Super‐simple holey Steiner pentagon systems and related designs

Abstract

A Steiner pentagon system of order v (SPS(v)) is said to be super-simple if its underlying (v, 5, 2)-BIBD is super-simple; that is, any two blocks of the BIBD intersect in at most two points. It is well known that the existence of a holey Steiner pentagon system (HSPS) of type T implies the existence of a (5, 2)-GDD of type T. We shall call an HSPS of type T super-simple if its underlying (5, 2)-GDD of type T is super-simple; that is, any two blocks of the GDD intersect in at most two points. The existence of HSPSs of uniform type h n has previously been investigated by the authors and others. In this article, we focus our attention on the existence of super-simple HSPSs of uniform type h n .

Super-Simple Twofold Steiner Pentagon Systems

“…Existence of HSPSs of uniform type h n has been investigated by Abel, Bennett and Zhang in [1] and subsequently by Abel, Bennett and Ge in [2]. The following known results are given in [1,2].…”

“…A (K , λ) group divisible design (or G D D) is a triple (X, G, A), which satisfies four properties: (1) G is a partition of X into subsets called groups; (2) A is a set of subsets of X (called blocks) with sizes from K ; (3) No two points in the same group appear in any block; (4) Any two points from distinct groups occur together in λ blocks. The parameter λ can be omitted if it equals 1.…”

“…Gronau and Mullin in [23] investigated the case for λ = 2; some mistakes in this paper were pointed out by Hartmann [24]; later proofs can be found in [25] and [2]. The case λ = 3 was solved independently by Khodkar [25] and Chen [14].…”

The Existence of Mixed Orthogonal Arrays with Four and Five Factors of Strength Two

“…, (3,6), (3,8), (4,5), (4, 9), (5, 0), (5,8), (6, 1), (6, 4), (7, 0), (7,1), (8,3), (8,7), (9, 4), (9, 5)}, Q 6 = {(0, 5), (0, 6), (1,4), (1, 7), (2, 1), (2,8), (3,4), (3,5), (4,3), (4,8), (5, 2), (5, 7), (6, 3), (6, 9), (7, 2), (7,9), (8,0), (8,6), (9, 0), (9, 1)}, (2,9), (3, 1), (3, 7), (4, 0), (4, 2), (5, 1), (5, 4), (6, 0), (6, 2), (7,5), (7,6), (8,4), (8,9), (9,6), (9,8)}.…”

“…When x = y = 0, there is an MOA(36; 2 3 3 (4x + 6) 1 6 1 , 2). For y > 0, since 4y ≡ 0 (mod 4), there is an MOA((4x + 6)(4y); 2 3 (4x + 6) 1 (4y) 1 , 2) as above. By applying Construction 2.2, we get an MOA(24x + 36 + (4x + 6)(4y); 2 3 (4x + 6) 1 (6 + 4y) 1 , 2), i.e., an MOA(bc; 2 3 bc, 2).…”