English

The number of polyiamonds is supermultiplicative

Combinatorics 2025-07-16 v2

Abstract

While the number of polyominoes is known to be supermultiplicative by a simple concatenation argument, it is still unknown whether the same applies to polyiamonds. This article proves that if ,m\ell,m are not both 11, then T(+m)T()T(m)T(\ell+m)\ge T(\ell)T(m), for which one can say that the number of polyiamonds T(n)T(n) is supermultiplicative. The method is, however, by concatenating, merging and adding cells at the same time. A corollary is an increment of the best known lower bound on the growth constant from 2.84232.8423 to 2.85782.8578.

Keywords

Cite

@article{arxiv.2304.10077,
  title  = {The number of polyiamonds is supermultiplicative},
  author = {Vuong Bui},
  journal= {arXiv preprint arXiv:2304.10077},
  year   = {2025}
}

Comments

10 pages, 11 figures; final version for publication