|
|
A341590
|
|
a(n) = (Sum_{j=1..3} StirlingS1(3,j)*(2^j-1)^n)/3!.
|
|
2
|
|
|
0, 0, 4, 44, 360, 2680, 19244, 136164, 957520, 6715760, 47049684, 329465884, 2306615480, 16147371240, 113034787324, 791253077204, 5538800238240, 38771687761120, 271402072608164, 1899815283098124, 13298709306209800
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
Number of 3-element subsets of the powerset P([n]) such that their union is equal to the universe [n] = {1,2, ... ,n}.
StirlingS1(k,j) is a signed Stirling number of the first kind (cf. A048994).
In general, the number of k-element subsets of P([n]) such that their union is equal to [n] is (Sum_{j=0..k} StirlingS1(k,j)*(2^j-1)^n)/k!. That can be expressed also as (-1)^n*(Sum_{j=0..n} binomial(n,j)*(-1)^j*binomial(2^j,k)). See the below link to Mathematics Stack Exchange for proofs. The case k = 2 is A003462.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = (Sum_{j=1..3} A048994(3,j)*(2^j-1)^n)/3!.
a(n) = (2 - 3^(1+n) + 7^n)/6.
a(n) = (-1)^n*(Sum_{j=0..n} binomial(n,j)*(-1)^j*binomial(2^j,3)).
G.f.: 4*x^2/(1 - 11*x + 31*x^2 - 21*x^3). - Stefano Spezia, Feb 15 2021
|
|
EXAMPLE
|
For n = 2 and [n] = [2] = {1,2} the a(2) = 4 solutions are {{},{1},{2}}, {{},{1},{1,2}}, {{},{2},{1,2}}, {{1},{2},{1,2}}.
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|