Elementary Proof of Basis of Order k

Question

Context

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Problem Statement

Let $N$ be the natural numbers.

$B \subseteq N$ is a basis of order $k$ if $N \setminus kB$ is finite.

I would like to show that there is a basis $B$ of order $k$ s.t.

$|B \cap [1,n]| = O(n^{1/2} \log^{1/k} n)$.

What I've tried

Suppose all we needed was $O(n^{1/2} \log^{1/2} n)$, then I would define $B$ by randomly sampling from $N$ s.t.

$$P(n \in B) = \frac{c\log^{1/2} n}{\sqrt{n}}$$

By the chernoff bound, with high probability we have $|B \cap [1,n]| = O(n^{1/2}\log^{1/2} n)$.

Furthermore, for any $n$, there does not exists $a,b\in B$ s.t. $a+b=N$ with probability at most $(1-\frac{c\log n}{n})^{n/2} \leq 1/n^2$, and we're done.

Unfortunately, however, I need to push this down to $O(n^{1/2}\log^{1/k} n)$.

What I'm stuck on

So far, I've only used $B$ as a order 2 base, rather than an order $k$ base.

Question:

What should I be looking at to go from order 2 to order $k$ and $\log^{1/2} n$ to $\log^{1/k} n$?

Answer 1

If $B$ is a basis of order $k$ such that every integer $n$ can be written as a sum of $k$ elements from $B$ in $\asymp n^{o(1)}$ ways, then a simple counting argument yields $|B \cap [1 , X]| \asymp X^{\frac{1}{k}+o(1)}$. Thus a stronger estimate $|B \cap [1 , X]| \asymp (X \log X)^{\frac{1}{k}}$ in your problem is certainly a more interesting goal.

Theorem 8.6.3 in "The Probabilistic Method" by Alon & Spencer gives precisely a set $B$ satisfying this estimate when $k=3$ (and the proof can be adapted in order to handle any value $k \geq 3$). They also give the following reference :

Erdos, P. and Tetali, P. (1990). Representations of integers as the sum of k terms, Random Structures Algorithms 1(3): 245-261.

EDIT : I answer here the comments below.

@Stanley Yao Xiao : I made an assumption on the number of representations of integers by $k$ elements from $B$ which essentially discards basis of smaller order.

@unknown : Writing $r(n)$ for the number of representations of $n$ as a sum $b_1 + \cdots + b_k$ with each $b_i \in B$, we have $$ |B \cap [1,X]|^k = \sum_{n \geq 1} \left( \sum_{b_1 + \cdots + b_k = n ;\\ b_i \leq X} 1 \right) \geq \sum_{1 \leq n \leq X} r(n) $$ and $$ |B \cap [1,X]|^k = \sum_{n \geq 1} \left( \sum_{b_1 + \cdots + b_k = n ;\\ b_i \leq X} 1 \right) \leq \sum_{1 \leq n \leq kX} r(n) $$ Under the assumption $r(n) \asymp n^{o(1)}$, both RHS are $X^{1 + o(1)}$, hence the result. Actually, Erdos & Tetali showed that some basis $B$ of order $k$ satisfies $r(n) \asymp \log n$. By the argument above, this implies $|B \cap [1,X]| \asymp (X \log X)^{\frac{1}{k}}$.

Answer 2

To elaborate on the above comment, the problem with order $k$ bases is precisely that Chernoff's inequality does not work. The joint independence assumption for Chernoff's inequality is essential; as seen by the following example taken from Tao and Vus' Additive Combinatorics:

Color the elements of $[1, N]$ either black or white independently and with equal probability. For each $A \subset [1, N]$ let $s_A$ denote the parity of black elements of $A$ (so say if $A$ contains 3 black elements then $s_A = 1$). One can check that the $s_A$'s are independent events. Write $X = \displaystyle \sum_{A \subset [1, N]} s_A$. One can check that $\mathbb{E}X = 2^N - 1/2$ and $\textbf{Var} X = 2^{N-2} - 1/4$. Further, $\mathbb{P}(X = 0) = 2^{-N}$. The upper-bound on Chernoff's inequality would be $2\exp(-2^{N-2})$, which is much smaller than $\mathbb{P}(X = 0)$, so the inequality fails.

The reason why a simple argument suffices for additive bases of order 2 is because we have

$$\displaystyle r_{2,B}(n) = \sum_{x < n/2} \mathbb{I}(x \in B) \mathbb{I}(n - x \in B) + E$$

where $E$ is a suitably small error, and $r_{2,B}(n)$ is the number of ways to write $n$ as a sum of two elements in $B$. The key here is that the events $\mathbb{I}(x \in B) \mathbb{I}(n - x \in B)$ are independent for $1 \leq x < n/2$. This is not the case when there are more summands. In the Erdos-Tetali paper cited above, this issue is circumvented via Janson's inequality. In particular, Erdos-Tetali showed that there are additive bases of order $k$ satisfying $| B \cap [1,N]| = \Theta(N^{1/k} \log^{1/k} N)$.

The main difficulty you have to circumvent is how to deal with the non-independence of the random variables $\mathbb{I}(x_1 \in B) \cdots \mathbb{I}(x_k \in B)$.