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A satanic prime is a prime number with $666$ in the decimal representation.

The smallest satanic prime is $6661$.

Prove that there are infinitely many satanic primes.


I used Dirichlet's theorem for the progression $10000n+6661$ and it is done.

I'm interested in solutions without Dirichlet's theorem.

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    $\begingroup$ Dirichlet is a big hammer, but sometimes you need a big hammer :) $\endgroup$ Sep 22, 2017 at 14:28
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    $\begingroup$ I suspect that, in a reasonable sense, almost all primes are satanic $\endgroup$
    – Henry
    Sep 22, 2017 at 14:43
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    $\begingroup$ I'd like to throw out here a (I think much harder) question: Are there infinitely many non-satanic primes? $\endgroup$
    – Nate
    Sep 22, 2017 at 15:33
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    $\begingroup$ @Nate Yes. Any prime with no 6 cannot be satanic. Maynard proved in 2016 that there are infinitely many primes with no 6 in their decimal expansions. $\endgroup$ Sep 22, 2017 at 17:16
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    $\begingroup$ Is this true in general for any natural number? $\endgroup$
    – jdods
    Sep 26, 2017 at 19:04

2 Answers 2

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Consider the set $S$ of all numbers without 666 in their base 10 expression. Here's a fun fact: the sum $\sum_{s\in S} \frac{1}{s}$ converges. It's actually pretty easy to prove, so I'll leave it as an exercise (or google "Kempner series").

On the other hand, a famous result of Euler says the sum of the reciprocals of the prime numbers diverges.

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    $\begingroup$ Most Elementary answer yet. Also consider numbers to base 1000 with the omitted digit being 666. $\endgroup$ Sep 22, 2017 at 15:29
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    $\begingroup$ "Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away." —Antoine de Saint-Exupéry $\endgroup$
    – Erick Wong
    Sep 22, 2017 at 15:54
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    $\begingroup$ Current Wikipedia articles with details: Kempner series and divergence of the sum of the reciprocals of the primes $\endgroup$ Sep 23, 2017 at 8:00
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Let $x=666\cdot10^n$; it has $n+3$ digits. Consider the interval $(x,(1+1/666)x)=(666\cdot10^n,667\cdot10^n)$. Then the prime number theorem says that there is at least one prime in this interval for sufficiently large $x$; such a prime must begin with 666 and is thus satanic.

Concretely, use Schoenfeld's 1976 result that says for every $x\ge2010760$ there is a prime in $(x,(1+1/16597)x)$; we extend this interval to the $1+1/666$ interval above. So there is at least one satanic prime with $n$ digits for $n\ge7$, and the result is proved.

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    $\begingroup$ Thanks! If I understand correctly, your argument can prove that for any fixed digits there is a prime with these digits at the beginning, am I right? $\endgroup$
    – tong_nor
    Sep 22, 2017 at 15:15
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    $\begingroup$ @tong_nor Yes. For any $\epsilon>0$ and sufficiently large $x$ the prime number theorem guarantees at least one prime in $(x,(1+\epsilon)x)$. $\endgroup$ Sep 22, 2017 at 15:17
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    $\begingroup$ I suspect Schoenfeld's result is a bigger hammer than Dirichlet, in a way. $\endgroup$
    – wythagoras
    Sep 23, 2017 at 11:06
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    $\begingroup$ @wythagoras These are Bertrand's postulate-type arguments. I happened to think of them first because I used them in my 14/21 interesting numbers answer... $\endgroup$ Sep 23, 2017 at 11:16

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