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I was wondering if this equation had any implications for prime research: $$3p-10$$ where $p$ is a prime.

I found this interesting and was curious if it had any backing through any prime proofs.

Here is my python script that will do the number crunching: https://github.com/Storms-Engineering/Prime-Scripts

Sample of output:

Skipped:5
Skipped:7
Skipped:11
Skipped:13
Skipped:17
Skipped:19
Skipped:23
Nope: 29
Output: 77
Factored: [7, 11]
Skipped:31
Skipped:37
Skipped:41
Nope: 43
Output: 119
Factored: [7, 17]
Skipped:47
Skipped:53
Skipped:59
Skipped:61
Skipped:67

Any thoughts?

Also if there is a better place to discuss this, I am all ears. This was really the only place that I could think of to ask this.

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    $\begingroup$ Your question seems complete ad hoc, random, and unmotivated. Why did you say $3\ x - 10$ and not $5 x - 11$? Or $7 x - 19$? Or $31 x - 458$? $\endgroup$ – David G. Stork Jul 16 at 1:09
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    $\begingroup$ It is because if $2, 3, 5$ does not divide $3x-10$ when $x$ is a prime. As the few smaller prime divisors are naturally ruled out, $3x-10$ should surely be close to a prime when $x$ is small. The conjecture is surely false: $3\times 337-10 = 1001 = 7\times11\times 13$, and should have lots of counterexamples. $\endgroup$ – Hw Chu Jul 16 at 1:30
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    $\begingroup$ Your "answer"..."I said $3𝑥−10$ because I made it up in my head" proves this is entirely ad hoc, random and unmotivated. $\endgroup$ – David G. Stork Jul 16 at 1:34
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    $\begingroup$ Generally speaking, with things like this, you may want to check up to large values of $x$ (I would think $6$ or $7$ digits is not too taxing on modern computers). $\endgroup$ – Clayton Jul 16 at 1:35
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    $\begingroup$ There are plenty of tools you can play with. For instance, go to magma.maths.usyd.edu.au/calc , type in something like Factorization(1234567); , and have fun ;) $\endgroup$ – Hw Chu Jul 16 at 1:39
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By the way, here is an "all-you-can-eat" approach to generate counterexamples. Find a series of prime $7 \leq p_1 < \cdots < p_l$, and let $N = p_1^{e_1} \cdots p_l^{e_l}$. $10$ is relatively prime to $N$, so there is a number $m$ such that $3m \equiv 10 \pmod N$.

Find a prime $x$ such that $x \equiv m \pmod N$. This is always possible by Dirichlet's Theorem on Primes in Arithmetic Progressions.

Then, $N \ | \ 3x -10$, which is very composite!

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    $\begingroup$ Thats awesome! Thanks so much for all the feedback! I will have to go refresh my memory on some of that math, and some of it I never learned in highschool and never went that far in college. Time for some self study. $\endgroup$ – StormsEngineering Jul 16 at 1:55
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For $3p-10>0$, it must be true that $p>3$. All primes greater than $3$ have the form $6n\pm 1$.

So $3p-10=3(6n\pm 1)-10=\{18n-7,18n-13\}$, either of which has the form $6n'-1$. Thus all numbers of the form $3p-10$ have a form that might possibly be a prime. For small numbers of the form $6n'-1$, the significant majority are prime: $5,11,17,23,29,41,47,53,59,71,83,89,101,107$ vs $35,65,77,95$, so more or less by coincidence, you find a lot of hits in the first (smaller) examples you look at, and it seems like there might be some kind of pattern. But if you look at enough larger examples of $p$, you will find plenty that do not produce a prime by this formula.

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