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

0,5

Links

Table of n, a(n) for n=0..17.

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

Example

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.

Mathematica

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

Prog

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

Crossrefs

Sequence in context: A280904 A210053 A128795 * A191510 A081232 A020548

Adjacent sequences: A323422 A323423 A323424 * A323426 A323427 A323428

Keyword

nonn

Author

Shreevatsa R, Jan 14 2019

Extensions

More terms from Vaclav Kotesovec, Jan 15 2019

Status

approved