Questions about prime numbers are notoriously hard. Let me ask one which may be easier:
QUESTION: are there infinitely many square-free Mersenne numbers
$$ M(n)\ := 2^n-1 $$
where $\ n\in\mathbb N\ $ are arbitrary natural numbers (not necessarily prime)?
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I am interested in baroque numbers $\ n,\ $ where $\ n\ $ is an $b$-baroque ($b\in\mathbb N)\ <=:=>\ S(n)/n=b\ $ and $\ S(n)\ $ is the sum of the divisors of $\ n.$
Of course, 2-baroque numbers are the same as perfect numbers.
We see that $\ s(n):=S(n)/n\ $ is a product of fractions of the form $\ s(p^t)/p^t $ for certain prime numbers $\ p\ $ (since $\ s\ $ is a multiplicative function). These fractions have to cancel out to result in $\ b.\ $
Since
$$ S(p^t)\ =\ \sum_{k=0}^t p^k $$
I am interested in the prime decomposition of such geometric series. The special case of $\ p=2\ $ is related to the above QUESTION:
$$ M(n)\ =\ S(2^{n-1}) $$
(All this, so far, is very elementary but... messy).
PS. I could ask about all $\ S(p^{n-1})\ $ but it goes without saying -- it's simpler to ask a simpler question.
