Are there too many 8-digit primes $p$ for Mersenne primes $M_p$?

Question

So it was recently announced that a new Mersenne Prime has been discovered:

https://www.mersenne.org/primes/press/M77232917.html

I was reading up a bit about Mersenne primes, and came across a conjecture of Lenstra–Pomerance–Wagstaff (LPW) on Wikipedia. The article says that a consequence of the conjecture is that there should be about

$$e^\gamma \log(10)/\log(2) \sim 5.92$$

primes $p$ of a given number of decimal digits such that $2^p-1$ is prime. When I look at the $p$ of this table of Mersenne primes, I get the following counts:

$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline 1 & 2& 3& 4& 5& 6& 7& 8& \text{Total}\\ \hline 4 & 6 & 4 & 8 & 6 & 5 & 5 & \color{blue}{12}? & 50\\ \hline \end{array} $$

I put $12?$ as there could still be some $8$-digit $p$ for which $M_p$ is prime. So my question is, isn't that $12$ out of place? Or does the LPW conjecture expect large deviations from $5.92$?

The person who discovered it got a nice (belated) Christmas present! — Tob Ernack, Jan 04 '18 at 03:13
Thanks for this post. Didn't know a new Mersenne prime was recently discovered. (Only $3$ have been found in this decade so far.) — Tito Piezas III, Jan 04 '18 at 06:05

Tito Piezas III · Answer 1 · 2018-01-04T13:07:10.383

(Too long for a comment.)

I was wondering if the phenomenon was an artifact of base-$10$, so I checked other bases $b$. Define

$$L(b)= e^\gamma \log(b)/\log(2)$$

I. Base-$12$ and $L(12)\sim6.4$

$$4, 8, 3, 9, 7, 5, 8, \color{red}6$$

II. Base-$10$ and $L(10)\sim5.9$

$$4, 6, 4, 8, 6, 5, 5, \color{blue}{12}$$

III. Base-$8$ and $L(8)\sim5.3$

$$4, 5, 3, 6, 8, 5, 4, 4, \color{blue}{11}$$

IV. Base-$6$ and $L(6)\sim4.6$

$$3, 5, 4, 3, 5, 7, 4, 4, 3, \color{blue}{10}, 2$$

V. Base-$4$ and $L(4)\sim3.6$

$$2, 3, 4, 3, 2, 4, 5, 4, 4, 2, 4, 2, \color{blue}9, 2$$

It is not verified if there are more $p$ between $3.7\times10^7$ and $7.7\times10^7$, so some of these counts may change.

The red count for $8$ digits in base-$12$ is still incomplete. Since this roughly overlaps with $10^7-10^9$, then $L(10)$ suggests there will be a total of about $12$ $p$ in that range, more or less. If not, then this nice increasing pattern will be ruined.

score 0 · Answer 2 · answered Sep 26 '19 at 13:10

First, an update. Since your question was asked another Mersenne prime has been discovered, with $p=82,589,933$ which is also 8 digits. So the last count should now be 13. We don't have a good expected error on the deviation on the LPW conjecture, but 13 in that range is now over twice the expected in that range. This does look weird and probably not what one would naively expect from the conjecture. Two notes: first, this is only a small number of data points, so reading too much into it may not be great. Second, strictly speaking the LPW conjecture implies a result that roughly says that the average number of such $p$ is around 5.92 in any such range. It turns out it is consistent with LPW that one will have some ranges of this sort which have much higher or much lower totals, and my guess would be that in fact the number of such primes in a given interval of $10^{m+1} - 10^m$ can get arbitrarily large if one goes out far enough, and that's still consistent with LPW. But, seeing such a large jump so fast does have to make one's credence in LPW go down somewhat.

Are there too many 8-digit primes $p$ for Mersenne primes $M_p$?

2 Answers2