A341590 a(n) = (Sum_{j=1..3} StirlingS1(3,j)*(2^j-1)^n)/3!.

0, 0, 4, 44, 360, 2680, 19244, 136164, 957520, 6715760, 47049684, 329465884, 2306615480, 16147371240, 113034787324, 791253077204, 5538800238240, 38771687761120, 271402072608164, 1899815283098124, 13298709306209800

Offset: 0

Author

Fabio Visonà, Feb 15 2021

Keywords

• nonn
• easy

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.

Examples

			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}}.
		

Links

• Fabio Visonà, Table of n, a(n) for n = 0..100
• Mathematics Stack Exchange, Answer to question about tuples of distinct sets in P([n]) such that their union is equal to [n]
• Index entries for linear recurrences with constant coefficients, signature (11,-31,21).

Crossrefs

Cf. A003462, A016212, A048994.

Programs

Mathematica

LinearRecurrence[{11,-31,21},{0,0,4},30] (* Harvey P. Dale, Mar 02 2025 *)

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

a(n) = 4 * A016212(n-2) for n >= 2. - Alois P. Heinz, Feb 15 2021