Really nice. Suppose that they are finitely many: $p_1, p_2, \cdots, p_k$. The statement says that in every block of $a$ consecutive numbers, at least one is in the sequence. Take $n \in [1;a]$ and large enough $M$, such that the number $x=(p_1p_2 \cdots p_k)^M+n$ is in the sequence. We have that $v_{p_i}(x)=v_{p_i}(n)$ for all $i=1,2,..., k$, and since $x$ doesn't have any other prime factors, we get a contradiction.

oVlad wrote: Let $a$ be a positive integer and $(a_n)_{n\geqslant 1}$ be a sequence of positive integers satisfying $a_n<a_{n+1}\leqslant a_n+a$ for all $n\geqslant 1$. Prove that there are infinitely many primes which divide at least one term of the sequence. Moldavia Olympiad, 1994 What do you mean by infinitely many primes dividing at least one term of the sequence? if there are really infinitely many primes dividing one term of the sequence the that term is definitely $0$.

PNT wrote: oVlad wrote: Let $a$ be a positive integer and $(a_n)_{n\geqslant 1}$ be a sequence of positive integers satisfying $a_n<a_{n+1}\leqslant a_n+a$ for all $n\geqslant 1$. Prove that there are infinitely many primes which divide at least one term of the sequence. Moldavia Olympiad, 1994 What do you mean by infinitely many primes dividing at least one term of the sequence? if there are really infinitely many primes dividing one term of the sequence the that term is definitely $0$. It means that there are infinitely many prime numbers $p$ such that $p$ divides $a_i$ for some index $i{}$.

oVlad wrote: PNT wrote: oVlad wrote: Let $a$ be a positive integer and $(a_n)_{n\geqslant 1}$ be a sequence of positive integers satisfying $a_n<a_{n+1}\leqslant a_n+a$ for all $n\geqslant 1$. Prove that there are infinitely many primes which divide at least one term of the sequence. Moldavia Olympiad, 1994 What do you mean by infinitely many primes dividing at least one term of the sequence? if there are really infinitely many primes dividing one term of the sequence the that term is definitely $0$. It means that there are infinitely many prime numbers $p$ such that $p$ divides $a_i$ for some index $i{}$. Still confusing. Do you mean that the set of primes dividing the terms of the sequence is infinite? Like $$S=\{p \text{ prime}| p\mid a_i \text{ for some } i\}$$is infinite?

@above I do not understand how it is "still" confusing. What you wrote is precisely what I said: the set of primes $p$ which divide some $a_i$ is infinite. In mathematical terms, $|S|=\infty$ where $S=\{p\text{ prime}:\exists \ i\text{ such that }p\mid a_i\}.$ I will not respond any further in order not to trash the thread.