Is there a theorem, lemma, or proof somewhere that proves an upper bound for the number of factors that a number can have?
If not, would it be fairly trivial to prove that it is $log_2 n$?
Asked
Active
Viewed 3,429 times
3
-
1The number of factors is $o(n^\epsilon)$ for every positive $\epsilon$, but it can be much larger than $\log n$. Or did you mean prime factors? – André Nicolas Mar 15 '16 at 21:56
-
I didn't mean prime factors, which means that something is off in my thinking. – Gavin D. Howard Mar 15 '16 at 22:11
-
Out of curiosity, how many prime factors are there? – Gavin D. Howard Mar 15 '16 at 22:13
-
Never mind. http://math.stackexchange.com/questions/938204/upper-bound-number-of-distinct-prime-factors – Gavin D. Howard Mar 15 '16 at 22:13
-
Also here: http://math.stackexchange.com/questions/409675/number-of-distinct-prime-factors – Friedrich Philipp Mar 15 '16 at 22:40
2 Answers
4
Let ${d(n)}$ denote the number of factors of $n$.
There is a useful bound which has applications in many area:
${d(n)\le n^{O(1/\log \log n)} = \exp(O(\frac{\log n}{\log\log n}))}$.
Maybe Tao's blog can help you understand the answer better.
https://terrytao.wordpress.com/2008/09/23/the-divisor-bound/
WangzhPP
- 41
- 2
2
given that log2(24) is approx. 4.5, but 24 has 8 factors. I have to say, that this is not an upper bound.
I personal looked for an upper bound recently to the number of factors function and I got sqrt(3*x), it might not be the tightest upper bound. but it works.
look at the image bellow
Sergio Fernandez
- 136
-
(39*x)^(1/3) is probably a tighter upper bound. again I got this from plotting the # of factors function, I did not do a formal proof. – Sergio Fernandez Oct 29 '16 at 02:43