A Kirkman square with index 2, latinicity #, block size k and v points, KSk(v; #, Z), is a t x t array (t = 3.(v -1)/#(k -1)) defined on a v-set V such that (1) each point of V is contained in precisely # cells of each row and column, (2) each cell of the array is either empty or contains a k subset of V, and (3) the collection of blocks obtained from the nonempty cells of the array is a (v, k, 3.)-BIBD. The existence question for KS2(v; #, has been completely selttled. We are interested in the next case k = 3. The ease k = 3 and # = 3. = 1 appears to be quite difficult, although some existence results are available. For 3.> 1 and # t> 1, the problem is more tractable. In this paper, we prove the existence of KS3(v; 2, 4) for v--3 (rood 12), v-= 6 (mod 60) and v -9 (rood 96).
Existence results for KS3(I3 ; 2, 4)s 199 Since x must occur/zt times in B we have the inequality 2/is + ~.w ~</zt. This is or 2/z,~(w-1) /~.(v-1) +~,w~< /z(k -1) /,(k-1) '
2w -2 + w(k -1) <~v -1.
Thus, we have w ~< (v + 1)/(k + 1). [] Corollary 2.2. If there is a KS3(v;2, 4) which contains as a subarray a KS3(3; 2, 4), then v >t 12.
It is possible to construct KS3(v;2, 4)s from special KS3(v;1, 2)s. A KS3(v; 1, 2), A, is called complementary if there exists a KS3(v; 1, 2), B, which can be written in the empty cells of A. The next result follows immediately from this definition. Lemma 2.3. If there exists a complementary KS3(v; 1, 2), then there is a KS3(1) ; 2, 4). KS3(I~ ; 1, 1)s can also be used to construct KS3(v; 2, 4)s.
Lemma 2.4 If there exists a KS3(13; 1, 1), then there exists a KS3(v; 2, 4) which contains as a subarray a KS3(3; 2, 4).
Proof. Let D be a KS3(v; 1, 1). The array [~ o ยฐ] is a KS3(v; 2, 4). It is clear that this array contains as a subarray a KS3(3; 2, 4). [] Theorem 2.5. There exist KS3(v; 2, 4) for v =-0 (mod 3) and 3 <-v <~ 33. For v = 15, 21, 27 and 33, the arrays contain as a subarray a KS3(3; 2, 4).
Proof. (i) v = 3, 6, 9, 12, 15 and 27. v Construction