A323425 Number of ways n people in a line can each choose two others both on the same side of them.

1, 0, 0, 0, 9, 648, 57600, 6615000, 972504225, 179499220992, 40789783609344, 11212877910528000, 3671848787797265625, 1413385410212064432000, 632129969391038455873536, 325176984737061807515098752, 190691488202627199302740850625, 126479088749202444199526400000000

Offset: 0

Author

Shreevatsa R, Jan 14 2019

Keywords

• nonn

Examples

			Example: For n = 4, with four people ABCD, A can choose any two of {B, C, D} (3 choices), B can choose {C, D} (1 choice), C can choose {A, B} (1 choice), and D can choose any two of {A, B, C} (3 choices), so there are 3*1*1*3=9 possible overall choices and a(4) = 9.
Example: For n = 5, with five people ABCDE, A can choose any two of {B, C, D, E} (6 choices), B can choose any two of {C, D, E} (3 choices), C can choose either {A, B} or {D, E} (2 choices), D can choose any two of {A, B, C} (3 choices), E can choose any two of {A, B, C, D} (6 choices), so a(5) = 6*3*2*3*6 = 648.
		

Programs

Haskell

a n = product [(k-1)*(k-2) `div` 2 + (n-k)*(n-k-1) `div` 2 | k<-[1..n]]

Mathematica

Table[Product[Binomial[k-1, 2] + Binomial[n-k, 2], {k, 1, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jan 15 2019 *)

Formula

a(n) = Product_{k = 1..n} ( binomial(k-1, 2) + binomial(n-k, 2) ).

a(n) ~ exp(Pi*(n/2 - 1) - 2*n) * n^(2*n) / 2^n. - Vaclav Kotesovec, Jan 15 2019

Extensions

More terms from Vaclav Kotesovec, Jan 15 2019