TITLE: Number of Divisors of N factorial
QUESTION [1 upvotes]: Say $d(N) =$ Number of factors of $N!$
Briefly: I wish to know if there is a Recurrence relation for this problem.
Now I wish to Know if there is a way to calculate $d(N)$ in terms of previously calculated values ...
I want to know this as a part of the problem on spoj (http://www.spoj.com/problems/EASYFACT/)
What I mentioned was a part of my approach to solve this.

REPLY [1 votes]: If you know three things, namely factorisation of $N=\prod_n p_n^{i_n}$,
value of $d(N-1)$ 
and $m_n$ maximal power of $p_n$ dividing $(N-1)!$ for each $n$ then I think you should be able to calculate $d(N)$.
In that case $(d(N)=d(N-1)/(\prod_n(m_n+1)) \times (\prod_n(m_n+i_n+1)) $ by standard number of divisors formula.
I don't think you can simplfy much more unless I misunderstood your question.
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