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Say $M$ is the number of divisors of an integer. Is there a simple formula for the minimal integer $n$ so that the number of divisors of $n$ is $M$?

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If $M$ is ordinary - yes (ordinary in sense defined here -…). I guess you found that already, but that's my 2¢. – Harun ┼áiljak Sep 4 '12 at 15:01
I guess the growth rate is all you could expect to say anything about. – i. m. soloveichik Sep 4 '12 at 15:14
3 - there are references to partial results – Julian Kuelshammer Sep 4 '12 at 15:21
Given n has prime factorization with exponents a,b,c..., M is (a+1)(b+1)((c+1)..., and the exponent bases are the first however many primes. You have an upper bound of 2^(M+1), which can be optimized quickly. If M itself has k prime factors, you get (p_k)(kf) as a quick upper bound, where p_k is the kth smallest prime and f is the largest of the k prime factors. However, even a greedy strategy may not yield the minimum, so you will still need to do some searching. Gerhard "Ask Me About System Design" Paseman, 2012.09.04 – Gerhard Paseman Sep 4 '12 at 15:28
Do you insist that the number of divisors is exatly $M$ or is at least $M$ what you are interested in. If you care about exact count note that this then depends quite a bit on the $M$ and not just its rough sizes (and this is what the comments refer to). To highligth something implictly in other comments. If $M$ is prime for example, it is clear that the only eligible $N$ are (M-1)th prime powers, and then clearly one needs to take a power of two. If you care just about at least $M$ this is a different question; then a keyword is "Highly composite number". – quid Sep 4 '12 at 15:45

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