The fourth positive element in the greedy $B_h$-set

TL;DR

This paper determines explicit formulas for the fourth element in the greedy $B_h$-set, showing how it depends on whether $h$ is odd or even, thus advancing understanding of these special integer sets.

Contribution

It provides the first explicit formula for the fourth element of the greedy $B_h$-set, extending known initial terms and clarifying the structure of these sets.

Findings

Explicit formulas for $a_4(h)$ depending on parity of $h$

Validation of initial terms $a_1(h)=1$, $a_2(h)=h+1$, $a_3(h)=h^2+h+1$

Enhanced understanding of the structure of greedy $B_h$-sets

Abstract

For $h \geq 1$, a $B_h$-set is a set of integers such that every integer $n$ has at most one representation in the form $n = a_{i_1} + \cdots + a_{i_h}$, where $a_{i_r} \in A$ for all $r = 1,\ldots, h$ and $a_{i_1} \leq \ldots \leq a_{i_h}$. The greedy $B_h$-set is the infinite set of nonnegative integers $\{a_0(h), a_1(h), a_2(h), \ldots \}$ constructed as follows: If $a_0(h) = 0$ and $\{a_0(h), a_1(h), a_2(h), \ldots, a_k(h) \}$ is a $B_h$-set, then $a_{k+1}(h)$ is the least positive integer such that $\{a_0(h), a_1(h), a_2(h), \ldots, a_k(h), a_{k+1}(h) \}$ is a $B_h$-set. Then $a_1(h) = 1$, $a_2(h) = h+1$, and $a_3(h) = h^2+h+1$ for all $h$. This paper proves that $a_4(h)$, the fourth term of the greedy $B_h$-set is $\left( h^3 + 3h^2 + 3h + 1\right) /2$ if $h$ is odd and $\left( h^3 + 2h^2 + 3h + 2\right) /2$ if $h$ is even.