A335344 Expansion of e.g.f. exp(x^2/(2*(1 - x)^2)).

1, 0, 1, 6, 39, 300, 2715, 28350, 335265, 4422600, 64298745, 1020477150, 17542820295, 324552648420, 6426708843555, 135573281994150, 3034400481137025, 71801374285040400, 1790523094644709425, 46923435009924823350, 1289032229351717425575

Offset: 0

Author

Ilya Gutkovskiy, Jun 02 2020

Keywords

• nonn

Comments

For n>0, a(n) is also the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select 2 people from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them). See example below. - Enrique Navarrete, Oct 01 2023

Examples

			For n = 5, using one table, there are 4! circular seatings and binomial(5,2) ways to select 2 persons, hence 240 ways. Using two tables, the only way we can select 2 persons from each one is seating 3 persons in one table and 2 in the other, which can be done in 20 ways; then choosing 2 persons from each table can be done in 3 ways, hence giving another 60 ways for a total of 300. - _Enrique Navarrete_, Oct 01 2023
		

Links

• Robert Israel, Table of n, a(n) for n = 0..436

Crossrefs

Cf. A001286, A052887, A082579, A185369, A335345.

Programs

Maple

f:= gfun:-rectoproc({a(n) = -(n-1)*(3*n-7)*a(n-2) + 3*(n-1)*a(n-1) + (n - 1)*(n - 2)*(n - 3)*a(n-3),a(0)=1,a(1)=0,a(2)=1},a(n),remember):
map(f, [$0..30]); # Robert Israel, Jun 04 2020

Mathematica

nmax = 20; CoefficientList[Series[Exp[x^2/(2 (1 - x)^2)], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = (1/2) Sum[Binomial[n - 1, k - 1] (k - 1) k! a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]

PARI

seq(n)=Vec(serlaplace(exp(x^2/(2*(1 - x)^2) + O(x*x^n)))) \\ Andrew Howroyd, Jun 02 2020

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * A001286(k) * a(n-k).

D-finite with recurrence a(n) = -(n - 1)*(3*n - 7)*a(n - 2) + 3*(n - 1)*a(n - 1) + (n - 1)*(n - 2)*(n - 3)*a(n - 3). - Robert Israel, Jun 04 2020

a(n) ~ n^(n - 1/6) * exp(1/6 - n^(1/3)/2 + 3*n^(2/3)/2 - n) / sqrt(3). - Vaclav Kotesovec, Jun 11 2020

a(n) = n! * Sum_{k=0..floor(n/2)} binomial(n-1,n-2*k)/(2^k * k!). - Seiichi Manyama, Mar 16 2023