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August 2nd, 2017, 10:49 AM   #11
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this is actually really good stuff! I modified the formula somewhat.

f(n) = 3(n-3) + 2(n-3) + (n+2)

Example 7 = 3*4 + 2*4 + 9 = 29

7,11,13,17,19,23,29

1, 2, 3, 4, 5 ,6 ,7

From 7 and upwards when the Prime starts with 1 or 3 you start counting at 0, else when it's 7 or 9 you start counting at 1.

Also if it starts at 3 you add 2 to the last number in the formula.

Last edited by HawkI; August 2nd, 2017 at 10:54 AM.
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August 2nd, 2017, 01:11 PM   #12
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From what you write, it looks like an attempt at a formula to produce the $n$th prime.
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From 7 and upwards when the Prime starts with 1 or 3 you start counting at 0, else when it's 7 or 9 you start counting at 1.
I can't say that means an awful lot to me. Do you mean $n$ starts with 1 or 3, or 7 or 9? And what are you counting?
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this is actually really good stuff!
I guarantee that it isn't.

What do you get for the 5000th prime (48,611)?
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August 2nd, 2017, 01:15 PM   #13
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the up loader doesn't seem like the kind of person who would make a new
apologetic video if they found out the formula was wrong.
What a person "seems like" is not considered material in math or the physical sciences.
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August 2nd, 2017, 03:01 PM   #14
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f(n) = 3(n-3) + 2(n-3) + (n+2)

48,611 = 3*48,608 + 2*48,608 + 48,613 = 291,653
_________145,824____97,216

Here is a Prime it is close to 291,649

Frustratingly I can't seem to find what number of Prime that is. But remember the author says that this is just their second draft.

I mean n ends in 1, 3, 7, 9. I'm counting the Primes in between n and the answer.

Perhaps the formula for nth Prime is the inspiration,
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August 3rd, 2017, 12:20 AM   #15
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48,611 is the 5,000th Prime

291,649 is the 25,359th Prime

So counting from 48,611 gets to the 53,611th Prime, 660,563

Ok that didn't work, I have noticed that near the lower numbers everything looks like patterns.
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August 3rd, 2017, 01:16 AM   #16
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Quote:
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48,611 is the 5,000th Prime

291,649 is the 25,359th Prime

So counting from 48,611 gets to the 53,611th Prime, 660,563

Ok that didn't work, I have noticed that near the lower numbers everything looks like patterns.
Once again the pattern is perfectly given by:

$z=n!/n^2$

or:

$z=(n-1)!/n$

where $n$ is your integer

if $z\in \mathbb{N}$ than $n$ is NOT a prime.

while if:

$z\in \mathbb{Q-N}$ than $n$ IS a prime

It works for all $n\in \mathbb{N}$ except for $n=4$

So, pls waste your time as you prefer, but no more looking for a more simple pattern
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August 3rd, 2017, 01:29 AM   #17
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Hey could you give me an example of that in action please?
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August 3rd, 2017, 09:25 AM   #18
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Hey could you give me an example of that in action please?
It's a (very inefficient) primality test. If you want to know whether a natural number n =/= 4 is prime, you can compute (n-1)!/n. This will be an integer if and only if n is not prime.

For example, take a non-prime n = 6: (6-1)!/6 = 20 is an integer.
As another example, take a prime n = 5: (5-1)!/5 = 4.8 is not an integer.

I invite you to have a think about why it works; it's not too hard to prove using only elementary results.

Last edited by cjem; August 3rd, 2017 at 09:28 AM.
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August 14th, 2017, 11:33 AM   #19
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I think I've sussed what this finished formula includes.

Apart from the Primes 2 and 3 the rest are either 1 more or less than a multiple of 6, and the use of imaginary numbers in the previous formulas means numbers are added to or taken away by 1.

It's possible this could have been really effective until Prime gaps started happening such as two Primes that are generally 6 times greater than 2.

But still, in a way I think Primes being close to multiples of 6 is an impressive achievement for the people who first discovered it.

Last edited by HawkI; August 14th, 2017 at 11:44 AM.
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August 14th, 2017, 02:15 PM   #20
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Not massively so if you think about why $6n \pm 1$ works. $6n$ is obviously divisible by $2$ and $3$, and therefore so are $6n \pm 2$ and $6n \pm 3$. $6n \pm 1$ are all that remain.
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