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Let $p_1,p_2,\ldots,p_n$ be distinct primes greater than $3$ and $k=p_1p_2 \ldots p_n$. It is a known result that $2^{k}+1$ has at least $4^n$ divisors (this was a shortlisted problem from IMO 2002 - N3).

Is it true that $2^{k}+1$ has at least $(n+1) \cdot 2^{2^n-n}$ divisors? Can this bound be improved?

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    $\begingroup$ Yes, it actually be improved to $2^{2^n}$ using Zsigmondy's theorm $\endgroup$
    – Antoine Labelle
    Commented Feb 13, 2021 at 14:14

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