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$131$ is a palindrome prime as well as $787$ , moreover $6\cdot 131+1=787$.

Are there further examples for a palindrome-prime $p$, such that $6p+1$ is a palindrome-prime as well ?

It is clear that $p$ must have an odd number od digits (since $11$ is the only palindrom-prime with an even number of digits) and the leading digit of $p$ must be $1$ , otherwise $6p+1$ has an even number of digits.

So, $p$ must have the form $1\cdots 1$ and $6p+1$ must have the form $7\cdots 7$. Upto $10^{10}$ , no further example exits.

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    $\begingroup$ Setting the issue of primality aside, how often is $6p+1$ a palindrome when $p$ is a palindrome? $\endgroup$ Commented Oct 30, 2018 at 13:43
  • $\begingroup$ @BarryCipra A good question, will already not appear often. $\endgroup$
    – Peter
    Commented Oct 30, 2018 at 13:44
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    $\begingroup$ For $6p+1$ to have the form $7\cdots7$, the palindrome $p$ must either have the form $12\cdots21$ or the form $13\cdots31$. $\endgroup$ Commented Oct 30, 2018 at 13:53
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    $\begingroup$ @Barry Cipra with Pari Gp we found p=131 and p=12130303121. First of all they both are congruent to -1 (mod 6) and $p+1$ in both cases is divisible by a prime ending with digits 11. Infact 132=2^2*3*11 and $12130303122=2*3^3*7*59*543911$. 543911=70^2*111+11 $\endgroup$
    – Enzo Creti
    Commented Oct 30, 2018 at 17:20
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    $\begingroup$ @EnzoCreti, if $s=6p+1$, then $(s+2)/3=2p+1$. I agree, it is interesting that this $p$ is a Sophie Germain prime (as is $131$). Have you checked any of the other $p$'s that you've found? $\endgroup$ Commented Oct 31, 2018 at 12:41

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This is just a long comment, but it might be of help for anyone wanting to go further.

It makes sense to begin by looking for palindromes $p$, prime or not, but with an odd number of digits (so that it might be prime), such that $6p+1$ is also a palindrome with an odd number of digits (so that it also might be prime). Let's let $p=a_0a_1\ldots a_{n-1}a_na_{n-1}\ldots a_1a_0$ with digits $a_0,a_1,\ldots,a_n\in\{0,1,2,\ldots,9\}$ and $6p+1=b_0b_1\ldots b_{n-1}b_nb_{n-1}\ldots b_1b_0$ with digits $b_0,b_1,\ldots,b_n\in\{0,1,2,\ldots,9\}$.

As the OP notes, for $6p+1$ to have an odd number of digits, we must have $a_0=1$, in which case $b_0=7$. But that requires a carry of exactly $1$ from the multiplication $6a_1$, which means $a_1\in\{2,3\}$. In particular we get the three-digit pairs

$$(121,727)\quad\text{and}\quad(131,787)$$

In general, in order for the carries to work to produce a palindrome, we must have $a_i\in\{0,1\}$ if $i$ is even and $a_i\in\{2,3\}$ if $i$ is odd; moreover, every such choice produces a palindromic pair $(p,6p+1)$. Thus, for example, the five-digit pairs are

$$(12021,72127),\quad(12121,72727),\quad(13031,78187),\quad(13131,78787)$$

Among these numbers, only $72727$ and $78787$ are prime; the $p$ values are all composite. (In particular, $12021$ and $13131$ are easily identified as multiples of $3$.)

The fact that there are only $2^n$ candidates for a palindromic pair of $2n+1$-digit primes should streamline the search for larger examples considerably. The $11$-digit pair $(12130303121, 72781818727)$ reported by Enzo Creti, for example, is one of only $32$ possible $11$-digit pairs, many of which can be ignored as obvious multiples of $3$. My guess would be that prime pairs will pop up periodically; perhaps someone can provide a heuristic estimate for their frequency (or an argument to the contrary, that they should eventually cease).

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