login
Number of different ways to select n elements from three sets of n elements such that there is at least one element from each set.
4

%I #16 Sep 08 2022 08:45:23

%S 0,0,27,288,2250,15795,105987,696864,4540968,29490750,191420427,

%T 1243565235,8091223647,52739879283,344402073027,2253045672480,

%U 14764068268068,96899123172708,636877933530303,4191430966219038,27617820628739718,182176855684869243

%N Number of different ways to select n elements from three sets of n elements such that there is at least one element from each set.

%H G. C. Greubel, <a href="/A115246/b115246.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = binomial(3n, n) - 3*binomial(2n, n) + 3.

%F From _G. C. Greubel_, Feb 08 2016: (Start)

%F E.g.f.: 3*exp(x) - 3*exp(2*x)*BesselI_{0}(2*x) + Hypergeometric2F2[1/3,2/3; 1/2,1; 27*x/4].

%F G.f.: (1/((x-1)sqrt(a*b)))*[3*sqrt(a)*(1-x) - 3*sqrt(a*b) - 2*(1-x)*sqrt(b)*cos(c/3)], where a = 4-27*x, b = 1-4*x, c = arcsin(3*sqrt(3*x)/2). (End)

%t Table[Binomial[3 n, n] - 3*Binomial[2 n, n] + 3, {n, 1, 100}] (* _G. C. Greubel_, Feb 08 2016 *)

%o (PARI) a(n) = binomial(3*n, n) - 3*binomial(2*n, n) + 3 \\ _Michel Marcus_, Jul 15 2013

%o (Magma) [Binomial(3*n, n)-3*Binomial(2*n, n)+3: n in [1..40]]; // _Vincenzo Librandi_, Feb 09 2016

%K nonn

%O 1,3

%A _Hieronymus Fischer_, Jan 21 2006