# Number of distinct prime divisors of three consecutive numbers

Let $\omega(n)$ be the number of distinct prime divisors of $n>1$. I could prove that: there are infinitely many positive integers $n$ such that $\omega(n+1)+\omega(n-1)>2\omega(n)$. Similarly, the reverse inequalities also hold for infinitely many positive integers.

I wish to study the case $\omega(n+1)+\omega(n-1)=2\omega(n)$. Is this equation has infinitely many solution in $\mathbb Z^+$? For example n=3.

• Can you show your claimed proof that there are infinitely many positive integers $n$ such that $\omega(n+1)+\omega(n-1)<2\omega(n)$? – quasi Nov 3 '17 at 5:01
• @quasi I think one way is via Chen primes: there are infinitely many primes $p$ such that $p+2$ is a product of $\leq 2$ primes. Then for $p>4$ $$\omega(p)+\omega(p+2)\leq 3 < 4 \leq 2\omega(p+1)$$ – Yong Hao Ng Nov 3 '17 at 6:36
• Ok, but that's not an elementary result. Not that it has to be, but the fact that the only proof you alluded to uses such a result suggests that your question is, most likely, a research-level question. – quasi Nov 3 '17 at 7:08
• There are large examples : $10^{50}+34,35,36$ all have $5$ distinct prime factors, so $\color\red {10^{50}+35}$ is an example. – Peter Nov 3 '17 at 23:48
• How to see that $10^{50}+34$ has 5 distinct prime factors, Peter? – Jie Fan Nov 4 '17 at 1:10