Let $a_n $ complex sequence prove that if $ a_n\to \infty$ then $|a_n|\to\infty$. Note that $a_n = x_n + y_ni$

Let $ a_n $ complex sequence prove that if $ a_n\to \infty$ then $ |a_n|\to\infty$ . Note that $ a_n = x_n + y_ni$ i dont know how to write that mathmatically.

trial :

Can i say that for every $ M>0$ there exist $ N$ such that for every $ n>N$ ,

$ ~~|x_n|>M~~ OR ~~~|y_n|>M$ ( At least one of them goes to $ \infty$ )

because of that $ |an| = \sqrt{(x_n)^2+(y_n)^2} > M$ and so $ |a_n|\to\infty$ .

What are the prime divisors of $\det(A_n)$, where $A_n$ is the $n\times n$ matrix given by $(A_n)_{i,j}={n\choose|i-j|}$?

For $ n\in\mathbb{Z}_{\ge 1}$ , let $ A_n$ be the $ n\times n$ matrix given by $ (A_n)_{i,j}={n\choose |i-j|}$ . From this post it is clear that $ $ \det(A_n)=\prod_{k=0}^{n-1}\left[\left(\exp\left(\frac{2\pi k i}{n}\right)+1\right)^n-1\right]=\prod_{k=0}^{n-1}\left(2^n(-1)^k\cos^n\left(\frac{\pi k}n\right)-1\right).$ $ Also, $ \det(A_n)$ is obviously integer for all $ n$ .

Question: What is known about the (prime) divisors of $ \det(A_n)$ ?

If this is too broad, I am particularly interested in pairs $ (p,d)$ where $ p$ is prime with $ d\mid p-1$ and $ p\nmid\det(A_{(p-1)/d})$ .

Asymptotic estimation of $A_n$

Let $ A_n$ represent the number of integers that can be written as the product of two element of $ [[1,n]]$ .

I am looking for an asymptotic estimation of $ A_n$ .

First, I think it’s a good start to look at the exponent $ \alpha$ such that :

$ $ A_n = o(n^\alpha)$ $

I think we have : $ 2 < alpha $ . To prove this lower bound we use the fact that the number of primer numbers $ \leq n$ is about $ \frac{n}{\log n}$ . Hence we have the trivial lower bound (assuming $ n$ is big enough) :

$ $ \frac{n}{\log n} \cdot \binom{ E(\frac{n}{\log n})}{2} = o(n^3)$ $

Now is it possible to get a good asymptotic for $ A_n$ and not just this lower bound ? Is what I’ve done so far correct ?

Thank you !