I have seen lots of posts on "exactly $n$ divisors" and understood the process as well, but I can't seem to find or come up with an algorithm, apart from brute force, for "at least $n$ divisors".
Since supposes we are looking for a number with number of divisors > 10;
11 is the first number that's > 10, but the smallest number that has 11 divisors (including 1 and itself) is 2^10 i.e 1024….
while if we consider 12, [12 is also > 10]… the smallest number that has 12 divisors is 60 , which is way less than 1024.
Is there an efficient algo apart from just looping and checking the num of divisors till we find a num whose #divisors > n ?
Best Answer
I am addressing the question of finding the smallest number with AT LEAST $N$ divisors. Trying to get exactly $N$ divisors is a different level of nightmare.
well, i cannot tell how much work you are willing to put into this. Guy Robin came up with a method for finding the highly composite numbers between two consecutitive superior highly composite numbers.
Assuming that is too ambitious, do this: first find the first primorial number $P =2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdots p_k$ that has more divisors than you need. That would mean $2^k$ is larger than you want.
Next, make a fairly complicated multiple loop to produce all the numbers $$ 2^{e_2} 3^{e_3} 5^{e_5} \cdots p_k^{e_{p_k}} \; \leq \; \; P $$ WITH $$ e_2 \geq e_3 \geq e_5 \geq \ldots \geq e_{p_k} \geq 0 $$ then calculate the number of divisors ( depends only on the exponents) and calculate the number itself.
The winner, smallest such number with at least some $N$ divisors, is likely to have a number of exponents zero at the end. The general pattern is that highly composite numbers are fairly similar to the least common multiple $L$ of $1,2,3,4,5,6,7,\ldots, w$ as far as the relative sizes of the prime exponents when the whole thing is factored. I should point out that the prime number theorem says $\log L \approx W.$
A little more efficiency: we can combine the two bounds. We can find the smallest LCM $L$ that has enough divisors. This will give a better (smaller) upper bound. Then, using the set of primes we used in producing the primorial $P,$ we can produce all the numbers $$ 2^{e_2} 3^{e_3} 5^{e_5} \cdots p_k^{e_{p_k}} \; \leq \; \; L $$ WITH $$ e_2 \geq e_3 \geq e_5 \geq \ldots \geq e_{p_k} \geq 0 $$
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