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I'm curious if anyone has noticed that all numbers that are primes must end in a 1, 3, 7, or 9, and that you can tell which ones don't by simply multiplying previous primes together and their exponents.

So p*p & p^n will never be prime
p*p always lines up with a potential prime
p^n doesn't but I'd bet there is some pattern to where it does, something like where n is odd or something like that.

This tells you where there wont be a prime in the 1, 3, 7, 9 thing and so I'd think that this can be inverted somehow to tell you where all the primes are. I didn't track this very far, cuz I was doing it manually so I only got to 300, but I see no reason why this wouldn't continue on like this. It is however a really simple thing so I would have expected someone to come up with this ages ago and thus there must be some problem with it... I'm just curious.

Oh also, this explains why there are ever bigger gaps in primes. It's somewhat like an avalanche Early you're only dealing with a few numbers that are being multiplied together but as you go further you're dealing with more primes multiplying together to create holes.

Durakken
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  • You can go through this https://www.youtube.com/watch?v=YVvfY_lFUZ8 – Kushal Bhuyan Sep 05 '16 at 01:14
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    $2$ and $5$ are prime and they don't end in $1$, $3$, $7$, or $9$. – Matt Samuel Sep 05 '16 at 01:38
  • @MattSamuel True, but I just dismissed them for what I was doing, but I would argue that they should be classified sperated than the other primes because they're transformative (making any number you multiply with them to even) and the others seem not to be so its obvious you wouldn't really consider them to be the exact same type... Funny thought thought is what if aliens thought this and this is why no aliens answer when we use the prime sequence as a show of intelligence lol. – Durakken Sep 05 '16 at 01:56
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    The only reason this is true is because $2$ and $5$ divide $10$. It's obvious that every prime other than $2$ and $5$ end with $1,3,7,9$ because if that isn't true then the number is divisible by $2$ or $5$. It's dependent on the base and would involve different last digits in a different base. – Matt Samuel Sep 05 '16 at 01:59
  • @MattSamuel Right and the only reason I mention the 1, 3, 7, and 9 is to show my thinking. I don't have to deal with, 0, 2, 4, 5, 6, and 8 because they will never be prime outside of that first part so I can just ignore them. We can then call 1,3,7,9 prime potentials and the question becomes is there any way to tell which ones will (and inversely which one won't) be knocked out. And there is all prime potentials that are knocked out that looked at seem to line up with primes being multiplied by other primes or powers of primes and the ones remaining should all be primes. – Durakken Sep 05 '16 at 02:09
  • Also, I looked at the numbers again and there seems to be another pattern that emerges that I didn't see before where it starts adding factors in the same way but I'd have to look more into it to tell... – Durakken Sep 05 '16 at 02:13
  • $$p^p + p^n = p(p^{p-1}+p^{n-1})\tag1$$ and $p^{p-1}+p^{n-1} > 1$ so this means that Eq. $(1)$ will never be prime. $\qquad\bigcirc$ – Mr Pie May 13 '18 at 12:06

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