we write, $n!= 2^{s} \times x \cdots (1)$
Here, $2^{s+1} \nmid n!$. Clearly, $x$ is the number that has all odd prime factors of $n!$ . As a condition of the problem, it is given that, $x=2^{s-2}-1$. Then - $$n!= 2^{s} \times (2^{s-2}-1) \cdots (2)$$
Definition:
We define, $s= (n-s_2(n))$ where $s_{2}(n) $ is the number of $1$'s in the binary representation of $n$.
For any integer $m$ let $P(m)$ denotes the greatest prime factor of $m$ .
Problem: Prove that there are only finitely many positive integers $n, s $ such that $$n!=2^s(2^{s−2}−1)$$
Reference : A work of Cam Stewart (the paper has appeared in Acta Mathematica), proving a conjecture of Erdos, Stewart shows that the largest prime factor of $2^n-1$ is at least $n \times \exp\Big( \frac{\log n}{104 \log \log n}\Big)$ , if $n$ is large enough.
Using this result we can say that, for large enough $n, P(x)>n$, but by definition, all prime factors of $x$ are less than $n$ , so the number of positive integers $n, s $ should be finite.
Question: Is this argument correct? Is there a simple derivation?
Related Post:$a$ has $90 \% $ primes less than $n$ If $n!= 2^s \times a \times b $ and $\lfloor{\frac{a}{b}}\rfloor = 2^{s-2}$
Thanks.
There are no such integers $n, s$. The argument is simple:
a) $n!=2^s(2^{s-2}-1) < 2^{2s-2}$
b) $s=\nu_2(n!) = n-s_2(n) < n\,\,$ (since $n>0$),$\,\,$ i.e. $\,\,s \le n-1$
Hence $n! < 2^{2n-4}$. But this isn't possible because $16\cdot n! \ge 4^n$ for all $n \ge 1$ (induction).