Proof of infinitude of primes whose reversal in base 10 is also prime - MathOverflow [closed]

Proof of infinitude of primes whose reversal in base 10 is also prime

2011-12-09T09:19:35Z

Is there any proof of infinitude of A007500 primes?

If you want to generate them here is trivial and naive python program.

def is_prime(n):
    i = 2
    while i*i <= n:
        if n%i == 0:
            return False
        i = i + 1
    else:
        return True

print [x for x in range(1,200) if is_prime(x) and is_prime(int(str(x)[::-1]))]

Answer by Igor Rivin for Proof of infinitude of primes whose reversal in base 10 is also prime

2011-12-09T10:54:26Z

The answer is: no proof is known at this time, for any base, but it is suspected that a proof exists (it should be reasonably easy to give a density under the "standard hypotheses").

Answer by Timothy Foo for Proof of infinitude of primes whose reversal in base 10 is also prime

2011-12-09T11:24:15Z

Hello all,

I must be overlooking something, but I wonder if the systems $\Psi:\mathbb{Z}^d\rightarrow \mathbb{Z}^t$ in the Green-Tao paper "Linear Equations in Primes" could apply to this question.

Answer by Timothy Foo for Proof of infinitude of primes whose reversal in base 10 is also prime

2011-12-09T14:00:40Z

Now, thinking about this a bit, let's say $f$ is the function that reverses the digits, so that $f(n)$ is the number that has the digits of $n$ in base 10 reversed. I think that when estimating $$|\{n \leq x: n,f(n) \mbox{ simultaneously prime}\}|,$$

then for each prime $p$, maybe you'll have to estimate the number of solutions to $$ nf(n) \equiv 0 \bmod p, $$ where $n \in \mathbb{Z}/p\mathbb{Z}$. This is similar to like when estimating the twin prime constant (but I'm not claiming that the whole thing goes through the same way). The problem is that $f(n)$ is not so straightforward like $n+2$ is. For $p=3$, at least $f(n) \equiv n \bmod 3$, so that is ok. $p=11$ is also not too bad. For other $p$, it doesn't seem so straightforward.

In fact, it's not even as straightforward as this, because for $n,f(n)\in \mathbb{Z}$, when one fixes $n \bmod p$, $f(n) \bmod p$ is not fixed in general. The point is just that I think one might have to estimate the probability that $p\nmid f(n)\in \mathbb{Z}$, given that $p\nmid n\in \mathbb{Z}$.