+m+k,m+k)/binomial(n+k,k)) \\ M. F. Hasler, Sep 26 2018
Formula
Product_{k=0..3} C(n+m+k, m+k)/C(n+k, k) gives the array as a square.
T(n,m,q) = c(n,q)/(c(m,q)*c(n-m,q)) with c(n,q) = Product_{i=1..n, j=0..q} (i + j), q = 3. - Roger L. Bagula, Mar 08 2010
From Peter Bala, Oct 13 2011: (Start)
T(n-1,k-1)*T(n,k+1)*T(n+1,k) = T(n-1,k)*T(n,k-1)*T(n+1,k+1).
Extensions
Edited by M. F. Hasler, Sep 26 2018
Comments