Is this a regularity in primes?

Question

For any prime $p$ subtract $24$ continuously. The last value before $0$ will always be one of these $8$ primes: $\{ 1, 5, 7, 11, 13, 17, 19, 23 \}$.

Prime Distribution Across Lengths of 24

Primes in Blue. Root Prime Path in Red

As can be observed above prime numbers do not deviated from the original host numbers. This suggests that primes have a geometric component underpinning their formation. Further, it should hold true that primes persist to appear in just $8$ or $24$ locations as observed above. Consider then:

$24^2 = 576 $

$576$ is the $24$th column of the images above. Subtracting $-1, -5, -7, -11, -13, -17, -19$ or $-23$ should then result in a prime hit or miss.

$-1$ is a miss

$-5$ is a hit: $571$ …a prime

$-7$ is a hit: $569$ …a prime

$-11$ is a miss

$-13$ is a hit: $563$ …a prime

$-17$ is a miss

$-19$ is a hit: $557$ …a prime

$-23$ is a miss

Now, let’s do this with arbitration:

Choose a large number: $1024169726$

The nearest prime to $1024169726$ is $1024169717$.

$1024169726 – 1024169717 = 9$.

This would seem to violate what I have stated above. However, $1024169726$ is not divisible by $24$.

$1024169726 – 14 = 1024169712$.

$1024169712$ is divisible by $24$.

$1024169717 - 1024169712 = 5$, one of the prime conditions stated above in this sieve.

Again, choosing from arbitration a number: $3248715756$. $3248715756 + 12 = 3248715768$ which is divisible by $24$.

$\dfrac{3248715768} {24} = 135363157$.

$3248715768 – 1, -5, -7, -11, -13, -17, -19, -23$ does not produce primes.

$3248715768 – 24 = 3248715744.$

$3248715744 – 13 = 3248715733$ which is prime.

Here we can see again that while there are not always primes within the gaps of $24$ the primes will nevertheless occur along the vector issued by adding $24$ to $\{ 1, 5, 7, 11, 13, 17, 19, 23 \}$ continuously.

This hypothesis was formed by myself and another professor at Penn State by investigating the mathematics behind old astrological sites such as Knowth. Unfortunately our network of Number Theorists from this campus is very small. If anyone can confirm this is indeed a pattern in primes it would be incredibly helpful.

Thank you, Iapyx

P.S. This website's formatting doesn't allow the original spacing in the numerical formatting. Sorry for the clutter.

Answer 1

This is a regularity to be found whatever the even interval you choose, be it 24, 70892, $17! - 8$, etc.

From Dirichlet's theorem, we know that $24m + n$ gives infinitely many primes provided $\gcd(24, n) = 1$. Without loss of generality, as the professional mathematicians like to say, we can assume $0 \leq n < 24$.

Now, what's $\gcd(24, 0)$? A vexing question, but it's not 1, I can tell you that much. Indeed we see that $24m + 0$ never gives a prime, because all those numbers are multiples of 24.

Then what about $24m + 1$? You have already gathered evidence that there are a few primes of this form. Dirichlet tells us there are infinitely many primes of this form. He gave a proof, which I have not seen, but I trust real mathematicians have reviewed it and found it to be in order.

$24m + 2$ will not work, though, for any value other than $m = 0$. All numbers of that form, aside from 2 itself, are even and divisible by some prime other than 2. We have a very similar situation with $24m + 3$, only $m = 0$ gives a prime (though it's possible for other numbers of that form to be divisible by no prime other than 3, e.g., if $m = 1$). Both 2 and 3 are divisors of 24, so obviously $\gcd(24, 3) > \gcd(24, 2) > 1$.

Let's skip ahead to $24m + 10$. Clearly 10 is not a divisor of 24, but $\gcd(24, 10) = 2$, and accordingly we find that all numbers of the form $24m + 10$ are even. And with $24m + 18$, we see that not only are all numbers of that form even, they are all divisible by 3 and by 6. That's because $\gcd(24, 18) = 6$.

What if we calculate $\gcd(24, n)$ for $0 < n < 24$? We get the sequence 1, 2, 3, 4, 1, 6, 1, 8, 3, 2, 1, 12, 1, 2, 3, 8, 1, 6, 1, 4, 3, 2, 1. So the symmetry that you're observing is merely the result of the symmetry of this sequence.

There are wondrous things in math. This is not one of them.

Answer 2

It seems to me that this is simply because $\{ 1, 5, 7, 11, 13, 17, 19, 23 \}$ are the only integers between $0$ and $24$ whose gcd (greatest common divisor) with $24$ is $1$. Since $24 = 2^3.3$, whenever $n = m\cdot 24 + k$, $k$ must not be divisible by $2$ or $3$ if $n$ is supposed to be prime. Removing all such numbers from $\{0,\ldots,23\}$ yields exactly the list you found...

Answer 3

The integers $\{1,5,7,11,13,17,19,23\}$ are natural numbers numbers less than $24$ that are co-prime to $24.$

Lets look at the compliment to this set.

$B=\{2,3,4,6,8,9,10,12,14,15,16,18,20,21,22\}$

For all $b\in B, \gcd(b,24) \ne 1$

If $k = 24n + b$ then $\gcd(k,24) \ne 1$ and $k$ is not prime.

Furthermore there are composite numbers, that pass your "test." e.g. $121$ is not prime, and $121 = 24\cdot5+1$