When is a numerical semigroup a quotient?
A natural operation on numerical semigroups is taking a quotient by a positive integer. If, the
first known family of numerical semigroups that cannot be written as a k-quotient. We also …
first known family of numerical semigroups that cannot be written as a k-quotient. We also …
On the genus of a quotient of a numerical semigroup
We find a relation between the genus of a quotient of a numerical semigroup S and the
genus of S itself. We use this identity to compute the genus of a quotient of S when S has …
genus of S itself. We use this identity to compute the genus of a quotient of S when S has …
The generating functions arise from the quotients of numerical semigroups
F Liu - arXiv preprint arXiv:2312.10889, 2023 - arxiv.org
In this paper, we propose a class generating functions $\textrm {RGF}(x) $ related to
Sylvester denumerant which arise from the quotients of numerical semigroups. Using …
Sylvester denumerant which arise from the quotients of numerical semigroups. Using …
GENERATING FUNCTIONS FOR THE QUOTIENTS OF NUMERICAL SEMIGROUPS
F Liu - Bulletin of the Australian Mathematical Society, 2024 - cambridge.org
GENERATING FUNCTIONS FOR THE QUOTIENTS OF NUMERICAL SEMIGROUPS Page 1
Bull. Aust. Math. Soc. (First published online 2024), page 1 of 12 ∗ doi:10.1017/S0004972724000054 …
Bull. Aust. Math. Soc. (First published online 2024), page 1 of 12 ∗ doi:10.1017/S0004972724000054 …
Numerical semigroups via projections and via quotients
We examine two natural operations to create numerical semigroups. We say that a
numerical semigroup S is k-normalescent if it is the projection of the set of integer points in …
numerical semigroup S is k-normalescent if it is the projection of the set of integer points in …
One half of almost symmetric numerical semigroups
F Strazzanti - Semigroup Forum, 2015 - Springer
Abstract Let S, TS, T be two numerical semigroups. We study when SS is one half of TT, with
TT almost symmetric. If we assume that the type of TT, t (T) t (T), is odd, then for any SS there …
TT almost symmetric. If we assume that the type of TT, t (T) t (T), is odd, then for any SS there …