ON SPARSE HOLES AND A FINITE-FOLD PROBLEM OF NATHANSON CONCERNING MINIMAL ADDITIVE COMPLEMENTS

Abstract

Let

be bounded below and normalised by

inf W = 1 $\inf W=1$ inf upper W equals 1

, and put

W ― = Z > 0 ∖ W $\overline W=\mathbb Z_{>0}\setminus W$ upper W overbar equals double struck upper Z Subscript greater than 0 Baseline minus upper W

. Nathanson asked whether infinite sets of integers admit minimal additive complements. Chen and Yang [‘On a problem of Nathanson related to minimal additive complements’, SIAM J. Discrete Math. 26 (2012), 1532–1536] proved that the two-sided case is positive and that sufficiently large consecutive gaps in

W ― $\overline W$ upper W overbar

force nonexistence in the one-sided case. Chen and Ding [‘On a problem of Nathanson on nonminimal additive complements’, Bull. Aust. Math. Soc. 114 , 6–13] isolated a sparse-hole condition under which no ordinary minimal complement exists. We prove a finite-fold version of this sparse-hole obstruction. For an integer

h ≥ 1 $h\geq 1$ h greater than or equals 1

, write

h C = C + ⋯ + C $hC=C+\cdots +C$ h upper C equals upper C plus midline horizontal ellipsis plus upper C

. We show that if

W ― $\overline W$ upper W overbar

contains a sequence

a 1 < a 2 < ⋯ $a_1<a_2<\cdots $ a 1 less than a 2 less than midline horizontal ellipsis

such that

a t + 1 − a t → ∞ $a_{t+1}-a_t\to \infty $ a Subscript t plus 1 Baseline minus a Subscript t Baseline right arrow infinity

and the four-neighbourhoods

( a t − 2 , a t + 2 ) $(a_{t-2},a_{t+2})$ left parenthesis a Subscript t minus 2 Baseline comma a Subscript t plus 2 Baseline right parenthesis

contain at most K elements of

W ― $\overline W$ upper W overbar

for all sufficiently large t , then there is no set

C ⊆ Z $C\subseteq \mathbb Z$ upper C subset of or equal to double struck upper Z

which is minimal subject to

h C + W = Z $hC+W=\mathbb Z$ h upper C plus upper W equals double struck upper Z

. This result recovers and broadens the Chen–Ding nonexistence theorem as the special case

h = 1 $h=1$ h equals 1

.