$2^{\omega (n)}\le \tau (n)\le 2^{\mathrm{\Omega }(n)}$

Throughout this entry, $\omega$, $\tau$, and $\mathrm{\Omega }$ 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 $n$, ${\mathrm{2}}^{\omega \mathit{}\mathrm{(}n\mathrm{)}}\mathrm{\le }\tau \mathit{}\mathrm{(}n\mathrm{)}\mathrm{\le }{\mathrm{2}}^{\mathrm{\Omega }\mathit{}\mathrm{(}n\mathrm{)}}$.

Proof.

Note that $2^{\omega (n)}$, $\tau (n)$, and $2^{\mathrm{\Omega }(n)}$ are multiplicative. Also note that, for any positive integer $n$, the numbers $2^{\omega (n)}$, $\tau (n)$, and $2^{\mathrm{\Omega }(n)}$ 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:

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

This theorem has an obvious corollary.

Corollary.

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