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Statement

Theorem 6 (p. 9). For each real $\varepsilon>0$, let $k(\varepsilon)$ be the maximum, over all sequences $A=\{a_1<a_2<a_3<\cdots\}$ with $a_{n+1}/a_n\ge1+\varepsilon$ for all $n$, of the length of the longest descending wave in $A$. Then

$$[1/\varepsilon]+1\le k(\varepsilon)\le(1/\varepsilon)+2.$$

The paragraph before the theorem (p. 9) considers sequences of real numbers with $a_{n+1}-a_n\ge1$ for all large $n$, and the lower-bound example in the proof has non-integer terms; the theorem's statement itself does not say whether the $a_n$ are integers.

Source. Brown, T. C., Erdős, P. and Freedman, A. R., Quasi-progressions and descending waves, J. Combin. Theory Ser. A 53 (1990), no. 1, 81--95, doi:10.1016/0097-3165(90)90021-N, read in the authors' copy identified on the source card, whose pages are numbered 1 to 13: the statement and proof on p. 9.

Read depth. Claims checked: the statement was read clause by clause on the print's page. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Section 4, p. 9. Upper bound: if $0<b_0<b_1<\cdots<b_t$ is a descending wave in such a sequence, then $b_t\ge t(b_t-b_{t-1})+b_0$, and the ratio bound $b_{t-1}/b_t\le1/(1+\varepsilon)$ turns this into $t<1+1/\varepsilon$. Lower bound, for $\varepsilon<1$: with $t=[1/\varepsilon]$, take $a_i=i$ for $i\le t$ and $a_{t+k}=t(1+\varepsilon)^k$ for $k\ge1$; then $1,2,\ldots,t,t(1+\varepsilon)$ is a descending wave of length $[1/\varepsilon]+1$.

Dependencies

None outside the paper.

Bears on

No problem page.