34

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.

tong_nor
  • 4,024

2 Answers2

109

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.

Nate
  • 11,785
40

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.

Parcly Taxel
  • 105,904
  • 3
    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? – tong_nor Sep 22 '17 at 15:15
  • 4
    @tong_nor Yes. For any $\epsilon>0$ and sufficiently large $x$ the prime number theorem guarantees at least one prime in $(x,(1+\epsilon)x)$. – Parcly Taxel Sep 22 '17 at 15:17
  • 7
    I suspect Schoenfeld's result is a bigger hammer than Dirichlet, in a way. – wythagoras Sep 23 '17 at 11:06
  • 2
    @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... – Parcly Taxel Sep 23 '17 at 11:16