OFFSET
1,3
COMMENTS
Row sums give A003465.
From Manfred Boergens, Apr 11 2024: (Start)
If more than half of the nonempty subsets of [n] are drawn their union covers [n] (see Formula). - The proof is based on 2^(n-1)-1 being the number of nonempty subsets of [n] with one fixed element of [n] missing.
For covers which may include one empty set see A163353.
For disjoint covers see A008277. (End)
REFERENCES
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 165.
LINKS
Alois P. Heinz, Rows n = 1..10, flattened
FORMULA
T(n,k) = Sum_{j=0..n} (-1)^j*C(n, j)*C(2^(n-j)-1, k), k=1..2^n-1.
From Vladeta Jovovic, May 30 2004: (Start)
T(n,k) = (1/k!)*Sum_{j=0..k} Stirling1(k+1, j+1)*(2^j-1)^n.
E.g.f.: Sum(exp(y*(2^n-1))*log(1+x)^n/n!, n=0..infinity)/(1+x). (End)
Also exp(-y)*Sum((1+x)^(2^n-1)*y^n/n!, n=0..infinity).
From Manfred Boergens, Apr 11 2024: (Start)
T(n,k) = C(2^n-1,k) for k>=2^(n-1).
T(n,k) < C(2^n-1,k) for k<2^(n-1).
(Note: C(2^n-1,k) is the number of all k-subsets of P([n])\{{}}.) (End)
EXAMPLE
Triangle begins:
[1],
[1,3,1],
[1,12,32,35,21,7,1],
...
There are 35 4-covers of a labeled 3-set.
MATHEMATICA
nn=5; Map[Select[#, #>0&]&, Transpose[Table[Table[Sum[(-1)^j Binomial[n, j] Binomial[2^(n-j)-1, m], {j, 0, n}], {n, 1, nn}], {m, 1, 2^nn-1}]]]//Grid (* Geoffrey Critzer, Jun 27 2013 *)
CROSSREFS
KEYWORD
easy,nonn,tabf
AUTHOR
Vladeta Jovovic, Jun 14 2000
STATUS
approved