User:CXTests/T245827

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). For example, 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.

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

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

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


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


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


 * $$\sum_{d\mid n} |\mu(d)| = 2^{\omega(n)} $$
 * $$\sum_{r\mid n} 2^{\omega(r)} = d(n^2) $$
 * $$\sum_{r\mid n} 2^{\omega(r)} d\left(\frac{n}{r}\right) = d^2(n) $$

\sum_{\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 )$$


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

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


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


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


 * $$\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} $$


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


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


 * $$\#\left\{n \leq x : \Omega(n) - \omega(n) > \sqrt{\log\log x}\right\} = O\left(\frac{x}{(\log\log x)^{1/2}}\right). $$
 * $$S_{\operatorname{odd}}(x) := \sum_{n \leq x} \omega(n) [n\text{ odd}]_{\delta}, $$


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



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


 * $$ \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} $$


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


 * $$\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.$$


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


 * $$ \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} $$


 * $$ \begin{align}

D_{\omega}(s) & := \sum_{n \geq 1} \frac{\omega(n)}{n^s} = \zeta(s) P(s) \\ D_{\Omega}(s) & := \sum_{n \geq 1} \frac{\Omega(n)}{n^s} = \zeta(s) \times \sum_{n \geq 1} P(ns), \end{align} $$

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

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

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


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


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


 * $$\sum_{d\mid n} |\mu(d)| = 2^{\omega(n)} $$
 * $$\sum_{r\mid n} 2^{\omega(r)} = d(n^2) $$
 * $$\sum_{r\mid n} 2^{\omega(r)} d\left(\frac{n}{r}\right) = d^2(n) $$

\sum_{\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 )$$


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

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


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


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


 * $$\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} $$


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


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


 * $$\#\left\{n \leq x : \Omega(n) - \omega(n) > \sqrt{\log\log x}\right\} = O\left(\frac{x}{(\log\log x)^{1/2}}\right). $$
 * $$S_{\operatorname{odd}}(x) := \sum_{n \leq x} \omega(n) [n\text{ odd}]_{\delta}, $$


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



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


 * $$ \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} $$


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


 * $$\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.$$


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


 * $$ \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} $$


 * $$ \begin{align}

D_{\omega}(s) & := \sum_{n \geq 1} \frac{\omega(n)}{n^s} = \zeta(s) P(s) \\ D_{\Omega}(s) & := \sum_{n \geq 1} \frac{\Omega(n)}{n^s} = \zeta(s) \times \sum_{n \geq 1} P(ns), \end{align} $$

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

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

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


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


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


 * $$\sum_{d\mid n} |\mu(d)| = 2^{\omega(n)} $$
 * $$\sum_{r\mid n} 2^{\omega(r)} = d(n^2) $$
 * $$\sum_{r\mid n} 2^{\omega(r)} d\left(\frac{n}{r}\right) = d^2(n) $$

\sum_{\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 )$$


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

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


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


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


 * $$\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} $$


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


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


 * $$\#\left\{n \leq x : \Omega(n) - \omega(n) > \sqrt{\log\log x}\right\} = O\left(\frac{x}{(\log\log x)^{1/2}}\right). $$
 * $$S_{\operatorname{odd}}(x) := \sum_{n \leq x} \omega(n) [n\text{ odd}]_{\delta}, $$


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



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


 * $$ \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} $$


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


 * $$\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.$$


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


 * $$ \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} $$


 * $$ \begin{align}

D_{\omega}(s) & := \sum_{n \geq 1} \frac{\omega(n)}{n^s} = \zeta(s) P(s) \\ D_{\Omega}(s) & := \sum_{n \geq 1} \frac{\Omega(n)}{n^s} = \zeta(s) \times \sum_{n \geq 1} P(ns), \end{align} $$