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  1. A prime number is a number larger than 1 which only positive divisors are itself and 1.

    Examples: 3,5,11.

  2. A number is palindromic in a base $b$ if when written with digits in that basis $d_1d_2\cdots d_n$, $$d_n=d_1\\ d_{n-1} = d_2\\ \cdots$$

For example : the number 191 (in base 10) is a palindromic prime, since it is a prime number and a palindromic number. Now, the sum of the digits, $1+9+1=11$ is also a palindromic prime and $1+1=2$ is also one. So 191 is example of a palindromic prime whose digit sum is a palindromic prime whose digit sum is a $\cdots$ ( and so on).

How many sequences of palindromic primes being preserved by digit sum (all the way down to a 1 digit prime) are there? Infinite or finite? If not infinite, can we calculate how many or give an upper bound?

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  • $\begingroup$ A prime number is a number divisible only by itself and 1- this allows 1 as a prime which goes against convention. Is this intentional or should we replace your definition with the more conventional notion of a prime? $\endgroup$ – Colm Bhandal Jun 25 '17 at 13:42
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    $\begingroup$ Wow, that seems like a really tough question. The one thing that bothers me is that it is dependent on our subjective base-10 number system, rather than something more "mathematical" so to speak. Since it's based on digits in base ten, a proof for this would probably heavily involve modulo 10 arithmetic... $\endgroup$ – Colm Bhandal Jun 25 '17 at 13:45
  • $\begingroup$ If $p$ is such a prime, we must have $p\mod 9\in\{2,3,5,7\}$, since $s(n)\equiv n\pmod 9$ for all $n\in\Bbb{N}$ and eventually when we reach a one-digit number it must be prime. $\endgroup$ – Mastrem Jun 25 '17 at 14:08
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    $\begingroup$ Proving that there's only finitely many would involve proving that there is only a finite amount of primes of the form $p=10^k+1$, which is very, very hard. $\endgroup$ – Mastrem Jun 25 '17 at 14:22
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    $\begingroup$ @mathreadler Whether there are infinitely many primes of the form $2^k+1$ is a long standing unsolved problem; I imagine it's the same for $10^k+1$; what we do know is that $k$ must be a power of $2$ $\endgroup$ – Mastrem Jun 25 '17 at 16:17
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There are the primes 2, 3, 5, 7, 11.

Then there are palindromic primes with a digit sum of 2, 5, 7, 11. The first ones are 101, 131, 151, 191, 313, 353, 10301, 10501, 11311, 13331, 30103.

Then there are palindromic primes with a digit sum of 101, 131, 151, 191, 313, 353, 10301, 10501, 11311, 13331, 30103. These would be rather large numbers.

Just an opinion: Heuristically, I expect that for any digit sum d, there is only a finite number of palindromic primes with a digit sum d (which is unproven even for the most trivial case d = 2). But also I expect that as d grows, the number of these primes will grow. So I'd expect there to be a rather large number of palindromic primes with a digit sum 30,103, just as an example. So in total, I'd expect the number to be infinite.

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  • $\begingroup$ Hmm... for the third sequence you list, the OEIS gives two results: A109830 and A222116. Well, they're slightly different. $\endgroup$ – Robert Soupe Apr 7 at 2:05

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