Define sum of digits (in base $10$) function as $sd_{10}(n)=\sum_{i=0}^ma_i$ where $n=\sum_{i=0}^ma_i \cdot 10^i$ and $0\le a_i\le 9$.

If we choose prime number $86423$ and sum its digits we obtain a number $8+6+4+2+3=23$ which is itself a prime number. Now if we repeat the same procedure on the number $23$ we obtain a number $2+3=5$ which is also a prime number.

So we have that $86423$ and $sd_{10}(86423)$ and $sd_{10}^2(86423)=sd_{10}(sd_{10}(86423))$ are all prime.

Let us call a number $p$ extremely prime if $p$ is prime and all the numbers in the set $\{sd_{10}(p),sd_{10}^2(p),...,sd_{10}^{n(p)}(p)\}$ are also prime, where $n(p)$ is the first number for which we have that $sd_{10}^{n(p)}(p)$ is a one digit number.

Is there an infinite number of extremely prime numbers?

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    $\begingroup$ If you only required the first digit sum to be prime, the answer is apparently yes, per oeis.org/A046704 Your requirement excludes numbers like 67 and 89, which gives us oeis.org/A070027 but no clue as to whether the sequence is infinite. $\endgroup$ – Robert Soupe May 2 '16 at 5:12

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