Just thought I’d let you know that I’ve proven Andrica’s Conjecture by re-examining the Sieve of Eratosthenes:
http://www.ugcs.caltech.edu/~kel/MPP/AndricaConjectureTrue.pdf
The proof shows that, at the n-th step of the sieve, the largest gap generated is at most $2 p_{n-1}$, from which Andrica follows. I call the state of elimination at the n-th step $\kappa_n$.

So, right off the bat, you can find a prime in $N^{1/2}$, even without the Riemann Hypothesis. (Note that this isn’t an asymptotic result, like $N^{0.525}$; it holds for all primes.)

However, I believe we can do a better search in practice. A naive generation of $\kappa_n$ up to $N$ requires $O(log N)$ computation and $O(N)$ storage. At that point, the density of primes is around
$M_n = \Pi_{i=1}^{n} \frac{p_i -1}{p_i} ~ \frac{1}{e^\gamma \log{p_n}}$
But a more sophisticated search would pre-calculate where in $\kappa_n$ $N$ lay and only keep around the requisite intervals, namely those that lie around a Biggest Resolution of $p_B = p_n$. Check out Theorems 2.8 and 2.9 for more details.

Btw, I’m firmly convinced Cramer’s Conjecture is true. I have developed a conditional proof that shows if Cramer’s Conjecture were true, then all the constellation infinity conjectures must be true simultaneously. Were Cramer true, then primes are packed incredibly tight and with much more regularity than we can currently show. The side effect of the conditional proof is that we could not only find _a_ prime after $N$ in log time, we could find the _exact next prime_ in log time.

I hope this helps you in your quest for deterministic prime finding!

(So the easy way to come up with a polynomial is to have x^n + bx^(n-1) + 2x^(n-2) + 2^(n-3) … +2 where b is even and 1 + 2*(n-2) + b is prime and so f(1) = a prime and f(x) is irreducible under Eisenstein’s Criterion).

Now suppose we have for each K at least one corresponding f(x). How do we know that for some constant m f(m) is prime for some not-so-general and very-integery m AND m does not change for each f(x)? Find the least prime number, q, which is greater than any coefficients for any of the f(x) AND q must also be greater than K. The coefficients of the f(x) when put together side by side should form a number base q that is at least pseudoprime.

Examples: x^2 + 2x + 2. f(1) = 5. 122 base 5 is 37. Also 122 base 3 is 17.

x^2 + 4x + 2. 142 base 5 is 47. 142 base 7 is 79.

x^10 + 4x^9 + 2x^8 + 2x^7 + 2x^6 + 2x^5 + 2x^4 + 2x^3 + 2x^2 + 2x + 2. Which according to Wolfram Alpha when x = 11 the expression is prime. 35840804903.

To summarize this problem has already been solved in the S adjoin x world if by prime we mean irreducible. For any K, f(x) = x^K + 3 works (with Eisenstein’s). I was hoping something really cool would pop out of the polynomials.

I apologize as this response was a bit fuzzy (as my brain was not feeling perfect). I have not checked what kind of numbers one would want to plug in for f(x) = x^K + p to spit out numeric primes.

I did plot a graph for prime number against whole number on y axis. The graph was initially exponential and had became bit linear to the X-axis more and more as we had moved away from origin and parallel to X-axis.
There fore i am sure one point will cone after which we can’t come up with any prime number and that point would be the very intresting as well because at that point we can understand the spliting of a digit into two.

Thanking you
Shail

http://www2.xp-dev.com/sc/browse/86755/

so you can _download_ the files at any time whenever you have internet access, but to upload anything you either need to install the Subversion software, or email me with any modified files etc. Either way should work fine, though the former way would of course be more direct.