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

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 4 '18 at 3:13
• 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 4 '18 at 6:05
• @MattSamuel: I've corrected his title. – Tito Piezas III Jan 4 '18 at 6:21

(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.