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Statement

Setting (p. 446). $A$ is a set of positive integers and $A(n)=\lvert A\cap\{1,2,\ldots,n\}\rvert$. $A$ is a $B_4$-sequence when every integer $n$ has at most one representation $n=a_1+a_2+a_3+a_4$ with $a_1\le a_2\le a_3\le a_4$ and $a_i\in A$. (The print's display (1) writes the sum with $k$ terms, $a_1+\cdots+a_k$; in a paper about $B_4$-sequences the intended $k$ is $4$.) For $n\ge1$, $nA=\{a_1+\cdots+a_n:a_i\in A\}$.

Main theorem (display (3), p. 446). The paper gives the result no theorem number. For every $B_4$-sequence $A$,

$$\liminf_{n\to\infty}\frac{A(n)\,(\log n)^{1/4}}{n^{1/4}}<\infty .$$

The print writes the lower limit as an underlined $\lim$. The abstract (p. 446) states the same bound.

The paper presents (3) as the analog of the bound

$$\liminf_{n\to\infty}\frac{A(n)\,(\log n)^{1/2}}{n^{1/2}}<\infty$$

for $B_2$-sequences (display (2), p. 446), which it attributes to Erdős, citing Stöhr's 1955 survey in J. reine angew. Math. 194.

A consequence not stated in the paper: since $(\log n)^{1/4}\to\infty$, the theorem gives $\liminf_{n\to\infty}A(n)/n^{1/4}=0$ for every infinite $B_4$-sequence.

Source. John C. M. Nash, On $B_4$-sequences, Canad. Math. Bull. 32 (4) (1989), 446--449, doi:10.4153/CMB-1989-064-2; the statement on p. 446, the proof on pp. 446--449. The edition read is identified on the source card.

Read depth. Claims checked: the definitions and the statement were read clause by clause on the printed pages. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pages 446--449. A $B_4$-sequence is also a $B_2$-sequence, and the paper notes $A(N)\ll N^{1/4}$. Since $(2A)(n)$ is at least of order $A(\lfloor n/2\rfloor)^2$, (3) follows from the $B_2$-type bound (6) for $C=2A$, and by Lemma 1 it suffices to prove the block condition (5) for $C=2A$, although $2A$ is not itself a $B_2$-sequence. With $D_l$ the number of elements of $2A$ in the $l$-th block of length $N$, $D_l^2\le4\binom{D_l}{2}$ unless $D_l=1$ (the print, p. 447, writes this inequality reversed, $4\binom{D_l}{2}\le D_l^2$, but uses it in the direction given here), so (5) reduces to $\sum_{l\le N}\binom{D_l}{2}\ll N$ (the paper's (7), p. 448), and the positive differences inside the blocks inject into the set $S$ of 4-tuples $(a_1,a_2,a_3,a_4)$ of elements of $A$ up to $N^2$ with $1\le a_1+a_2-a_3-a_4\le N$. It suffices that $\lvert S\rvert\ll N$ (the paper's (8)). The 4-tuples with $a_1,a_2$ both distinct from $a_3,a_4$ contribute at most $4N$, by the $B_4$ property; the others contribute at most $4A(N^2)\lvert T\rvert$, where $T$ is the set of pairs $(a_2,a_4)$ of elements of $A$ up to $N^2$ with $1\le a_2-a_4\le N$ (p. 449). Then $A(N^2)\ll N^{1/2}$ and $\binom{\lvert T\rvert}{2}\le\lvert S\rvert$ (the paper's (13)) give $\lvert T\rvert^2\ll N+N^{1/2}\lvert T\rvert$, hence $\lvert T\rvert\ll N^{1/2}$ and $\lvert S\rvert\ll N$.

Dependencies

Lemma 1 (p. 447), the paper's form of Erdős's argument for $B_2$-sequences; the bound $A(N)\ll N^{1/4}$ for $B_4$-sequences, which the paper uses without proof.

Bears on

• Problem 41: the problem asks, for infinite sets with all triple sums distinct ($B_3$-sequences), whether $\liminf\lvert A\cap\{1,\ldots,N\}\rvert/N^{1/3}=0$. The theorem treats $B_4$-sequences, the even case $h=4$ of the corresponding $B_h$ question, and gives $\liminf A(n)/n^{1/4}=0$ there with a logarithmic factor to spare; it says nothing about $B_3$-sequences.