A276890
Number A(n,k) of ordered set partitions of [n] such that for each block b the smallest integer interval containing b has at most k elements; square array A(n,k), n>=0, k>=0, read by antidiagonals.
Original entry on oeis.org
1, 1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 3, 6, 0, 1, 1, 3, 10, 24, 0, 1, 1, 3, 13, 44, 120, 0, 1, 1, 3, 13, 62, 234, 720, 0, 1, 1, 3, 13, 75, 352, 1470, 5040, 0, 1, 1, 3, 13, 75, 466, 2348, 10656, 40320, 0, 1, 1, 3, 13, 75, 541, 3272, 17880, 87624, 362880, 0
Offset: 0
Square array A(n,k) begins:
1, 1, 1, 1, 1, 1, 1, 1, ...
0, 1, 1, 1, 1, 1, 1, 1, ...
0, 2, 3, 3, 3, 3, 3, 3, ...
0, 6, 10, 13, 13, 13, 13, 13, ...
0, 24, 44, 62, 75, 75, 75, 75, ...
0, 120, 234, 352, 466, 541, 541, 541, ...
0, 720, 1470, 2348, 3272, 4142, 4683, 4683, ...
0, 5040, 10656, 17880, 26032, 34792, 42610, 47293, ...
Columns k=0-10:
A000007,
A000142,
A240172,
A276893,
A276894,
A276895,
A276896,
A276897,
A276898,
A276899,
A276900.
-
b:= proc(n, m, l) option remember; `if`(n=0, m!,
add(b(n-1, max(m, j), [subsop(1=NULL, l)[],
`if`(j<=m, 0, j)]), j={l[], m+1} minus {0}))
end:
A:= (n, k)-> `if`(k=0, `if`(n=0, 1, 0),
`if`(k=1, n!, b(n, 0, [0$(k-1)]))):
seq(seq(A(n, d-n), n=0..d), d=0..12);
-
b[n_, m_, l_List] := b[n, m, l] = If[n == 0, m!, Sum[b[n - 1, Max[m, j], Append[ReplacePart[l, 1 -> Nothing], If[j <= m, 0, j]]], {j, Append[l, m + 1] ~Complement~ {0}}]]; A[n_, k_] := If[k==0, If[n==0, 1, 0], If[k==1, n!, b[n, 0, Array[0&, k-1]]]]; Table[A[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Jan 06 2017, after Alois P. Heinz *)
A129847
a(n) = number of set partitions of {1, 2, ..., n} whose blocks consist only of elements that differ by two or less (that is, have only the forms {i}, {i,i+1}, {i,i+2}, or {i,i+1,i+2}).
Original entry on oeis.org
1, 1, 2, 5, 10, 20, 42, 87, 179, 370, 765, 1580, 3264, 6744, 13933, 28785, 59470, 122865, 253838, 524428, 1083466, 2238435, 4624595, 9554390, 19739321, 40781336, 84254032, 174068400, 359624425, 742982225, 1534997482, 3171296957, 6551883314, 13536157460
Offset: 0
Rhodes Peele (rpeele(AT)mail.aum.edu), May 22 2007
a(4) = 10, with set partitions {{1},{2},{3}, {4}}; {{1,2}, {3}, {4}}; {{1,3}, {2}, {4}}; {{1}, {2,3}, {4}}; {{1}, {2,4}, {3}}; {{1}, {2}, {3,4}}; {{1,2,3}, {4}}; {{1}, {2,3,4}}; {{1,2}, {3,4}} and {{1,3}, {2,4}}.
- Herbert S. Wilf, Generatingfunctiononogy (2nd Edition), Academic Press 1990, pp. 1 - 10 and pp. 20 - 23.
- Arthur T. Benjamin and Jennifer J. Quinn, Proofs that Really Count, Dolciani Mathematical Expositions (MAA), (2003), p. 1. (A relevant combinatorial interpretation of Fibonacci numbers.)
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Kassie Archer, Ethan Borsh, Jensen Bridges, Christina Graves, and Millie Jeske, Cyclic permutations avoiding patterns in both one-line and cycle forms, arXiv:2312.05145 [math.CO], 2023. See p. 2.
- B. Duncan and R. Peele, Bell and Stirling numbers for graphs, JIS 12 (2009), Article 09.7.1.
- W. Kuszmaul, Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations, arXiv preprint arXiv:1509.08216 [cs.DM], 2015.
- W. Kuszmaul, Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations, Mathematics of Computation 87 (2018), 987-1011.
-
a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <1|2|1|1>>^n)[4, 4]:
seq(a(n), n=0..35); # Alois P. Heinz, Sep 15 2016
-
a[1] = 1; a[2] = 2; a[3] = 5; a[4] = 10
a[n_] := a[n] = a[n-1] + a[n-2] + 2 a[n-3] + a[n-4]
Table[a[n],{n,1,30}]
Showing 1-2 of 2 results.
Comments