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A055154
Triangle read by rows: T(n,k) = number of k-covers of a labeled n-set, k=1..2^n-1.
16
1, 1, 3, 1, 1, 12, 32, 35, 21, 7, 1, 1, 39, 321, 1225, 2919, 4977, 6431, 6435, 5005, 3003, 1365, 455, 105, 15, 1, 1, 120, 2560, 24990, 155106, 711326, 2597410, 7856550, 20135050, 44337150, 84665490, 141118250, 206252550, 265182450, 300540190
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.
For disjoint covers which may include one empty set see A256894 (amendment by Manfred Boergens, Mar 09 2025). (End)
REFERENCES
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 165.
LINKS
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
Cf. A369950 (partial row sums).
Cf. 256894.
Sequence in context: A209424 A129619 A094573 * A373168 A338875 A015112
KEYWORD
easy,nonn,tabf
AUTHOR
Vladeta Jovovic, Jun 14 2000
STATUS
approved