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(Read the whole thing) So, I was playing around with prime numbers and found a pattern whith them. It goes like this: First, you write down all numbers ending with 1, 3, 7, and 9, like this:

\begin{align} 1001 \\ 1003 \\ 1007 \\ 1009 \\ \\ 1011 \\ 1013 \\ 1017 \\ 1019 \\ \\ 1021 \\ 1023 \\ 1027 \\ 1029 \end{align}

Then we check if the number can be divided by 3(without remainder) and if so we will mark it with "Yes" \begin{align} 1001:Yes\\ 1003:Yes\\ 1007:Yes\\ 1009:Yes\\ \\ 1011:No\\ 1013:Yes\\ 1017:No\\ 1019:Yes\\ \\ 1021:Yes\\ 1023:No\\ 1027:Yes\\ 1029:No \end{align} Now the weird thing is that if we keep going, it repeats itself: \begin{align} 1001:Yes\\ 1003:Yes\\ 1007:Yes\\ 1009:Yes\\ \\ 1011:No\\ 1013:Yes\\ 1017:No\\ 1019:Yes\\ \\ 1021:Yes\\ 1023:No\\ 1027:Yes\\ 1029:No \\ \\ 1031:Yes\\ 1033:Yes\\ 1037:Yes\\ 1039:Yes\\ \\ 1041:No\\ 1043:Yes\\ 1047:No\\ 1049:Yes\\ \\ 1051:Yes\\ 1053:No\\ 1057:Yes\\ 1059:No \end{align}

Now, if we test every number marked with "yes" to see if it's primes, we observe that 56.25% of them are, in fact, prime. From what I've tested, this is one of the more inaccurate examples(We would need more for a more accurat one). This theory is based on a post by David Shulman on Quora, and he states that there is a 73.3% chance of it being prime: One method which works most of the time (I’ll clarify that in a moment) is to think of a moderately large number whose ones digit is 1, 3, 7, or 9. You already know that this number cannot be a multiple of either 2 or 5, so perform a quick mental test for divisibility by 3 (is the “digital sum” divisible by 3?). If the test fails, add 2, skipping multiples of 5 of course, until it passes. Now you’ve got a number which is not a multiple of 2, 3, or 5 — not “5-smooth”, as a number theorist might say. It turns out that a number like this is fairly likely to be prime: the probability 73.3¯%

Now, hear me out: This is so inaccurate because it's only testing for multiples of 3. But the main point is that it is repeating. So, if we add another testing layer (for 7, 11, etc.), it should increase the accuracy (if they also repeat). And in the end, we don't even have to do something in our head because we just have to remember the layers. For example, if we start at 3803, we only have to figure out where 3803 and its 'to be tested' number (in this case, 3087) are on the layers, and then we know the other primes onward.

Please confirm if this is correct, keeping in mind that I have never done something like this with math, and I only discovered it by accident while I was looking for a way to count primes upwards in my head to keep myself busy

Etscharntmin
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    The relative density of primes out of the numbers who are $7$-rough (you are using the word "smooth" incorrectly here) is zero, and would be asymptotic to $\frac{1}{\log x}$ as a corollary of the Prime Number Theorem. There should be no reason why one should believe it to be 73.3% or 56.25% or anything like that unless we tested a truly insignificant number of small examples in comparison to the infinitely many out there. – JMoravitz Sep 20 '23 at 14:48
  • The probability that there is a pattern in the primes not vanishing for larger primes is indistinguishable from $0$. – Peter Sep 20 '23 at 15:07

1 Answers1

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Considering the numbers modulo $n$ causes repeating patterns because the remainders form a repeating pattern. For $n=3$ we see that $7-1=9-3=6$ which is divisible by $3$. This means I should expect numbers that end with $1$ and $7$ to be included or excluded together by the divisibility test since they will have the same remainder. The same is true for $3$ and $9$.

The second pattern comes from following the second digit. Here we can see that the pattern repeats on $1031$ and $1001$ and again since $1031-1001=30$ which is a multiple of $3$ we should expect them to have the same remainder modulo $3$. So the pattern you're seeing is exactly the pattern caused by excluding multiples of $3$.

CyclotomicField
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