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This question was asked at MSE but recieved no attention at all. Here it is:

Are there finitely many $(n,k) \in \mathbb{N}^2$ with $2^n-1=p_1p_2\cdots p_k$ ?

$p_1=3,p_2=5 , ...,p_k$ are consecutive odd primes in ascending order.

An example is when $n=4, k=2$:

$2^4-1=3\cdot 5=p_1p_2$

Are there finitely many $n$?

I tried to use Zsigmondy's theorem
without success.

Thanks in advance!

anyconsecutive primes (unless it itself is a (Mersenne) prime) $\endgroup$ – მამუკა ჯიბლაძე Oct 10 '15 at 10:52