A283424 Number T(n,k) of blocks of size >= k in all set partitions of [n], assuming that every set partition contains one block of size zero; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
1, 2, 1, 5, 3, 1, 15, 10, 4, 1, 52, 37, 17, 5, 1, 203, 151, 76, 26, 6, 1, 877, 674, 362, 137, 37, 7, 1, 4140, 3263, 1842, 750, 225, 50, 8, 1, 21147, 17007, 9991, 4307, 1395, 345, 65, 9, 1, 115975, 94828, 57568, 25996, 8944, 2392, 502, 82, 10, 1
Offset: 0
Examples
T(3,2) = 4 because the number of blocks of size >= 2 in all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 1+1+1+1+0 = 4. Triangle T(n,k) begins: 1; 2, 1; 5, 3, 1; 15, 10, 4, 1; 52, 37, 17, 5, 1; 203, 151, 76, 26, 6, 1; 877, 674, 362, 137, 37, 7, 1; 4140, 3263, 1842, 750, 225, 50, 8, 1; 21147, 17007, 9991, 4307, 1395, 345, 65, 9, 1; ...
Links
- Alois P. Heinz, Rows n = 0..140, flattened
- Wikipedia, Partition of a set
Crossrefs
Programs
-
Maple
T:= proc(n, k) option remember; `if`(k>n, 0, binomial(n, k)*combinat[bell](n-k)+T(n, k+1)) end: seq(seq(T(n, k), k=0..n), n=0..14);
-
Mathematica
T[n_, k_] := Sum[Binomial[n, j]*BellB[j], {j, 0, n - k}]; Table[T[n, k], {n, 0, 14}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 30 2018 *)
Formula
T(n,k) = Sum_{j=0..n-k} binomial(n,j) * Bell(j).
T(n,k) = Bell(n+1) - Sum_{j=0..k-1} binomial(n,j) * Bell(n-j).
Sum_{k=0..n} T(n,k) = n*Bell(n)+Bell(n+1) = A124427(n+1).
Sum_{k=1..n} T(n,k) = n*Bell(n) = A070071(n).
T(n,0)-T(n,1) = Bell(n).
Sum_{k=0..n} (-1)^k * T(n,k) = A224271(n+1). - Alois P. Heinz, May 17 2023
Comments