Mersenne primes and lowest divisor

Let $p_1,p_2,..p_k$ be distinct Mersenne primes let $N = 2^{p_1 \cdot p_2 \cdot..p_k}-1$. Let LD(n) be the lowest divisor of n greater than 1.

My question is: Is $LD(N)$ always a Mersenne prime?

• Unless I am misunderstanding something here, it seems that we always have $7|N$, since $(2^3-1)$ is always going to be a factor of $N$. (Are we assuming that $p_1$ is the first Mersenne prime, etc.?) – Old John Dec 9 '13 at 23:56
• @OldJohn I think it's for any collection of Mersenne primes. I'm not sure whether the $p_i$ are assumed distinct, however. – Daniel Fischer Dec 9 '13 at 23:59
• @DanielFischer: Yes, they should be distinct. – Adam Dec 10 '13 at 0:05
• @OldJohn: no $p_1$ does not have to be the first Mersenne prime. – Adam Dec 10 '13 at 0:16
• I doubt it. If you pick a Mersenne prime $M_p$ such that $2^{M_p}-1$ is not prime ($M_{13}$ for example), I would be surprised if its smallest prime factor would necessarily be a Mersenne prime. If you compose with large Mersenne primes, there's a chance that the smallest prime factor of the rogue $M_p$ is the smallest overall. – Daniel Fischer Dec 10 '13 at 0:20

The answer is that no, in general LD(N) does not need to be a Mersenne prime. $2^{8191}-1$ is not divisible by any Mersenne primes smaller than itself (I checked).
If 3 is among the $p_is$, then as OldJohn points out 7 is a factor.
• Yes, I had worked out about $2^{8191}-1$ not being divisible by any smaller Mersenne prime, but - I did not post it, because I could not prove that the same would be true if we have more than one of the $p_i$s. The numbers become far too big to try hunting for factors, and I have not got a good argument yet for $2^{p_1p_2}-1$ where $p_1$ is 8191 and $p_2$ is bigger. Excellent question, though! – Old John Dec 10 '13 at 18:17