cp's OEIS Frontend

A365520 Number of 1-factorizations of complete graph K_{2n} that all share one arbitrary pairing in common.

%I A365520 #27 Oct 19 2023 07:35:58
%S A365520 1,1,2,416,11672064,266965735243776,9500592190171594780311552
%N A365520 Number of 1-factorizations of complete graph K_{2n} that all share one arbitrary pairing in common.
%C A365520 Consider a round robin tournament with 2n teams competing. For the first round, select a set of pairings arbitrarily. a(n) is the number of possible round robin tournaments after choosing one of those sets of pairings.
%C A365520 a(n) can be obtained by dividing A000438(n) by the product (2n-3)*(2n-5)*...*3*1 when n > 2. For n = 1 or n = 2, a(n) is simply 1, since there is only one way to set up those rounds with 2 or 4 teams.
%C A365520 To show how this works, consider the number of possible round robin tournaments given 8 teams (A000438(4) = 6240) identified by the letters A through H.
%C A365520 Each of these tournaments can be constructed from a set of matchups of this form:
%C A365520    (AB)(XX)(XX)(XX)
%C A365520    (AC)(YY)(YY)(YY)
%C A365520    (AD)(YY)(YY)(YY)
%C A365520    (AE)(YY)(YY)(YY)
%C A365520    (AF)(YY)(YY)(YY)
%C A365520    (AG)(YY)(YY)(YY)
%C A365520    (AH)(YY)(YY)(YY)
%C A365520 Note that the 6240 tournaments can be divided evenly by the number of ways the "X" teams can be paired up to fill the arbitrary round. This is why A001147(3) = 15 is a factor, since there are 5 ways to pick the first pair, followed by 3 ways to pick the second pair.
%C A365520 For larger numbers of teams, there are more ways to pair up; e.g., for 10 teams, there will be a factor of 7*5*3, and for 12 teams, a factor of 9*7*5*3, and so on.
%F A365520 a(n) = A000438(n)/A001147(n-1).
%e A365520 For n = 3, given teams A through F (2n), the only two round robin tournaments that share the pairing (AB)(CD)(EF) are:
%e A365520    (AB)(CD)(EF)
%e A365520    (AC)(BE)(DF)
%e A365520    (AD)(BF)(CE)
%e A365520    (AE)(BD)(CF)
%e A365520    (AF)(BC)(DE)
%e A365520 and
%e A365520    (AB)(CD)(EF)
%e A365520    (AC)(BF)(DE)
%e A365520    (AD)(BE)(CF)
%e A365520    (AE)(BC)(DF)
%e A365520    (AF)(BD)(CE)
%e A365520 which agrees with a(3) = 2.
%Y A365520 Cf. A000438, A001147.
%K A365520 nonn,more
%O A365520 1,3
%A A365520 _Brian Lathrop_, Sep 08 2023