$(\left\lfloor\frac{n}{3}\right\rfloor)_{n \in \mathbb{N}}$ is a subsequence of $(\left\lfloor\frac{n}{5}\right\rfloor)_{n \in \mathbb{N}}$

Question

How to prove that $\left(\left\lfloor\frac{n}{3}\right\rfloor\right)_{n \in \mathbb{N}}$ is a subsequence of $\left(\left\lfloor\frac{n}{5}\right\rfloor\right)_{n \in \mathbb{N}}$?

Answer 1

Given a sequence $(a_n)_{n\in \Bbb N}$ and a strictly increasing sequence of natural numbers $\alpha \colon \Bbb N\to \Bbb N$, the composition $a\circ \alpha$, often denoted by $(a_{\alpha _n})_{n\in \Bbb N}$, is a subsequence of $(a_n)_{n\in \Bbb N}$.

For $\left(\left\lfloor\frac{n}{3}\right\rfloor\right)_{n \in \mathbb{N}}$ to be a subsequence of $\left(\left\lfloor\frac{n}{5}\right\rfloor\right)_{n \in \mathbb{N}}$ you need to find $\alpha$ (as above) such that $\left(\left\lfloor\frac{\alpha _n}{5}\right\rfloor\right)_{n \in \mathbb{N}}=\left(\left\lfloor\frac{n}{3}\right\rfloor\right)_{n \in \mathbb{N}}$.

Can you find $\alpha$?

Hint: $\alpha$ is the floor function composed with something else.

Answer:

Let $\alpha _n=\left\lfloor \dfrac{5n}3\right\rfloor$, for all $n\in \Bbb N$.