#density of primes in an interval

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granite valeBOT
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tall harbor
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The maximum number of prime numbers that can be found in the interval n < p < n + 10, where n is a multiple of 10 and p is prime, is four.

To see why this is the case, note that the possible values of n modulo 10 are 0, 2, 4, 6, and 8. If n ends in 0 or 5, then n+2, n+4, n+6, and n+8 are all even and therefore not prime. If n ends in 2 or 8, then either n+2 or n+8 is divisible by 5 and therefore not prime. Similarly, if n ends in 4 or 6, then either n+4 or n+6 is divisible by 5 and therefore not prime. This leaves only the possibility that n ends in 1, 3, 7, or 9, in which case at most four primes can be found in the interval n < p < n + 10.

There is no known conjecture or theorem that states that for any interval of consecutive integers that is a multiple of 10, the maximum number of prime numbers in that interval is four. However, the behavior of primes in arithmetic progressions is a well-studied area of number theory, and there are several results that give bounds on the number of primes in such intervals. The specific case of intervals of the form n < p < n + 10 has been extensively studied and is well understood.

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Hope this helps @chilly heath

nova pumice
tall harbor
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Yes, you are correct. I apologize for the error in my previous response. If n is a multiple of 10, then n mod 10 is equal to 0, and the possible values of n modulo 10 are 0, 2, 4, 6, and 8. The maximum number of primes in the interval n < p < n + 10 is four, and this is achieved when n ends in 1, 3, 7, or 9. Thank you for pointing out my mistake.

nova pumice
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...what? You just made the same mistake again.

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If n mod 10 = 0, then n == 0 (mod 10).

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Like, that's almost literally saying exactly the same thing.

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It's not 0, 2, 4, 6, or 8, it's 0, 0, 0, 0, or 0.

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Because it's 0.

tall harbor
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Yes.

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Hmm.

nova pumice
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I don't think it's that complicated. It's 4 because primes become less dense the higher you count, so they're most dense on the interval (0, 10), which has 4; {2, 3, 5, 7}.

tall harbor
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I apologise for being incorrect. I can see where I went wrong I will refrain myself from helping people.

nova pumice
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Obviously that's not formally proved, but it's a much more intuitive reason to suspect that's the answer than whatever you were trying to do.

nova pumice
tall harbor
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Thank you.

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For understanding.

chilly heath
pine pond
tall harbor
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Felt like it.

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Ban me.

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But wait.

pine pond
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No, refrain from doing so.

tall harbor
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Will do, won’t happen again.

pine pond
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Thanks

chilly heath
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Can't you just use Euler's toty function?

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Oh nvm actually.

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Anyways

chilly heath
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n+2
n+4
n+6
n+8 are obviously not primes

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If n ≥ 10,

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Then

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Among

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n+1
n+3
n+5 (divisible by 5)
n+7
n+9

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At least 1 number should be divisible by 3

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Was missing a number there

chilly heath
chilly heath
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1 2 3 4 5 6 7 8 9

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Literally how number 3 works

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See

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its okay you dont have to explain it i already got the answer thanks though

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Ah alright.

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The answer is 3/9?

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its 4

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Ah yeah 4

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Maximum so yeah 4

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why would you get a fraction though

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You asked for density 💀

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Isn't density frequency/total?

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But yeah, the frequency would be 4.

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i meant density as in how much primes exist in that interval not the literal density

chilly heath
# chilly heath how did you come to that conclusion though

n = k (mod 3) hence 0 ≤ integer k ≤ 2
If k = 0,
n+3 and n+9 mod k = 0
If k = 1,
n+5 = 6 = 0 (mod 3) [update: this is dumb because n+5 is always divisible by 5 BUT this is also the case when you will get the maximum number of primes]
If k = 2,
n+1 and n+7 = 0 (mod 3)

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Proof by exhaustion

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We can always remove the even numbers when n ≥ 10

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And based on the n mod 3, we can remove other numbers

fluid sun
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@tall harbor @pine pond Super fast my ass

Screenshot_2023-04-15-23-45-47-274-edit_com.discord.jpg
pine pond
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Actually

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we should ban him already

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I'll ask the admins

fluid sun
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Eh, ok i don't mind either way, the shame is enough