Prime omega function

In number theory, the prime omega functions $$\omega(n)$$ and $$\Omega(n)$$ count the number of prime factors of a natural number $$n.$$ Thereby $$\omega(n)$$ (little omega) counts each distinct prime factor, whereas the related function $$\Omega(n)$$ (big omega) counts the total number of prime factors of $$n,$$ honoring their multiplicity (see arithmetic function). That is, if we have a prime factorization of $$n$$ of the form $$n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$$ for distinct primes $$p_i$$ ($$1 \leq i \leq k$$), then the respective prime omega functions are given by $$\omega(n) = k$$ and $$\Omega(n) = \alpha_1 + \alpha_2 + \cdots + \alpha_k$$. These prime factor counting functions have many important number theoretic relations.

Properties and relations
The function $$\omega(n)$$ is additive and $$\Omega(n)$$ is completely additive.

$$\omega(n)=\sum_{p\mid n} 1$$

If $$p$$ divides $$n$$ at least once we count it only once, e.g. $$\omega(12)=\omega(2^2 3)=2$$.

$$\Omega(n) =\sum_{p^\alpha\mid n} 1 =\sum_{p^\alpha\parallel n}\alpha$$

If $$p$$ divides $$n$$ $$\alpha \geq 1$$ times then we count the exponents, e.g. $$\Omega(12)=\Omega(2^2 3^1)=3$$. As usual, $$ p^\alpha\parallel n $$ means $$\alpha $$ is the exact power of $$p$$ dividing $$n $$.

$$\Omega(n) \ge \omega(n)$$

If $$\Omega(n)=\omega(n) $$ then $$n$$ is squarefree and related to the Möbius function by


 * $$\mu(n) = (-1)^{\omega(n)} = (-1)^{\Omega(n)}$$

If $$ \omega(n) = 1 $$ then $$ n $$ is a prime power, and if $$\Omega(n)=1 $$ then $$n$$ is a prime number.

It is known that the average order of the divisor function satisfies $$2^{\omega(n)} \leq d(n) \leq 2^{\Omega(n)}$$.

Like many arithmetic functions there is no explicit formula for $$\Omega(n)$$ or $$\omega(n) $$ but there are approximations.

An asymptotic series for the average order of $$\omega(n)$$ is given by


 * $$\frac{1}{n} \sum\limits_{k = 1}^n \omega(k) \sim \log\log n + B_1 + \sum_{k \geq 1} \left(\sum_{j=0}^{k-1} \frac{\gamma_j}{j!} - 1\right) \frac{(k-1)!}{(\log n)^k}, $$

where $$B_1 \approx 0.26149721$$ is the Mertens constant and $$\gamma_j$$ are the Stieltjes constants.

The function $$\omega(n)$$ is related to divisor sums over the Möbius function and the divisor function including the next sums.


 * $$\sum_{d\mid n} |\mu(d)| = 2^{\omega(n)} $$
 * $$\sum_{d\mid n} |\mu(d)| k^{\omega(d)} = (k+1)^{\omega(n)} $$
 * $$\sum_{r\mid n} 2^{\omega(r)} = d(n^2) $$
 * $$\sum_{r\mid n} 2^{\omega(r)} d\left(\frac{n}{r}\right) = d^2(n) $$
 * $$\sum_{d\mid n} (-1)^{\omega(d)} = \prod\limits_{p^{\alpha}||n} (1-\alpha)$$

\sum_{\stackrel{1\le k\le m}{(k,m)=1}} \gcd(k^2-1,m_1)\gcd(k^2-1,m_2) =\varphi(n)\sum_{\stackrel{d_1\mid m_1} {d_2\mid m_2}} \varphi(\gcd(d_1, d_2)) 2^{\omega(\operatorname{lcm}(d_1, d_2))},\ m_1, m_2 \text{ odd}, m = \operatorname{lcm}(m_1, m_2) $$
 * $$\sum_\stackrel{1\le k\le n}{\operatorname{gcd}(k,m)=1} \!\!\!\! 1 = n \frac {\varphi(m)}{m} + O \left ( 2^{\omega(m)} \right )$$

The characteristic function of the primes can be expressed by a convolution with the Möbius function:


 * $$ \chi_{\mathbb{P}}(n)

= (\mu \ast \omega)(n) = \sum_{d|n} \omega(d) \mu(n/d). $$

A partition-related exact identity for $$\omega(n)$$ is given by


 * $$\omega(n) = \log_2\left[\sum_{k=1}^n \sum_{j=1}^k \left(\sum_{d\mid k} \sum_{i=1}^d p(d-ji) \right) s_{n,k} \cdot |\mu(j)|\right], $$

where $$p(n)$$ is the partition function, $$\mu(n)$$ is the Möbius function, and the triangular sequence $$s_{n,k}$$ is expanded by


 * $$s_{n,k} = [q^n] (q; q)_\infty \frac{q^k}{1-q^k} = s_o(n, k) - s_e(n, k), $$

in terms of the infinite q-Pochhammer symbol and the restricted partition functions $$s_{o/e}(n, k)$$ which respectively denote the number of $$k$$'s in all partitions of $$n$$ into an odd (even) number of distinct parts.

Continuation to the complex plane
A continuation of $$\omega(n)$$ has been found, though it is not analytic everywhere. Note that the normalized $$\operatorname{sinc}$$ function $$ \operatorname{sinc}(x) = \frac{\sin(\pi x)}{\pi x} $$ is used.


 * $$\omega(z) = \log_2\left(\sum_{n=1}^{\lceil Re(z) \rceil} \operatorname{sinc} \left(\prod_{m=1}^{\lceil Re(z) \rceil+1} \left( n^2+n-mz \right) \right) \right) $$

This is closely related to the following partition identity. Consider partitions of the form
 * $$a= \frac{2}{c} + \frac{4}{c} + \ldots + \frac{2(b-1)}{c} + \frac{2b}{c} $$

where $$ a $$, $$ b $$, and $$ c $$ are positive integers, and $$ a > b > c $$. The number of partitions is then given by $$ 2^{\omega(a)} - 2 $$.

Average order and summatory functions
An average order of both $$\omega(n)$$ and $$\Omega(n)$$ is $$\log\log n$$. When $$n$$ is prime a lower bound on the value of the function is $$\omega(n) = 1$$. Similarly, if $$n$$ is primorial then the function is as large as $$\omega(n) \sim \frac{\log n}{\log\log n}$$ on average order. When $$n$$ is a power of 2, then $$\Omega(n) \sim \frac{\log n}{\log 2}$$ .

Asymptotics for the summatory functions over $$\omega(n)$$, $$\Omega(n)$$, and $$\omega(n)^2$$ are respectively computed in Hardy and Wright as


 * $$\begin{align}

\sum_{n \leq x} \omega(n) & = x \log\log x + B_1 x + o(x) \\ \sum_{n \leq x} \Omega(n) & = x \log\log x + B_2 x + o(x) \\ \sum_{n \leq x} \omega(n)^2 & = x (\log\log x)^2 + O(x \log\log x) \\ \sum_{n \leq x} \omega(n)^k & = x (\log\log x)^k + O(x (\log\log x)^{k-1}), k \in \mathbb{Z}^{+}, \end{align} $$

where $$B_1 \approx 0.2614972128$$ is the Mertens constant and the constant $$B_2$$ is defined by


 * $$B_2 = B_1 + \sum_{p\text{ prime}} \frac{1}{p(p-1)} \approx 1.0345061758.$$

Other sums relating the two variants of the prime omega functions include


 * $$\sum_{n \leq x} \left\{\Omega(n) - \omega(n)\right\} = O(x), $$

and


 * $$\#\left\{n \leq x : \Omega(n) - \omega(n) > \sqrt{\log\log x}\right\} = O\left(\frac{x}{(\log\log x)^{1/2}}\right). $$

Example I: A modified summatory function
In this example we suggest a variant of the summatory functions $$S_{\omega}(x) := \sum_{n \leq x} \omega(n)$$ estimated in the above results for sufficiently large $$x$$. We then prove an asymptotic formula for the growth of this modified summatory function derived from the asymptotic estimate of $$S_{\omega}(x)$$ provided in the formulas in the main subsection of this article above.

To be completely precise, let the odd-indexed summatory function be defined as


 * $$S_{\operatorname{odd}}(x) := \sum_{n \leq x} \omega(n) [n\text{ odd}], $$

where $$[\cdot]$$ denotes Iverson bracket. Then we have that


 * $$S_{\operatorname{odd}}(x) = \frac{x}{2} \log\log x + \frac{(2B_1-1)x}{4} + \left\{\frac{x}{4}\right\} - \left[x \equiv 2,3 \bmod{4}\right] + O\left(\frac{x}{\log x}\right).$$

The proof of this result follows by first observing that



\omega(2n) = \begin{cases} \omega(n) + 1, & \text{if } n \text{ is odd; } \\ \omega(n), & \text{if } n \text{ is even,} \end{cases} $$

and then applying the asymptotic result from Hardy and Wright for the summatory function over $$\omega(n)$$, denoted by $$S_{\omega}(x) := \sum_{n \leq x} \omega(n)$$, in the following form:


 * $$ \begin{align}

S_\omega(x) & = S_{\operatorname{odd}}(x) + \sum_{n \leq \left\lfloor\frac{x}{2}\right\rfloor} \omega(2n) \\ & = S_{\operatorname{odd}}(x) + \sum_{n \leq \left\lfloor\frac{x}{4}\right\rfloor} \left(\omega(4n) + \omega(4n+2)\right) \\ & = S_{\operatorname{odd}}(x) + \sum_{n \leq \left\lfloor\frac{x}{4}\right\rfloor} \left(\omega(2n) + \omega(2n+1) + 1\right) \\ & = S_{\operatorname{odd}}(x) + S_{\omega}\left(\left\lfloor\frac{x}{2}\right\rfloor\right) + \left\lfloor\frac{x}{4}\right\rfloor. \end{align} $$

Example II: Summatory functions for so-termed factorial moments of &omega;(n)
The computations expanded in Chapter 22.11 of Hardy and Wright provide asymptotic estimates for the summatory function


 * $$\omega(n) \left\{\omega(n)-1\right\},$$

by estimating the product of these two component omega functions as


 * $$\omega(n) \left\{\omega(n)-1\right\} = \sum_{\stackrel{pq\mid n} {\stackrel{p \neq q}{p,q\text{ prime}}}} 1 =

\sum_{\stackrel{pq\mid n}{p,q\text{ prime}}} 1 - \sum_{\stackrel{p^2\mid n}{p\text{ prime}}} 1.$$

We can similarly calculate asymptotic formulas more generally for the related summatory functions over so-termed factorial moments of the function $$\omega(n)$$.

Dirichlet series
A known Dirichlet series involving $$\omega(n)$$ and the Riemann zeta function is given by


 * $$\sum_{n \geq 1} \frac{2^{\omega(n)}}{n^s} = \frac{\zeta^2(s)}{\zeta(2s)},\ \Re(s) > 1. $$

We can also see that
 * $$ \sum_{n \geq 1} \frac{z^{\omega(n)}}{n^s} = \prod_p \left(1 + \frac{z}{p^s-1}\right), |z| < 2, \Re(s) > 1,$$
 * $$ \sum_{n \geq 1} \frac{z^{\Omega(n)}}{n^s} = \prod_p \left(1 - \frac{z}{p^s}\right)^{-1}, |z| < 2, \Re(s) > 1,$$

The function $$\Omega(n)$$ is completely additive, where $$\omega(n)$$ is strongly additive (additive). Now we can prove a short lemma in the following form which implies exact formulas for the expansions of the Dirichlet series over both $$\omega(n)$$ and $$\Omega(n)$$:

Lemma. Suppose that $$f$$ is a strongly additive arithmetic function defined such that its values at prime powers is given by $$f(p^{\alpha}) := f_0(p, \alpha)$$, i.e., $$f(p_1^{\alpha_1} \cdots p_k^{\alpha_k}) = f_0(p_1, \alpha_1) + \cdots + f_0(p_k, \alpha_k)$$ for distinct primes $$p_i$$ and exponents $$\alpha_i \geq 1$$. The Dirichlet series of $$f$$ is expanded by
 * $$\sum_{n \geq 1} \frac{f(n)}{n^s} = \zeta(s) \times \sum_{p\mathrm{\ prime}} (1-p^{-s}) \cdot \sum_{n \geq 1} f_0(p, n) p^{-ns},

\Re(s) > \min(1, \sigma_f). $$ Proof. We can see that
 * $$ \sum_{n \geq 1} \frac{u^{f(n)}}{n^s} = \prod_{p\mathrm{\ prime}} \left(1+\sum_{n \geq 1} u^{f_0(p, n)} p^{-ns}\right). $$

This implies that


 * $$ \begin{align}

\sum_{n \geq 1} \frac{f(n)}{n^s} & = \frac{d}{du}\left[\prod_{p\mathrm{\ prime}} \left(1+\sum_{n \geq 1} u^{f_0(p, n)} p^{-ns}\right)\right] \Biggr|_{u=1} =     \prod_{p} \left(1 + \sum_{n \geq 1} p^{-ns}\right) \times \sum_{p} \frac{\sum_{n \geq 1} f_0(p, n) p^{-ns}}{ 1 + \sum_{n \geq 1} p^{-ns}} \\ & = \zeta(s) \times \sum_{p\mathrm{\ prime}} (1-p^{-s}) \cdot \sum_{n \geq 1} f_0(p, n) p^{-ns}, \end{align} $$

wherever the corresponding series and products are convergent. In the last equation, we have used the Euler product representation of the Riemann zeta function. $$\boxdot$$

The lemma implies that for $$\Re(s) > 1$$,


 * $$ \begin{align}

D_{\omega}(s) & := \sum_{n \geq 1} \frac{\omega(n)}{n^s} = \zeta(s) P(s) \\ & \ = \zeta(s) \times \sum_{n \geq 1} \frac{\mu(n)}{n} \log \zeta(ns) \\ D_{\Omega}(s) & := \sum_{n \geq 1} \frac{\Omega(n)}{n^s} = \zeta(s) \times \sum_{n \geq 1} P(ns) \\ & \ = \zeta(s) \times \sum_{n \geq 1} \frac{\phi(n)}{n} \log\zeta(ns) \\ D_{\Omega\lambda}(s) & := \sum_{n \geq 1} \frac{\lambda(n) \Omega(n)}{n^s} = \zeta(s) \log \zeta(s), \end{align} $$ where $$P(s)$$ is the prime zeta function and $$\lambda(n) = (-1)^{\Omega(n)}$$ is the Liouville lambda function.

The distribution of the difference of prime omega functions
The distribution of the distinct integer values of the differences $$\Omega(n) - \omega(n)$$ is regular in comparison with the semi-random properties of the component functions. For $$k \geq 0$$, define


 * $$N_k(x) := \#(\{n \in \mathbb{Z}^{+}: \Omega(n) - \omega(n) = k\} \cap [1, x]).$$

These cardinalities have a corresponding sequence of limiting densities $$d_k$$ such that for $$x \geq 2$$


 * $$N_k(x) = d_k \cdot x + O\left(\left(\frac{3}{4}\right)^k \sqrt{x} (\log x)^{\frac{4}{3}}\right).$$

These densities are generated by the prime products


 * $$\sum_{k \geq 0} d_k \cdot z^k = \prod_p \left(1 - \frac{1}{p}\right) \left(1 + \frac{1}{p-z}\right).$$

With the absolute constant $$\hat{c} := \frac{1}{4} \times \prod_{p > 2} \left(1 - \frac{1}{(p-1)^2}\right)^{-1}$$, the densities $$d_k$$ satisfy


 * $$d_k = \hat{c} \cdot 2^{-k} + O(5^{-k}).$$

Compare to the definition of the prime products defined in the last section of in relation to the Erdős–Kac theorem.