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As there is much freedom in ways in which we can form number sequences and ask questions about them I defined the sequence of numbers associated to every natural number in this way:

Choose some natural number $n$. Add its digits in base $10$ and concatenate them with chosen number so to put them to the left to form a new number. Do the same with new number. And so on and so on...

To clarify let us add an example:

Choose, for example, $n=13$. Then $DS_{10}(13)=1+3=4$ and we get a new number $413$. Now we get $DS_{10}(413)=4+1+3=8$ and we obtain $8413$ and after that $168413$ and so on... It is obvious that we can do this with every natural number to obtain an infinite sequence $d(n)$ associated to every $n \in \mathbb N$.

The question is:

Is there at least one $n_0 \in \mathbb N$ such that $d(n_0)$ has an infinite number of prime numbers in itself?

Also, computational efforts are highly welcomed.

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    $\begingroup$ Some observation: as soon as a number in your sequence is divisible by $2$, $3$ or $5$, all following number will be too. So no more primes can occur. $\endgroup$
    – M. Winter
    Commented Sep 18, 2017 at 12:08
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    $\begingroup$ Regarding @M.Winter observation, if you start with a number which is not divisible by $2$ or $5$, then you will never end up in a number divisible by one of those. $\endgroup$ Commented Sep 18, 2017 at 13:55
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    $\begingroup$ @DarielRudt Same seems to holds for $3$. The digit sum of a number $n$ is $\equiv n$ modulo $3$. So the sequence $d(n)$ will look like $1,2,1,2,1,2,...$ modulo $3$. $\endgroup$
    – M. Winter
    Commented Sep 18, 2017 at 14:19

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I'm afraid that, as these numbers grow super fast, it's going to be difficult to simulate the process for large input. However, I did some computational effort, and this is the result for all the primes below $25000$ (it took 2.5 minutes to compute this). The longest chain was of six numbers:

We try with 4021
We try with 74021
We try with 1474021
We try with 191474021
We try with 29191474021
We try with 40029191474021
4021 lasted 6

Despite I didn't use the best (most efficient) algorithm to do the primality test, the difference would not be worth the effort.

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  • $\begingroup$ +1. Can you check, for example, every number below 25000 by calculating its first, say, 20 concatenations, and then see which one will hold the record for the number of primes in first 20 concatenations? See the comment of @M. Winter : "Some observation: as soon as a number in your sequence is divisible by 2, 3 or 5, all following number will be too. So no more primes can occur" so that you do not need to check all the numbers. $\endgroup$
    – user480281
    Commented Sep 18, 2017 at 13:26

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