A366155 Expansion of e.g.f. exp(x^3/(3*(1-x)^3)).

1, 0, 0, 2, 24, 240, 2440, 26880, 329280, 4518080, 69148800, 1168675200, 21564188800, 430048819200, 9195964377600, 209593877292800, 5068718054400000, 129599032442880000, 3492894468128665600, 98968805893769011200, 2940975338620999680000, 91452266705317726208000, 2969664371124258103296000

Offset: 0

Author

Enrique Navarrete, Oct 01 2023

Keywords

• nonn
• changed

Comments

For n>0, a(n) is the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select 3 people from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them).

2*A001754(n) is the number of ways to seat n persons around a circular table and select 3 of them if only one table is used.

A335344 is the corresponding sequence if 2 persons are selected from each table, and A000262 if only one person is selected from each table.

Examples

			a(7)=26880 since, using one table, there are 6! circular seatings and binomial(7,3) ways to select 3 persons, hence 25200 ways. Using two tables, the only way we can select 3 persons from each one is seating 4 persons in one table and 3 in the other, which can be done in 420 ways; then choosing 3 persons from each table can be done in 4 ways, for a total of 1680 ways; hence 25200 + 1680 = 26880.
		

Crossrefs

Cf. A000262, A001754, A335344.

Programs

Mathematica

CoefficientList[Series[Exp[x^3/(3*(1-x)^3)],{x,0,22}],x]Table[n!,{n,0,22}] (* Stefano Spezia, Oct 02 2023 *)
nmax = 25; Join[{1}, Table[n!*Sum[Binomial[n-1, 3*k-1]/(3^k*k!), {k, 1, n}], {n, 1, nmax}]] (* Vaclav Kotesovec, Aug 28 2025 *)

Formula

From Vaclav Kotesovec, Aug 28 2025: (Start)

For n > 0, a(n) = n! * Sum_{k=1..n} binomial(n-1, 3*k-1) / (3^k * k!).

a(n) = 4*(n-1)*a(n-1) - 6*(n-2)*(n-1)*a(n-2) + (n-2)*(n-1)*(4*n-11)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4).

a(n) ~ exp(-1/12 + 5*n^(1/4)/24 - n^(1/2)/2 + 4*n^(3/4)/3 - n) * n^(n - 1/8) / 2 * (1 - 563/(2560*n^(1/4))). (End)