--- On Sat, 11/8/08, amin B <

aminb_6829@...> wrote:

> Hello friends

> Being shocked when I have heard such this amazing news

> that an Iranian professor named MR Moosavi has

> explored the prime numbers formula .For more information and

> viewing the formula visit the following site:

>

> http://www.primenumbersformula.com/

>

> I guess this claim based on finding the prime numbers

> formula is just a lie since none valuated source talks about

> so.And this event just announced by Iran news.

> Please leave your opinions about it.

"Strange" is indeed the word. This is quite an interesting case. His claim about the production of prime numbers and only prime numbers using his formula is indeed correct, it's not one I've seen before, and it's remarkably simple, in particular it doesn't involve multiply-nested sums. So from the outset, it's worth a little bit of investigation.

The function in question is:

H(m) = 2 * ((2m+1)/2) ^ floor(((2m+1)/((2m)!+1))*floor(((2m)!+1)/(2m+1)))

It took me about 2 minutes to work out how and why it works, and it might be fun for the easily amused to also cut their teeth on working it out themselves. There are a few clues on the above webpage, if you can't get it immediately. Here are also some spoilers ....

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Perform a change of variable n = 2m+1 to get H as a function of n.

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Look only at the value of the exponent, and ignore what is being raised to that power.

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We know for natural numbers a,b, (a/b)*(b/a) is always 1. However, when is (a/b)*floor(b/a) not equal to 1? (Assume a>=b for simplicity.)

That should be enough, I hope. If not, you did notice Wilson's Theorem being mentioned on the webpage, I trust?

In essense, the formula's very close to

h(n) = p^(prime_p(p))

Which would be p^1 = p for prime p, and p^0 = 1 for composite p.

In order to not generate all those 1s, he's tweaked it to be

h'(n) = 2*(p/2)^(prime_p(p))

Phil