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a(n) = (Sum_{j=1..3} StirlingS1(3,j)*(2^j-1)^n)/3!.
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%I #29 Aug 20 2022 02:22:19

%S 0,0,4,44,360,2680,19244,136164,957520,6715760,47049684,329465884,

%T 2306615480,16147371240,113034787324,791253077204,5538800238240,

%U 38771687761120,271402072608164,1899815283098124,13298709306209800

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

%C Number of 3-element subsets of the powerset P([n]) such that their union is equal to the universe [n] = {1,2, ... ,n}.

%C StirlingS1(k,j) is a signed Stirling number of the first kind (cf. A048994).

%C 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.

%H Fabio VisonĂ , <a href="/A341590/b341590.txt">Table of n, a(n) for n = 0..100</a>

%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/a/3972609/573047">Answer to question about tuples of distinct sets in P([n]) such that their union is equal to [n]</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (11,-31,21).

%F a(n) = (Sum_{j=1..3} A048994(3,j)*(2^j-1)^n)/3!.

%F a(n) = (2 - 3^(1+n) + 7^n)/6.

%F a(n) = (-1)^n*(Sum_{j=0..n} binomial(n,j)*(-1)^j*binomial(2^j,3)).

%F G.f.: 4*x^2/(1 - 11*x + 31*x^2 - 21*x^3). - _Stefano Spezia_, Feb 15 2021

%F a(n) = 4 * A016212(n-2) for n >= 2. - _Alois P. Heinz_, Feb 15 2021

%e 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}}.

%Y Cf. A003462, A016212, A048994.

%K nonn,easy

%O 0,3

%A _Fabio VisonĂ _, Feb 15 2021