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For a definition, see the wikipedia page: http://en.wikipedia.org/wiki/Strictly_non-palindromic_number

So according to the wikipedia page, under properties, all strictly non-palindromic numbers with three exceptions are primes.

So I have some questions about these numbers. First, what is their relative density in the primes? Does the sum of their reciprocals converge? If not, then do they contain arbitrarily long arithmetic progressions? Are they known to be an additive basis of any order?

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    $\begingroup$ Based on divisibility considerations, one only needs to consider representing a prime in a base that yields an odd number of digits. Since for each base a, roughly a^(d/2) out of a^d numbers are palindromic, I would think that a positive proportion of primes survive the sifting process. Gerhard "Ask Me About System Design" Paseman, 2011.05.25 $\endgroup$
    – Gerhard Paseman
    Commented May 26, 2011 at 5:41
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    $\begingroup$ Not really on topic, but why should we care about these numbers? (I'm genuinely curious) $\endgroup$
    – Woett
    Commented May 26, 2011 at 10:39
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    $\begingroup$ @Stanley Yao Xiao: do you know this reference : Banks W., Hart D., Sakata M., Almost all palindromes are composite. Math. Res. Lett. 11 (2004), no. 5-6, 853–868. Maybe this could be used to prove that almost all prime numbers are non-palindromic (at least in a given base). On the other hand I'm not sure it has been proved (or even is easy to show) that there are infinitely many strictly non-palindromic numbers... $\endgroup$
    – François Brunault
    Commented May 31, 2011 at 8:54
  • $\begingroup$ @Woett. Possible answer : Proving that there are infinitely many palindromic primes would be some kind of weak version of the difficult conjecture that there are infinitely many primes of the form $n^2+1$ (not sure however this is any easier). $\endgroup$
    – François Brunault
    Commented May 31, 2011 at 8:57
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    $\begingroup$ @Stanley Yao Xiao : There are some heuristic arguments predicting the asymptotic number of primes of the form $n^2+1$. See for example the Bateman-Horn conjecture en.wikipedia.org/wiki/Bateman%E2%80%93Horn_conjecture Maybe the same kind of ideas could be used to at least guess the asymptotic number of strictly non-palindromic numbers. $\endgroup$
    – François Brunault
    Commented May 31, 2011 at 9:08

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Some numbers and graphs about your question. https://experimentalmath.wordpress.com/2015/07/20/expmath2-strictly-non-palindromic-numbers/

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    $\begingroup$ You should make this answer more self-contained. What is behind the link? $\endgroup$
    – András Bátkai
    Commented Jul 24, 2015 at 19:22
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    $\begingroup$ The following heuristic gives a different prediction, and I'd be interested to understand why: the "probability" that a number $n$ is not a three-digit palindrome in any base is about $\prod_{\sqrt[3]n<b<\sqrt n} (1-\frac1b)$, since $1$ out of every $b$ three-digit numbers in base $b$ are palindromes. That product works out to $n^{-1/6}$. So maybe at most $x^{5/6}$ integers up to $x$, including primes, are not three-digit palindromes. $\endgroup$
    – Greg Martin
    Commented Aug 23, 2015 at 19:09
  • $\begingroup$ @GregMartin how did you deduce product is $n^{-1/6}$? $\endgroup$
    – user94040
    Commented Dec 16, 2016 at 4:36
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    $\begingroup$ It's a telescoping product: for integers $A<B$, we have $\prod_{A<b\le B} (1-\frac1b) = \frac A{A+1} \frac{A+1}{A+2} \cdots \frac{B-2}{B-1} \frac{B-1}B = \frac AB$. $\endgroup$
    – Greg Martin
    Commented Dec 16, 2016 at 8:21

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