

The Set of Prime Numbers (2007) [pdf] - anacleto
http://arxiv.org/pdf/0709.1539v1.pdf

======
wbhart
Summary of paper:

All n = 0, 2, 4 mod 6 are divisible by 2. All n = 0, 3 mod 6 are divisible by
3.

Thus all primes are either 1 or 5 mod 6.

If n is composite and 1 mod 6, n = pq with p, q > 1 such that both p and q are
both 1 or both 5 mod 6.

If n is composite and 5 mod 6, n = pq with p, q > 1 such that (wlog) p is 1
mod 6 and q is 5 mod 6.

What remains are the primes.

You have to look a little bit deeper than arithmetic modulo 6 if you want to
say something interesting about the primes!

------
Chinjut
This seems like... maybe "crankery" isn't the right word, but certainly
something overblown in that ballpark.

The claim is made "We obtain for the first time an explicit relation for
generating the full set P of prime numbers smaller than n or equal to n", but
the rule ultimately boils down to just "A number is non-prime if it is either
divisible by 2, or divisible by 3, or the product of two numbers which are
neither divisible by 2 nor divisible by 3", which is completely trivial.

(Also, the author was apparently attempting to patent this insight, as noted
at the bottom of the first page. You know what? I've changed my mind. I'm
going to happily call this "crankery". Well-LaTeXed crankery, but crankery
nonetheless...)

~~~
wbhart
They even threw in a reference to a fields medallist on page 1 just to alert
everyone that it was crankery.

------
phkahler
So if you have some lookup tables you can determine if a given number is
prime. Up to some maximum based on the size of the tables. And apparently the
time to generate the tables is quadratic in the size of n? Why is this
relevant? AKS shows that determining if n is prime can be done in polynomial
time - polynomial in log(n), not n. Is there something significant about these
2 sets of primes they talk about?

~~~
wbhart
You mean O(sqrt(n)), not quadratic. And given that the most obvious algorithm
there is (check every possible divisor up to sqrt(n)) is O(sqrt(n)) arithmetic
operations, I think it is fair to say this is not an important result.

I suppose it is a constant factor better than sqrt(n). But tricks along these
lines go back to Gauss, at least.

Edit: ah I see, _they_ claim O(n^2). Well I've never had an O(n^2) primality
test before. I will put that in my kit of "do-not-ever-use-this" tools, just
in case all my other ones break.

