Construction of sequence satisfying some bound

Question

Can we construct a sequence of integers $(a_n)_{n\in\mathbb{N}}$ such that, $$ \frac{1}{4}< \frac{a_n}{n}<\frac{1}{2}, \;\;\;\;\;\;\;\;\forall n\ge n_0 $$ where $n_0$ some integer.

Answer 1

Note that,

$$\frac 14\lt\frac{a_n}n\lt\frac 12\iff\frac n4\lt a_n\lt\frac n2\tag 1$$

Then, an integer sequence $\{a_n\}$ satisfying $(1)$ for all $n\geq 8$ can be given by, $$a_n=\left\lfloor\frac{\frac n4+\frac n2}2\right\rfloor=\left\lfloor\frac{3n}8\right\rfloor$$

where $\lfloor\cdot\rfloor$ denotes the floor function.

Answer 2

It is easy to check that we must have $n_0 \ge 5$.

If $n_0 \ge 5$, and $n \ge n_0$, then the points ${p \over n}$ are spaced apart by multiples of ${1 \over n} \le {1 \over n_0} \le {1 \over 5}$ and hence any closed interval of length $\ge {1 \over 5}$ will contain one of these points.

Now note that $[{11 \over 40}, {19 \over 40}] \subset ({1 \over 4} , {1 \over 2})$ and has length ${ 1\over 5}$.

Hence if we let $a_n = \lceil {11 n \over 40} \rceil$, then ${a_n \over n} \in [{11 \over 40}, {19 \over 40}]$.