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 $n$?
Inequality involving $\omega(n)$
0
$\begingroup$
number-theory
-
0One 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