Published online by Cambridge University Press: 20 November 2018
Let ${{\left( {{n}_{k}} \right)}_{k\ge 1}}$ be an increasing sequence of positive integers, and let $f\left( x \right)$ be a real function satisfying
1
If ${{\lim }_{k\to \infty }}\frac{{{n}_{k+1}}}{{{n}_{k}}}=\infty $ the distribution of
2
converges to a Gaussian distribution. In the case
there is a complex interplay between the analytic properties of the function $f$, the number-theoretic properties of ${{\left( {{n}_{k}} \right)}_{k\ge 1}}$, and the limit distribution of (2).
In this paper we prove that any sequence ${{\left( {{n}_{k}} \right)}_{k\ge 1}}$ satisfying $\lim {{\sup }_{k\to \infty }}{{n}_{k+1}}/{{n}_{k}}=1$ contains a nontrivial subsequence ${{\left( {{m}_{k}} \right)}_{k\ge 1}}$ such that for any function satisfying (1) the distribution of
converges to a Gaussian distribution. This result is best possible: for any $\varepsilon >0$ there exists a sequence ${{\left( {{n}_{k}} \right)}_{k\ge 1}}$ satisfying lim $\underset{k\to \infty }{\mathop{\sup }}\,\frac{{{n}_{k+1}}}{{{n}_{k}}}\le 1+\varepsilon$ such that for every nontrivial subsequence ${{\left( {{m}_{k}} \right)}_{k\ge 1}}$ of ${{\left( {{n}_{k}} \right)}_{k\ge 1}}$ the distribution of (2) does not converge to a Gaussian distribution for some $f$.
Our result can be viewed as a Ramsey type result: a sufficiently dense increasing integer sequence contains a subsequence having a certain requested number-theoretic property.