20090303, 15:16  #1 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
7×13×47 Posts 
"prime numbers formula" crankery
http://www.primenumbersformula.com/
In classic crank fashion, it's hard to understand what he's trying to say, but I'm pretty sure he claims to have formulas for generating primes and Mersenne primes, has proved the infinitude of twin primes, and has solved the Riemann hypothesis. 
20090303, 18:47  #2  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Quote:
Quote:
I do see references to "the solution of Riemann Zeta equation" and "solution to Riemann zeta equation (http://www.primenumbersformula.com/a...s/image038.gif)", but those are not at all the same as claiming a proof of the Riemann Hypothesis. (The Riemann Hypothesis is a conjecture about a common property of all nontrivial zeros (solutions) of the zeta function.) At least he credits GIMPS properly, though without a single link (even in his "Favorite Links" section!) to us: Quote:
Last fiddled with by cheesehead on 20090303 at 19:46 Reason: removed quibble about adaptation. 

20090303, 20:11  #3 
Aug 2006
1011101011011_{2} Posts 
His claim that generating a formula for the prime is a "2300years old unsolvable problem" is a little wacky. It's been solved... over and over and over.
Last fiddled with by CRGreathouse on 20090303 at 20:17 
20090304, 00:15  #4 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{3}·3·269 Posts 
But the formula is posted there on the page in plain view. I would think it would by easy to find a nonprime generated by the formula.
Although strange how it says "unsolvable problem" and then goes on to give a solution! 
20090304, 01:27  #5 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2^{2}·3·877 Posts 
Correct me if I am wrong, but, doesn't this simplify to:
H(m)=(2m+1)^{1} ? 
20090304, 01:38  #6 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{3}·3·269 Posts 
Well, at this stage I am too lazy to program anything up to check it, but in the website it shows this sequence "3,5,7,2,11,13,2,17,19, ..." So assuming they applied their formula correctly (I Haven't checked), then it can't be a simple 2m+1.
Last fiddled with by retina on 20090304 at 01:38 
20090304, 02:10  #7  
"Ben"
Feb 2007
3,617 Posts 
Quote:
Here's excel code: Code:
=2*((2*B16+1)/2)^(FLOOR((2*B16+1)/(FACT(2*B16)+1)*FLOOR((FACT(2*B16)+1)/(2*B16+1),1),1)) 

20090304, 02:36  #8 
Feb 2006
Denmark
346_{8} Posts 
You probably have rounding errors. In PARI/GP the formula is:
H(m) = 2*((2*m+1)/2)^floor((2*m+1)/((2*m)!+1)*floor(((2*m)!+1)/(2*m+1))) The formula is correct but useless. It uses Wilson's theorem and the floor function to give: H(m) = 2m+1, if 2m+1 is prime = 2, otherwise Wilson's theorem says 2m+1 is prime iff (2m)!+1 is divisible by 2m+1. Then floor((2*m+1)/((2*m)!+1)*floor(((2*m)!+1)/(2*m+1))) becomes 1 when 2m+1 is prime, and 0 when it's composite. 
20090304, 02:59  #9 
Aug 2006
175B_{16} Posts 
The formula looks fine to me  it should generate the odd primes in order, with 2 replacing each odd composite.
Edit: Jens Anderson beat me to it. Last fiddled with by CRGreathouse on 20090304 at 02:59 
20090304, 02:59  #10  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Quote:
Quote:


20090304, 10:42  #11 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{3}·3·269 Posts 
It seems to me that this is more of prime testing formula than a function. They have taken a prime testing formula and manipulated the algebra to give you back the original prime you tested if it passes or 2 if it fails. It would be easy to change the '2' to any other number (like pi or e or i) to give a fixed value when the test fails.
Hmm, that has given me an idea ... Last fiddled with by retina on 20090304 at 10:44 
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