# Tag Archives: asymptotics

## An asymptotic property of the divisor counting function

Let $\tau(n)$ be the number of positive divisors of the positive integer $n$ and let $\phi(n)$ be Euler’s function. Then it is easily seen that

$\phi(n)+\tau(n)\le n+1$

for each $n$. So

$\displaystyle\frac{\phi(n)}{n}+\frac{\tau(n)}{n}\le 1+\frac 1n$

holds for all $n$.

$\displaystyle\limsup_{n\to\infty}\frac{\phi(n)}{n}=\lim_{n\to\infty}\frac{n-1}{n}=1$.

So it follows that

$\displaystyle\limsup_{n\to\infty}\frac{\tau(n)}{n}=0$.

Therefore, by the sandwich theorem,

$\displaystyle\lim_{n\to\infty}\frac{\tau(n)}{n}=0$,

i.e., $\tau(n)=o(n)$ as $n\to\infty$. In layman’s terms, this is saying that the number of divisors of $n$ is small compared to $n$ when $n$ is large. The following plot of $\tau(n)/n$ against $n$ for $1\le n\le 500$ captures this nicely.

So this means, for example, that there are only finitely many positive integers $n$ with more than $n/2015$ divisors, which I think is pretty neat!

Filed under Number theory

## A nice result on arithmetic progressions

Let $\mathbb N_0$ denote the set of the non-negative integers. For a subset $S\subseteq\mathbb N_0$, we can define the ‘counting function’

$\pi_S(x):=|\{n\in\mathbb N_0\cap S:n\le x\}|$

i.e. $\pi_S(x)$ is the number of elements in $S$ that are at most $x$. Then we can define the (asymptotic) density of $S$ as

$\displaystyle \delta(S):=\lim_{x\to\infty}\frac{\pi_S(x)}{x}$

when this limit exists.

For example, the set of the even (resp. odd) numbers has density $1/2$. More generally, any arithmetic progression $a+d\mathbb N_0:=\{a+dn:n\in\mathbb N_0\}$ has density $1/d$. Another example is the set of prime numbers, which has density $0$ by the prime number theorem. From this it also follows that the set of composite numbers has density $1$.

Theorem. If a finite union of arithmetic progressions of positive integers has density $1$, then it contains all sufficiently large positive integers.

Proof. Let

$\displaystyle A=\bigcup_{i=1}^k(a_i+d_i\mathbb N_0)$

be a finite union of arithmetic progressions such that $\delta(A)=1$. Note that (this is the main trick) we can write:

$\displaystyle A=\bigcup_{i=1}^k(a_i+d_i\mathbb N_0)=\bigcup_{j=1}^l(b_i+d\mathbb N_0)$

for some $b_1,\dots,b_l$, where $d$ is the least common multiple of $d_1,\dots,d_k$.

Why can we do this? Consider for example $(1+2\mathbb N_0)\cup(2+3\mathbb N_0)$. We can write this as

$\displaystyle \underbrace{(1+6\mathbb N_0)\cup (3+6\mathbb N_0)\cup (5+6\mathbb N_0)}_{1+2\mathbb N_0}\cup \underbrace{(2+6\mathbb N_0)\cup (5+6\mathbb N_0)}_{2+3\mathbb N_0}$

$\displaystyle =(1+6\mathbb N_0)\cup (2+6\mathbb N_0)\cup (3+6\mathbb N_0)\cup (5+6\mathbb N_0)$

And this idea works for any finite union of arithmetic progressions.

It is then clear that $\{b_1,\dots,b_l\}$ has to be a complete set of residues modulo $d$, because otherwise $A$ would have density strictly less than $1$. Therefore $A$ has to contain all positive integers $\ge\max\{b_1,\dots,b_l\}$. $\square$

We get the following nice corollary:

Corollary. The set of all composite numbers cannot be expressed as a finite union of arithmetic progressions of positive integers.

Proof. Otherwise by the above theorem all sufficiently large positive integers would be composite, contradicting the fact that there are infinitely many primes. $\square$