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Let $d_{3}(n)$ denote the number of ordered 3-tuples of positive integers whose product is $n$ and $\omega(n)$ the number of distinct primes dividing $n$. How does one show that $3^{\omega(n)} \geq d_{3}(n)$ for every $n$?

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    One doesn't. If $n$ is, say, $2^{10}$, then $\omega(n)=1$, so $3^{\omega(n)}=3$, but $d_3(2^{10})$ is a lot bigger than $3$.2012-11-29

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