# $2^{\omega(n)}\leq\tau(n)\leq 2^{\Omega(n)}$

Throughout this entry, $\omega$, $\tau$, and $\Omega$ denote the number of distinct prime factors function, the divisor function, and the number of (nondistinct) prime factors function (http://planetmath.org/NumberOfNondistinctPrimeFactorsFunction), respectively.

###### Theorem.

For any positive integer $n$, $2^{\omega(n)}\leq\tau(n)\leq 2^{\Omega(n)}$.

###### Proof.

Note that $2^{\omega(n)}$, $\tau(n)$, and $2^{\Omega(n)}$ are multiplicative. Also note that, for any positive integer $n$, the numbers $2^{\omega(n)}$, $\tau(n)$, and $2^{\Omega(n)}$ are positive integers. Therefore, it will suffice to prove the inequality for prime powers.

Let $p$ be a prime and $k$ be a positive integer. Thus:

$\begin{array}[]{rl}\displaystyle 2^{\omega(p^{k})}&=2\\ \\ \tau(p^{k})&=k+1\\ \\ \displaystyle 2^{\Omega(p^{k})}&=2^{k}\end{array}$

Hence, $2^{\omega(p^{k})}\leq\tau(p^{k})\leq 2^{\Omega(p^{k})}$. It follows that $2^{\omega(n)}\leq\tau(n)\leq 2^{\Omega(n)}$. ∎

This theorem has an obvious corollary.

###### Corollary.

For any squarefree positive integer $n$, $2^{\omega(n)}=\tau(n)=2^{\Omega(n)}$.

Title $2^{\omega(n)}\leq\tau(n)\leq 2^{\Omega(n)}$ 2omeganletaunle2Omegan 2013-03-22 16:07:28 2013-03-22 16:07:28 Wkbj79 (1863) Wkbj79 (1863) 14 Wkbj79 (1863) Theorem msc 11A25 NumberOfDistinctPrimeFactorsFunction TauFunction NumberOfNondistinctPrimeFactorsFunction DisplaystyleYOmeganOleftFracxlogXy12YRightFor1LeY2