Using the prime number theorem find an asymptotic expression for $d(n!)$ where $d$ is the number of divisors.
number-theoryprime-numbersanalytic-number-theory
asked 2011-11-29
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What do you mean by $d(n!)$? – 2011-11-29
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corrected, thanks. – 2011-11-29
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You might want to have a look [here (OEIS)](http://oeis.org/A027423). – 2011-11-29
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$d$ is multiplicative so write $n!=\prod_{p\le n}p^{v_p(n)}$ where each $v_p(n)=\sum_{l=1}\lfloor n/p^l\rfloor$, then $$d(n!)=\prod_{p\le n}\left(1+\sum_{l=1}\left\lfloor\frac{n}{p^l}\right\rfloor\right)\le \prod_{p\le n}\left(1+\frac{n}{p-1}\right)$$ which is bounded below by $n^{\pi(n)}/n\#$ and above by $n^{\pi(n)}/\prod(1-1/p)$. I'll see if I can flesh this out later. – 2011-11-29