# “Mixed” Waring bases of fixed order

Recall that a subset $B \subset \mathbb{N}$ is called an additive basis of order $h$ if there exists a finite set $C \subset \mathbb{N}$ such that $\mathbb{N} \setminus C = hB$, where $hB = B + \cdots + B$ $h$ times. Waring's problem asks whether $\mathbb{N}^k = \{n^k : n \in \mathbb{N}\}$ is an additive basis of finite order, and if it is, find the order. This is usually denoted $G(k)$ and the precise value of $G(k)$ is not known except for $k=2,4$, and the upper bound $G(k) \leq k \log k + k \log \log k + ck$ for some constant $c > 0$ is known (due to Wooley).

My question is the following. Suppose that we fix an $h$. Certainly if $h \geq G(k)$ for some $k$, then $\mathbb{N}^k$ is an additive basis of order $h$. Hence choose $k$ so that $h < G(k)$. What is the minimum number $l$ of integers from the set $\{k, k+1, \cdots \}$, say $m_1, \cdots, m_l$ that can be chosen so that $\mathbb{N}^{m_1} \cup \cdots \cup \mathbb{N}^{m_l}$ is an additive basis of order $h$? Is $l$ guaranteed to be finite? For cases when $l$ is finite, is there a canonical choice for the $m_1, \cdots, m_l$ (with respect to $h$) that can be made?

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It's customary (and basic politeness) to acknowledge responses to questions you ask. – Anthony Quas Mar 30 '14 at 23:42

$\ell$ is certainly not guaranteed to be finite in general for essentially trivial (counting) reasons. Set $\mathbb N^{k+}=\bigcup_{j\ge k}\mathbb N^j$ and let $F(N)=|\mathbb N^{k+}\cap [1,N]|$. We have $F(N)\le \sum_{k\le j\le \log_2 N}N^{1/j}\lesssim N^{1/k}$. Hence if you count the number of $[1,N]$ that can be represented as the sum of $k-1$ elements of $\mathbb N^{k+}$, it's at most $F(N)^{k-1}\ll N$.