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Is there a formula that can tell us how many distinct prime factors a number has? We have closed form solutions for the number of factors a number has and the sum of those factors but not the number of distinct prime factors.

So for example: \begin{array}{rr} \text{number} & \text{distinct}\atop \text{prime factors} \\ 1&0 \\ 2&1 \\ 3&1 \\ 4&1 \\ 5&1 \\ 6&2 \\ 7&1 \\ 8&1 \\ 9&1 \\ 10&2 \end{array}

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  • $\begingroup$ What closed form solutions are you referring to? If you mean $d(n) = \prod_{p^r \| n} (r+1)$ then what is wrong with $\omega(n) = \sum_{p \mid n} 1$? $\endgroup$
    – Erick Wong
    Jun 2, 2013 at 23:25

4 Answers 4

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This is a historically interesting question as it led Hardy and Ramanujan to lay the foundation to probabilistic number theory in course of their solution to this problem. Given $n$ there is no non-trivial deterministic closed form formula for the number of distinct prime factors of $n$. However we have very good probabilistic formula for the same.

Hardy and Ramanujan proved that for almost all integers, the number is distinct primes dividing a number $n$ is formula

$$ \omega(n) \sim \log\log n. $$

We can do much better than the Hardy-Ramanujan estimate and find and estimate of $\omega(n)$ which can be bounded by normal distribution. Erdos and Kac imporved the estimate of $\omega(n)$ and proved that

$$ \lim_{x \to \infty} \frac{1}{x}\# \Big\{n\le x, \frac{\omega(n) - \log\log n}{\sqrt{\log\log n}} \le t \Big\} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{t}e^{-\frac{u^2}{2}}du $$

This formula says that if $n$ is a large number, we can estimate the distribution of the number of prime factors for numbers of this range. For example we can show that around 12.6% of 10,000 digit numbers are constructed from 10 distinct prime numbers and around 68% (±$\sigma$) are constructed from between 7 and 13 distinct primes.

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    $\begingroup$ "for almost all integers, the number is distinct primes dividing a number $n$ is given by the asymptotic formula" does not make sense. Asymptotic formulas do not apply to individual instances; you wouldn't be able to cite a single number $n$ for which the formula gives its number of distinct primes. $\endgroup$
    – Marc van Leeuwen
    Jun 3, 2013 at 7:44
  • $\begingroup$ Yeah asymptotic is not the right word, but the result remains the same. This is a probabilistic result. $\endgroup$
    – Nilotpal Sinha
    Jun 3, 2013 at 8:16
  • $\begingroup$ I see nothing random at all in the formulation. $\endgroup$
    – Marc van Leeuwen
    Jun 3, 2013 at 8:21
  • $\begingroup$ Well then for a more precise formulation, you might want to refer to mathworld.wolfram.com/Hardy-RamanujanTheorem.html and mathworld.wolfram.com/Erdos-KacTheorem.html $\endgroup$
    – Nilotpal Sinha
    Jun 3, 2013 at 8:33
  • $\begingroup$ And that statement involves an expression $\sqrt{\ln\ln x}$, which cannot be taken out without the statement losing its meaning. $\endgroup$
    – Marc van Leeuwen
    Jun 3, 2013 at 8:37
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This is commonly known as $\omega(n)$. See Wikipedia.

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We do not have the closed form solutions you claim. We have formulas that assume that we know the prime factorization.

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  • $\begingroup$ I am interested in counting $\sum_{i \le N} |\omega(i)=4|$ . was wondering if you could help. $\endgroup$
    – sibillalazzerini
    Mar 27 at 9:02
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If you want to calculate $\omega(n)$ for a large range of $n$, it's simple to modify the Sieve of Eratosthenes to count distinct prime factors. Instead of marking multiples of each prime, keep a counter for every number and increment instead of marking. SO code example

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