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For what odd integers $n \geq 3$ does there exist a self-complementary $(2n,8n−4,4n−2,n,2n−2)$ balanced incomplete block design?

By "self-complementary" I mean that the complement of each block is a block itself. A (trivial) example of such a design is given by the set of all twenty 3-element subsets of a set of size 6 ($n=3$).

Motivating problem: One wants to divide $2n$ players into 2 equal-sized teams in $4n−2$ different ways in such a way that each player is on the same team as each other player exactly $2n−2$ times. See my earlier MO question Self-complementary block designs , which asked about dividing the players into teams in $2n-1$ ways so that each player is on the same team as each other player exactly $n-1$ times. When $n$ is odd, no such scheme exist, as can be shown by an easy parity argument; here I am asking whether, by doubling the number of games played, one can get around the parity obstacle and obtain a design of the desired kind.

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  • $\begingroup$ They exist for $n=5,7,9$. In fact, resolvable such designs exist for odd $n$ up to $9$. A general construction seems elusive at the moment, but I wonder if conference matrices can be used. $\endgroup$ – Peter Dukes Apr 29 '16 at 23:15

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