The fact that each prime number (greater than $9$) ends with one of the four digits $1,3,7,9$, allows us to classify the tens in which the primes are found according to which of these four digits, added to the tens, yields to a prime number.

For example, for the first ten we have $1 \rightarrow \{1,3,7,9\}$. In fact, $10+1$, $10+3$, $10+7$ and $10+9$ are all primes. Conversely, for the twentieth ten the association reads $20 \rightarrow \{\}$, since there are no primes between $200$ and $209$.

It is easy to see that each ten is associated to one (and only one) group of symbols, chosen among the following $16$ distinct alternatives: $\{\}$, $\{1\}$, $\{3\}$, $\{7\}$, $\{9\}$, $\{1,3\}$, $\{1,7\}$, $\{1,9\}$, $\{3,7\}$, $\{3,9\}$, $\{7,9\}$, $\{1,3,7\}$, $\{1,3,9\}$, $\{1,7,9\}$, $\{3,7,9\}$, $\{1,3,7,9\}$.

For the sake of simplicity, we can identify each of these $16$ distinct groups of symbols with a single symbol, or with a single color, as illustrated below:

enter image description here

Arranging the tens in a Pascal's triangle, we find (*) the following structure (omitting the first ten on the edges of the triangle):

enter image description here

(*) I hope that my code is correct! It would be great if someone, more skilled than me, could confirm the emergence of such structure. In case you are interested in double-checking, please have a look to this post for details.

However, assuming that I did not mess up too much with the code, my conjecture is that

For very big tens, there cannot be colored squares other than on the outer diagonal of the triangle.

In other words, big primes $p$ must be in the form $p=10^{\binom{n}{k}}+1$, or $p=10^{\binom{n}{k}}+3$, or $p=10^{\binom{n}{k}}+7$, or $p=10^{\binom{n}{k}}+9$, and $k=1$. Clearly, a weaker version of such conjecture is that $k$ can oscillate among some little integer $2,3,4,5\ldots$ (which ones?).

This is probably an obvious result for the experts (I apologize, in case), nevertheless I would be glad to understand the connections between this approach and others, and also to know if there is some technique to attack such problem.

Sorry for possible naivety, and thank you very much for your comments and suggestions!

EDIT: The conjecture was based on the assumption that the plot of the triangle was correct. But, as Ross has shown, that was not the case. An improved version of the code producing that plot, in fact, results in this picture:

enter image description here

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    $\begingroup$ Sorry...you think that sufficiently large primes must be all be of the form $10^N+d$ where $d\in \{1,3,7,9\}$? But this is clearly false. I expect I have misunderstood what you have written...can you clarify? $\endgroup$ – lulu Aug 28 '18 at 19:04
  • $\begingroup$ +1 Nice question. $\endgroup$ – Karl Aug 28 '18 at 19:07
  • $\begingroup$ The top square looks to be one of the blues. I don't understand what you mean be omitting the first ten on the edges. Should the top square be ${5 \choose 2}={5\choose 3}=10?$ Why are the next two purple? I don't understand how you came up with your picture. Does the big white area in the middle claim that if I take a binomial coefficient in that region and replace the ones digit with $1,3,7,9$ I will not get a prime? It would be helpful to have only the top ten rows in a larger picture with the coefficients shown in colored squares. $\endgroup$ – Ross Millikan Aug 28 '18 at 19:08
  • $\begingroup$ @lulu That's definitely possible. There can be some mistake in the code, and the triangle I obtained is an artifact. However, there is still the weaker version of the conjecture. Your observation is anyway very interesting. Why this is false? Thanks! $\endgroup$ – user559615 Aug 28 '18 at 19:08
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    $\begingroup$ Another problem with your claim (and, I think, with any natural weakening of it) is that $\sum \frac 1p$ diverges...where the sum is taken over the primes...but $\sum \frac 1{10^n}$ converges as does any fixed multiple of it. $\endgroup$ – lulu Aug 28 '18 at 19:19

The large white area in the middle claims that if $N$ is any number in that area of the triangle then none of $10N+1, 10N+3, 10N+7, 10N+9$ are prime. The claim is false as ${37 \choose 13}=3562467300$ and $35624673001$ is prime.

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  • $\begingroup$ Yes. I understood the problem. Please, can you suggest me a way to efficiently code this structure? $\endgroup$ – user559615 Aug 28 '18 at 20:08
  • $\begingroup$ You just need to compute $n \choose k$, which you can just do by summing the two entries above, then check the four numbers for being prime. Clearly one of these steps is not working properly. I suspect it is your primality test. Maybe you are getting overflow of the integers you are using, because that region of the triangle has reasonably large numbers in it. $\endgroup$ – Ross Millikan Aug 28 '18 at 20:12
  • $\begingroup$ Yes, This is the problem. Also, do you know a good repository for primes? Thanks, Ross! I should delete this post, but I can't. $\endgroup$ – user559615 Aug 28 '18 at 20:15
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    $\begingroup$ There are a few prime repositories mentioned in this question. $\endgroup$ – Peter Phipps Aug 28 '18 at 20:27
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    $\begingroup$ You can't hope to look up numbers in a list to see if they are prime. There are just too many primes, even small ones like only $15$ digits as mentioned in the linked question, for anybody to have bothered listing them. For your purposes you only need to check some thousands or maybe millions of numbers, That can be quite quick. You need to use a package that works with arbitrarily large integers, but there are many of those. The Fermat pseudoprime test will be good enough for what you need. It is quick to code. $\endgroup$ – Ross Millikan Aug 28 '18 at 20:35

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