How prove this such $\frac{w(n+k)}{w(n)}>\alpha,\frac{\Omega{(n+k)}}{\Omega{(n)}}<\beta$ For any positive integer $n$, write $$n=p^{a_{1}}_{1}\cdots p^{a_{l}}_{l},$$ where the $p_{i}$ are prime numbers. Define
$$w(n)=l,\quad\Omega{(n)}=a_{1}+a_{2}+\cdots+a_{l} \, .$$
For any given positive integer $k$ and positive real numbers $\alpha,\beta$, show that there exists an integer $n>1$ such that
$$\dfrac{w(n+k)}{w(n)}>\alpha,\dfrac{\Omega{(n+k)}}{\Omega{(n)}}<\beta \, .$$
This problem is 2013-2014 China Mathematical Olympaid P4 problem.
 A: outline of the proof 
Let $N$ be the set of positive integers.
Lemma 1    $m,n\in\Bbb N$, then 
1) $\omega(m)\leqslant\omega(mn)\leqslant\omega(m)+\omega(n)$;
2) $\Omega(mn)=\Omega(m)+\Omega(n)$.
Lemma 2    $a,n\in\Bbb N, a\gt1$, $q\gt2$ is a prime factor of $a^{2^n}+1$, then $q\equiv1\pmod{2^{n+1}}$.
Lemma 3 $a,m,n\in\Bbb N$,  $m,n$ are odd numbers, then $(a^m+1, a^n+1)=a^{(m, n)}+1$
Proof of the Olympaid problem  By Lemma 1 and lemma 3, we get that
$$\frac{\omega(k2^{2^np_1p_2\dotsb p_m}+k)}{\omega(k2^{2^np_1p_2\dotsb p_m})}\ge\frac{\omega(2^{2^np_1p_2\dotsb p_m}+1)}{\omega(k)+1}\ge\frac{m}{\omega(k)+1}$$
By Lemma 1 , we have
$$\frac{\Omega(k2^{2^np_1p_2\dotsb p_m}+k)}{\Omega(k2^{2^np_1p_2\dotsb p_m})}\le \frac{\Omega(k)+\Omega(2^{2^np_1p_2\dotsb p_m}+1)}{2^np_1p_2\dotsb p_m}$$
By Lemma 2, 
$$\Omega(2^{2^np_1p_2\dotsb p_m}+1)\le \frac{2^np_1p_2\dotsb p_m}{n+1}$$
so
$$\frac{\Omega(k2^{2^np_1p_2\dotsb p_m}+k)}{\Omega(k2^{2^np_1p_2\dotsb p_m})}\le \frac1{n+1}+\frac{\Omega(k)}{2^n}$$
The complete answer, Please refer to Solutions to Chinese Mathematical Olympiad(CMO) 2014, in Chinese.
