We has a result: If $ \tau(k)\geq n + 1$ then k is product of $ n$ positive integer great than 1. Now we prove that for each n there exist an $ k$ such that $ \tau(k + 1),\tau(k)\geq {n + 1}$ Consider the reside $ x\equiv 0(\mod p_i),\forall i = 1,..,n + 1$ $ x\equiv - 1(\mod p_i),\forall i = n + 2,..,2n + 2$ Where $ p_i$ is distint prime . More we can prove for a sequence contain k term with the same method.

What is $ \tau$ in this post?

$ \tau(k)$ is the number of different prime divisors of k.