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Prove that $f(n) = 2^{\omega(n)}$ is multiplicative where $\omega(n)$ is the number of distinct primes.

My attempt:

Let $a = p_1p_2\cdots p_k$ and $b = q_1q_2\cdots q_t$ where $p_i$ and $q_j$ are prime factors, and $p_i \neq q_j$ for all $1 \leq i \leq k$ and $1 \leq j \leq t$. We will show that $2^{\omega(ab)} = 2^{\omega(a)} \times 2^{\omega(b)}$

Indeed, $\omega(a) = k$ and $\omega(b) = t$. Then $2^{\omega(a)} \times 2^{\omega(b)} = 2^{k + t}$
Where $2^{\omega(ab)} = 2^{k + t}$
$\therefore 2^{\omega(ab)} = 2^{\omega(a)} \times 2^{\omega(b)}$

Am I in the right track?

Thanks,

Grigory M
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roxrook
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2 Answers2

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Hint: Notice $$2^{\omega(a)}\times 2^{\omega(b)}=2^{\omega(a)+\omega(b)}$$ so try to relate $\omega(a)+\omega(b)$ and $\omega(ab)$.

To simplify things, you need only show it holds for $a=p^r$, $b=q^t$ where $q,p$ are prime numbers.

Eric Naslund
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  • Thanks for your hint. So what my attempt was incorrect? Could you show where did I miss? Thank you. – roxrook Mar 01 '11 at 05:50
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    @Chan: Everything looks correct to me. The only thing I suggest in the future is notice that multiplicative functions are given completely by the prime powers, and it suffices to show that a function is multiplicative by looking only at the prime powers. (Consequence of FTA) This makes solutions quicker I find. – Eric Naslund Mar 01 '11 at 05:53
  • got it. Thank you. By the way would you mind taking a look at this thread again http://math.stackexchange.com/questions/24143/prove-that-two-distinct-number-of-the-form-a2n-1-and-a2m-1-ar – roxrook Mar 01 '11 at 05:57
  • Done, hope that helps. – Eric Naslund Mar 01 '11 at 06:18
  • I am interested in counting $\sum_{i \le N} |\omega(i)=4|$ . was wondering if you could help. – sibillalazzerini Mar 27 '23 at 09:00
  • @sibillalazzerini: Look into $\pi_{k}(x)$, which is the number of integers less than $x$ with exactly $k$ prime factors. In particular, your sum is closely related to $\pi_4(x)$. You'll have to be careful about prime powers though. – Eric Naslund Mar 28 '23 at 15:56
  • @EricNaslund I guess I am talking about small omega and you are talking about its relation with big omega https://en.wikipedia.org/wiki/Prime_omega_function ? was wondering why big omega counting is more popular. even wikipedia has an asymptotic for it. – sibillalazzerini Mar 28 '23 at 17:17
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Note that if $f : \mathbb{N} \to \mathbb{C}$ is additive:

$$ \gcd(m,n) = 1 \quad \Longrightarrow \quad f(mn) = f(m) + f(n), $$

then for any $t > 0$, $g(n) := t^{f(n)}$ is multiplicative:

$$ \gcd(m,n) = 1 \quad \Longrightarrow \quad g(mn) = g(m) g(n), $$

since

$$ g(mn) = t^{f(mn)} = t^{f(m) + f(n)} = t^{f(m)}t^{f(n)} = g(m) g(n). $$

Note also that this $g$ will never be the zero function. Therefore, it suffices to show that $\omega(n)$ is additive (which has indeed already been shown).

JavaMan
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