A349740 Number of partitions of set [n] in a set of <= k noncrossing subsets. Number of Dyck n-paths with at most k peaks. Both with 0 <= k <= n, read by rows.
1, 0, 1, 0, 1, 2, 0, 1, 4, 5, 0, 1, 7, 13, 14, 0, 1, 11, 31, 41, 42, 0, 1, 16, 66, 116, 131, 132, 0, 1, 22, 127, 302, 407, 428, 429, 0, 1, 29, 225, 715, 1205, 1401, 1429, 1430, 0, 1, 37, 373, 1549, 3313, 4489, 4825, 4861, 4862, 0, 1, 46, 586, 3106, 8398, 13690, 16210, 16750, 16795, 16796
Offset: 0
Examples
For n=4 the T(4,3)=13 partitions are {{1,2,3,4}}, {{1,2,3},{4}}, {{1,2,4},{3}}, {{1,3,4},{2}}, {{2,3,4},{1}}, {{1,2},{3,4}}, {{1,4},{2,3}}, {{1,2},{3},{4}}, {{1,3},{2},{4}}, {{1,4},{2},{3}}, {{1},{2,3},{4}}, {{1},{2,4},{3}}, {{1},{2},{3,4}}. The set of sets {{1,3},{2,4}} is missing because it is crossing. If you add the set of 4 sets, {{1},{2},{3},{4}}, you get T(4, 4) = 14 = A000108(4), the 4th Catalan number. Triangle begins: 1; 0, 1; 0, 1, 2; 0, 1, 4, 5; 0, 1, 7, 13, 14; 0, 1, 11, 31, 41, 42; 0, 1, 16, 66, 116, 131, 132; 0, 1, 22, 127, 302, 407, 428, 429; 0, 1, 29, 225, 715, 1205, 1401, 1429, 1430; 0, 1, 37, 373, 1549, 3313, 4489, 4825, 4861, 4862; ...
Links
- Alois P. Heinz, Rows n = 0..200, flattened
- David Callan, Sets, Lists and Noncrossing Partitions, Journal of Integer Sequences, Vol. 11 (2008), Article 08.1.3; also on arXiv, arXiv:0711.4841 [math.CO], 2007-2008.
Crossrefs
Programs
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Maple
b:= proc(x, y, t) option remember; expand(`if`(y<0 or y>x, 0, `if`(x=0, 1, add(b(x-1, y+j, j)* `if`(t=1 and j<1, z, 1), j=[-1, 1])))) end: T:= proc(n, k) option remember; `if`(k<0, 0, T(n, k-1)+coeff(b(2*n, 0$2), z, k)) end: seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Nov 28 2021
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Mathematica
T[n_, k_] := If[n == 0, 1, Sum[j Binomial[n, j]^2 / (n - j + 1), {j, 0, k}] / n]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Peter Luschny, Nov 29 2021 *)
Formula
T(n,k) = Sum_{j=0..k} A090181(n,j), the partial sum of the Narayana numbers.
T(n,n) = A000108(n), the n-th Catalan number.
G.f.: (1 + x - x*y - sqrt((1-x*(1+y))^2 - 4*y*x^2))/(2*x*(1-y)).
T(n,k) = (1/n)*Sum_{j=0..k} j*binomial(n,j)^2 / (n-j+1) for n >= 1. - Peter Luschny, Nov 29 2021
Comments