A Steiner triple system of order v (briefly STS(v)) is 1-rotational under G if it admits G as an automorphism group acting sharply transitively on all but one point. The spectrum of values of v for which there exists a 1-rotational STS(v) under a cyclic, an abelian, or a dicyclic group, has been established in Phelps and Rosa (Discrete Math 33:57–66, 1981), Buratti (J Combin Des 9:215–226, 2001) and Mishima (Discrete Math 308:2617– 2619, 2008), respectively. Nevertheless, the spectrum of values of v for which there exists a 1-rotational STS(v) under an arbitrary group has not been completely determined yet. This paper is a considerable step forward to the solution of this problem. In fact, we leave as uncertain cases only those for which we have $v = (p^3 − p)n + 1 \equiv 1$ (mod 96) with p a prime, $n \not\equiv 0$ (mod 4), and the odd part of $(p^3 − p)n$ that is square-free and without prime factors congruent to 1 (mod 6).

Some progress on the existence of 1-rotational Steiner triple systems / S., Bonvicini; Buratti, Marco; G., Rinaldi; T., Traetta. - In: DESIGNS, CODES AND CRYPTOGRAPHY. - ISSN 0925-1022. - 62:1(2012), pp. 63-78. [10.1007/s10623-011-9491-3]

Some progress on the existence of 1-rotational Steiner triple systems

BURATTI, Marco;
2012

Abstract

A Steiner triple system of order v (briefly STS(v)) is 1-rotational under G if it admits G as an automorphism group acting sharply transitively on all but one point. The spectrum of values of v for which there exists a 1-rotational STS(v) under a cyclic, an abelian, or a dicyclic group, has been established in Phelps and Rosa (Discrete Math 33:57–66, 1981), Buratti (J Combin Des 9:215–226, 2001) and Mishima (Discrete Math 308:2617– 2619, 2008), respectively. Nevertheless, the spectrum of values of v for which there exists a 1-rotational STS(v) under an arbitrary group has not been completely determined yet. This paper is a considerable step forward to the solution of this problem. In fact, we leave as uncertain cases only those for which we have $v = (p^3 − p)n + 1 \equiv 1$ (mod 96) with p a prime, $n \not\equiv 0$ (mod 4), and the odd part of $(p^3 − p)n$ that is square-free and without prime factors congruent to 1 (mod 6).
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Utilizza questo identificativo per citare o creare un link a questo documento: https://hdl.handle.net/11573/1654620
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