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If we position all the natural numbers into a 'periodic table' with the period equal to $32$, we get the following pattern for primes.

The primes are colored according to their last digit. I did not color the primes past the first 'full diagonal', but the pattern continues indefinitely. The numbers on the 'prime diagonals' that are composite are in bold.

How can this pattern be explained? I'm genuinely surprised by it.

Edit The so called 'pattern' was explained to me as trivial in the answers but what about the bold numbers $[49,77,91,119,121,133,143,...]$? Is there something special about this sequence? Some of them are squares of primes, some of them are the product of two primes, but maybe there is something else?

Prime numbers table

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    $\begingroup$ The diagonals represent congruence classes mod $30$. In order to contain primes, the smallest residue must be prime to $30$. Those resides are $\{7,11,13,17,19,\dots\}$. $\endgroup$
    – lulu
    Feb 19, 2016 at 22:56
  • $\begingroup$ If I have understood your question correctly, this has nothing to do with primes. If you colour all the integers according to their last digit, and then display them in columns of 32, you will get the same pattern, only more so. You might like to work out why. $\endgroup$
    – TonyK
    Feb 19, 2016 at 22:57
  • $\begingroup$ It's not about their last digits (that's just in base 10 representation anyway). That's about them being on separate diagonales. $\endgroup$
    – Yuriy S
    Feb 19, 2016 at 23:02
  • $\begingroup$ The are on seperate diagonals because the last digits are on separate diagonals. This pattern is ONLY about base 10 representation. You'd get a stronger pattern if you used 31 or 33. If you used base 8 and had 26 ($= 32_8$) in the column the result would be exactly the same. $\endgroup$
    – fleablood
    Feb 19, 2016 at 23:30
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    $\begingroup$ Explain exactly what it is you think you are seeing. If it's that primes with the same last digits appear in diagonals, that isn't surprising because all numbers with the same last digits appear in diagonals. If it's something to do with the frequency in which the appear.... well, what are you seeing? I'm not seeing anything. $\endgroup$
    – fleablood
    Feb 20, 2016 at 0:11

4 Answers 4

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Actually, this is a little misleading. If you ignore the primes and mark off every number by its last digit (but only consider those that end in 1, 3, 7, or 9) you will get the exact same pattern. Except now the lines will be solid without breaks in them.

This isn't a mystery. Each column we add 32 to each coresponding cell so by going over 1 column and up 2 rows you will always get a digit that has the same last digit and the lines with the same last digits will always form by going over 1 and up 2. (If we colored all digits we'd get a confusing checkerboard like pattern but by restricting to odd values we can see lines.)

Filtering by primes (or any other condition) will put blank spots in the lines but will not change the basic existence of the lines.

Basically the lines exist because $32 \equiv 2 \mod 10$. If we used 31 the lines would go over one up one. If we used 30, the lines would be horizontal. If we used 33 the lines would slant down. Etc.

NOTHING to do with primes.

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But what is surprising, is Ulam's spiral. Google it. (Which has NOTHING to do with what you have done--- which is utterly trivial.)

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  • $\begingroup$ Thank you - I used Mathematica code to highlight the primes and there is no mistery here. $\endgroup$
    – Yuriy S
    Feb 20, 2016 at 17:15
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Primes do follow a basic pattern, but it is still not completely fixed as subsequent Composite numbers increasingly interrupt the pattern P-P---P-P---P-P etc etc. If you use a grid based on the multiple of Prime numbers (such as $2\times3=6$; or $6\times5=30$; or $30\times7=210$; or $210\times11=2310$ etc etc.) you will see that there is a pattern of Composite numbers and to some extent Prime numbers. This is useful in determining what is NOT a Prime number, rather than finding what is a Prime.

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Concerning your edit, all these numbers are numbers such that all their prime divisors are greater or equal than $7$ (A038510). This is the only apparent property they share, and this property can be easily proven:

Obviously, since you're looking at things modulo 10, and that you want to locate prime numbers, you didn't look at the diagonals ending with $0, 2, 4, 5, 6, 8$ as they will not contain any prime. Hence, no number on the diagonals you consider will be a multiple of $2$ or $5$.

As for $3$, it's a bit more subtle. If $N$ is such that $N = 3k$, then the next element $N'$ on the diagonal is such that $N' = N + 30$, so $N'$ is also a multiple of $3$. Hence, all multiples of $3$ will be on the same diagonals. By consequent, on a diagonal with primes (integers not divisible by $3$), there will be no integer divisible by $3$.

Finally, all non-prime integers on the diagonals you considered cannot be divided by $2, 3, 5$, QED.

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There's what you talking about https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

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  • $\begingroup$ I don't think it's that complicated! $\endgroup$
    – TonyK
    Feb 19, 2016 at 22:57
  • $\begingroup$ Um, no, that's not it $\endgroup$
    – Yuriy S
    Feb 19, 2016 at 23:07

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