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Can all three numbers $\ n\ \ n^2-1\ \ n^2+1\ $ be fine (as opposed to coarse)?

Let $$ m\ n\ \in\ \mathbb N_{_{>1}}\ :=\ \{x\in\mathbb Z: x>1\} $$ be arbitrary.   Let $\ P(n)\ $ be the largest prime divisor of $n$.

Definition:   Molecularity of $n$ is $$ M(n)\ :=\ \log_{P(n)}(n) $$

Instantly,

Theorem

• $ M(n) \ge 1;$
• $ M(n)=1\quad\Leftrightarrow\quad p\ $ is a prime;
• $ M(n^k)\ =\ k\cdot M(n)\qquad $ (for every $\ k=1\ 2\ \ldots);$
• $ M(m\cdot n)\ \le\ M(n)+M(n);$
• $ M(m\cdot n)=M(n)+M(n)\quad\Leftrightarrow\quad P(\gcd(m\ n))\ =\ P(m\cdot n). $

For instance: $$ n>3\quad\Rightarrow\quad M(n^2-1)\ <\ M(n-1)+M(n+1) $$

Question 1:   What is $$ \sup_{n>2}\ \min(M(n-1)\,\ M(n)\,\ M(n+1))\quad ?$$ Question 2:   What is $$ \inf_{n>2}\ \frac1{M(n-1)}+\frac1{M(n)}+\frac1{M(n+1)} \quad? $$

The ever-sharper bounds would be greatly appreciated.

I dare, this time with a greater probability, that the above sup is $\ \le 4,\ $ and that $4$ cannot be actually attained.