August 2nd, 2017, 10:49 AM  #11 
Senior Member Joined: Mar 2015 From: England Posts: 201 Thanks: 5  this is actually really good stuff! I modified the formula somewhat. f(n) = 3(n3) + 2(n3) + (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. 
August 2nd, 2017, 01:11 PM  #12  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra 
From what you write, it looks like an attempt at a formula to produce the $n$th prime. Quote:
I guarantee that it isn't. What do you get for the 5000th prime (48,611)?  
August 2nd, 2017, 01:15 PM  #13 
Senior Member Joined: May 2016 From: USA Posts: 1,047 Thanks: 430  
August 2nd, 2017, 03:01 PM  #14 
Senior Member Joined: Mar 2015 From: England Posts: 201 Thanks: 5 
f(n) = 3(n3) + 2(n3) + (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, 
August 3rd, 2017, 12:20 AM  #15 
Senior Member Joined: Mar 2015 From: England Posts: 201 Thanks: 5 
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. 
August 3rd, 2017, 01:16 AM  #16  
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24  Quote:
$z=n!/n^2$ or: $z=(n1)!/n$ where $n$ is your integer if $z\in \mathbb{N}$ than $n$ is NOT a prime. while if: $z\in \mathbb{QN}$ 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  
August 3rd, 2017, 01:29 AM  #17 
Senior Member Joined: Mar 2015 From: England Posts: 201 Thanks: 5 
Hey could you give me an example of that in action please?

August 3rd, 2017, 09:25 AM  #18 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 202 Thanks: 60 Math Focus: Algebraic Number Theory, Arithmetic Geometry  It's a (very inefficient) primality test. If you want to know whether a natural number n =/= 4 is prime, you can compute (n1)!/n. This will be an integer if and only if n is not prime. For example, take a nonprime n = 6: (61)!/6 = 20 is an integer. As another example, take a prime n = 5: (51)!/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. 
August 14th, 2017, 11:33 AM  #19 
Senior Member Joined: Mar 2015 From: England Posts: 201 Thanks: 5 
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. 
August 14th, 2017, 02:15 PM  #20 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra 
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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