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(Theorem)
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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, respectively.
Theorem For any positive integer $n$ , $2^{\omega(n)} \le \tau(n) \le 2^{\Omega(n)}$ .
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:
Hence, $2^{\omega(p^k)} \le \tau(p^k) \le 2^{\Omega(p^k)}$ . It follows that $2^{\omega(n)} \le \tau(n) \le 2^{\Omega(n)}$ . 
This theorem has an obvious corollary.
Corollary For any squarefree positive integer $n$ , $2^{\omega(n)}=\tau(n)=2^{\Omega(n)}$ .
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Cross-references: squarefree, obvious, theorem, prime, inequality, numbers, multiplicative, integer, positive, divisor function, number of distinct prime factors function
This is version 11 of  , born on 2006-07-29, modified 2008-01-01.
Object id is 8194, canonical name is 2omeganLeTaunLe2Omegan.
Accessed 1640 times total.
Classification:
| AMS MSC: | 11A25 (Number theory :: Elementary number theory :: Arithmetic functions; related numbers; inversion formulas) |
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Pending Errata and Addenda
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