Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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]))] 
share|improve this question

closed as too localized by Mark Sapir, Felipe Voloch, Dmitri Pavlov, Benjamin Steinberg, quid Dec 9 '11 at 14:59

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

11  
Mark Sapir: that page is about a subset of the set OP asks about, and does not say whether that subset is infinite. Thus, I don't see how this answers the question. –  Zsbán Ambrus Dec 9 '11 at 9:49
1  
It says that "the largest known ... is ...". What else do you want? All such problems are open. The maximal result about a natural set of numbers with infinite subset of primes is the Dirichlet theorem. Everybody with access to Google can learn it in a few minutes. –  Mark Sapir Dec 9 '11 at 10:01
10  
The OP is not asking about palindromes which are primes, and that is why the linked mathworld page is irrelevant. The OP is asking about primes whose reverses are also primes, regardless of whether the number and its reverse are the same. –  Sridhar Ramesh Dec 9 '11 at 11:01
4  
@Sridhar Ramesh: These questions are the same: this does not follow from Dirichlet (or Chebotarev density), hence the answer is unknown. One can easily cook up infinitely many questions like that. For example: are there infinitely many primes of the form $x^2+1$ ($x^2+2$, $x^2+3, $2^n+1, 2^n+3$, etc.). –  Mark Sapir Dec 9 '11 at 13:04
4  
@Mark Sapir: while I agree with the general sentiment, that there are inifnitely many primes of the form x^2 + y^4 does not follow from Dirichlet but is still known; also there are some results on primes with given Hamming weight and some other things. –  quid Dec 9 '11 at 15:19
show 5 more comments

3 Answers 3

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").

share|improve this answer
add comment

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.

share|improve this answer
add comment

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 $p\nmid n\in \mathbb{Z}$.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.