66th Belarusian Mathematical Olympiad

a) Let the sequence $(a_n)$, $n\in \mathbb{N}$, be lacunar. Then there exists a number $q>1$ such that $$a_{n+1}\ge q{a}_n\forall n\in \mathbb{N}.(1)$$ In particular, any lacunar sequence is increasing. From (1) it follows that any half-interval $(x,qx]$ contains at most one term of this sequence. Indeed, if we assume that $a_n$ and $a_{n+1}$ belong to this half-interval, then $\frac{a_{n+1}}{a_n}<\frac{qx}{x}=q$, contrary to (1).

Consider any positive integer $k$ such that $q^k>2$ (e.g., $k=[{\log }_{2}2]+1$, where $[y]$ stands for the integer part of $y$). It is evident that $$(x,2x)\subset \bigcup _{i=1}^k(q^{i-1}x,q^{i}x].(2)$$ Since at most one term of $(a_n)$ belongs to the half-interval $(q^{i-1}x,q^{i}x]$, from inclusion (2) it follows that at most $k$ terms of $(a_n)$ belong to the interval $(x,2x)$.

b) We define the sequence $(a_n)$, $n\in \mathbb{N}$, as $$a_{2n-1}=2^n,a_{2n}=2^n\left(1+\frac{1}{n+1}\right),n\in \mathbb{N}.(3)$$ This sequence is increasing since from (3) it follows that $a_{2n-1}<a_{2n}<a_{2n+1}$ for all $n\in \mathbb{N}$. Moreover, this sequence is rare since at most one number of the form $2^m$, $m\in \mathbb{N}$, belongs to any interval $(x,2x)$, and so at most three terms of this sequence belong to this interval.

On the other hand, we have $$\frac{a_{2n}}{a_{2n-1}}=1+\frac{1}{n+1},n\in \mathbb{N},$$ but the numbers $1+\frac{1}{n+1}$ may be arbitrary close to $1$ (for $n$ large enough), hence, the sequence $(a_n)$, $n\in \mathbb{N}$, is not lacunar.