There’s a nifty number $\theta$ in the following equation $$p_n=\lfloor\theta^{3^n}\rfloor$$ where $p_n$ is a prime number and $n$, as usual, is the sequence of natural numbers (0 doesn’t work—the result is 1, which is not prime). $\theta$ in fact represents the smallest such number, and is approximately equal to 1.3063778838630806904686. This number is known as Mills’ constant, and the prime numbers that it generates are Mills’ primes. The difficulty is, though, that these numbers grow very, *very* rapidly. The first is $2$, the second $11$, then $1361$, $2521008887$, and each number after that roughly triples in the number of digits it contains.

So how would one go about finding the value of $\theta$? Through some simple rearranging, $$\theta=exp(\frac{\log{p_n}}{3^n})$$, but you have to know the “largest” prime number in this (by definition infinite) sequence to get more decimals of theta.

Okay, so given the sequence, how many digits of $\theta$ are known from each Mills’ prime?

Mills' Prime | Prime Digits | Mills' Constant | Constant Digits (correct after decimal point) |
---|---|---|---|

2 | 1 | 1.259... | 0 |

11 | 2 | 1.30529... | 2 |

1361 | 4 | 1.3063778838246... | 10 |

2521008887 | 10 | 1.306377883863080690468614492575... | 27 |

p_{5} | 28 | ~87 | |

p_{6} | 84 | ~255 | |

p_{7} | 253 | ~763 | |

skip a few... | |||

p_{12} | 61683 | ~185051 | |

p_{13} | 185051 | ~555153 | |

p_{14} | 555153 | ~1665460 |

So since the 14th Mills’ prime is known with a fairly high degree of confidence, nearly 1665460 digits of Mills’ constant are known. Conservatively, one would chop off a few digits to be sure of the numbers’ accuracy.

Oddly, however, this detail is not included in the OEIS (related sequences A051254, A051021, A224845, A108739).

The last of the related sequences listed above is a useful one for Mills’ primes, because the sequence of [Mills’] primes can be written far more succinctly with these numbers:

$$p_n=p_{n-1}^3+b_{n-1}$$, where $b_n$ is the nth number in A108739.

In other words

$$p_2=p_1^3+b_1$$

$$11=2^3+3$$

The sequence $b_n$ I’ll reproduce here:

$$[3, 30, 6, 80, 12, 450, 894, 3636, 70756, 97220, 66768, 300840, 1623568]$$

I’d like to find $p_{15}$, but since it’s likely 1665460 digits in length, it takes a long time to test compositeness using even one witness with Miller-Rabin, much less the several it would probably take. And forget factoring it! The biggest number I’ve managed to factor on my home computer is a mere 133 digits long, and that took 4 days (it’s now crunching on a 126 digit number, and may have that cracked by the time I get home).

Speaking of which, the sequence I’m working on is the aliquot sequence starting with 499080. My contributions began with entry 2040.

Now I need to figure out why $$\sum_{n=1}^{\infty} n=-\frac{1}{12}$$. I’m still not convinced…

Finally, a Numberphile video on Mills’ constant: