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 A323425 Number of ways n people in a line can each choose two others both on the same side of them. 7
 1, 0, 0, 0, 9, 648, 57600, 6615000, 972504225, 179499220992, 40789783609344, 11212877910528000, 3671848787797265625, 1413385410212064432000, 632129969391038455873536, 325176984737061807515098752, 190691488202627199302740850625, 126479088749202444199526400000000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 LINKS 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

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Last modified October 19 15:05 EDT 2021. Contains 348091 sequences. (Running on oeis4.)