Prove this conjecture or disprove it if couldn't then determ...

Prompt

Prove this conjecture or disprove it if couldn't then determine the limitations that why it cannot be proved or Disproved answer in detail with every information: Conjecture: No Consecutive Members of \(S\) For every natural number \(n\ge2\), let \[ f(n)=\text{the largest prime factor of }n \] and let \[ \Omega(n)=\text{the total number of prime factors of }n, \] counted with multiplicity. Define \[ S=\{n\ge2:f(n)=\Omega(n)\}. \] Conjecture. There do not exist two consecutive natural numbers \(n\) and \(n+1\) such that both belong to \(S\). Equivalently, \[ \boxed{ \forall n\ge2,\qquad f(n)=\Omega(n)\ \Longrightarrow\ f(n+1)\ne\Omega(n+1). } \] In other words: > No two consecutive integers can simultaneously have their largest prime factor equal to their total number of prime factors (counted with multiplicity). For example, \[ 4,12,18,27,80,120,180,200,270,300,\ldots \] This conjecture is distinct from the earlier conjecture concerning a prime between consecutive members of \(S\).