- Here's another one:

4x^2+12x-1583

is prime from x=0 to x=34, a span of 35 unique primes increasing

from -1583 to 3449. It too met its end at the hands of factor 37,

which is still at large.

Mark

--- In primenumbers@yahoogroups.com, "Mark Underwood"

<mark.underwood@s...> wrote:>

it

>

> Here's one that won't be a winner but is fairly long. I can't find

> in the Integer Sequences, so maybe it's 'new':

end

>

> 4x^2 - 212x + 2411

>

> Generates 27 distinct primes from x=0 (p=2411) to x=26 (p=-397)

> before doubling back. So it's prime from x=0 to x=53. It met its

> at the hands of a factor of 37.

>

> Mark

>

>

>

> --- In primenumbers@yahoogroups.com, ed pegg <ed@m...> wrote:

> >

> > The next Al Zimmermann Programming contest might be about

> > Prime Generating Polynomials. It's currently running neck

> > and neck with Protein Folding.

> > http://www.recmath.com/contest/votes.php

> >

> > Under the rules I'm proposing for the contest, here are the

> > current known record holders for Prime Generating Polynomials,

> > orders 1 to 4. Orders 5 and up seem to be unexplored.

> >

> > 1) 44546738095860 n + 56211383760397 Score:23 n=0..22 Frind

> > 2) 36 n^2 - 810 n + 2753 Score:45 n=0..44 Ruby

> > 3) 3 n^3 - 183 n^2 + 3318 n - 18757 Score:43 n=0..46 Ruiz

> > 4) n^4 + 29 n^2 + 101 Score:20 n=0..19 Pegg

> >

> > The scoring rules:

> > 1. Polynomial f(k) must produce primes from 0 to n.

> > 2. The score will be the number of *distinct* primes

> > when |f(k)| is evaluated from 0 to n.

> > 3. In case of a tie, the lower value (tighter value) of n wins.

> > 4. In case of a tie, the product of non-zero coefficients will

> > be evaluated, and the lowest product wins.

> >

> > My own result for order-4 polynomials will likely be surpassed

> > very easily. If you would like to vote for this contest, please

> > visit http://www.recmath.com/contest/ . The winners will split

> > up $500.

> >

> > Ed Pegg Jr

> >

> - Dick wrote:

> A quadratic expression that yields no prime factors less than a given

Yes. Here is PARI/GP code:

> value is easy to generate arbitrarily.

f(n)=local(c,p,i,r);c=1;r=2;forprime(p=3,n,i=1;

while(issquare(Mod(i,p)),i++);c+=lift((Mod(1-i,p)/4-c)/r)*r;r*=p);c

x^2+x+f(n) never has a factor <= n.

f(n) is a little below n# (which is near e^n).

f(277) has 113 digits:

52211040781253690937101509813868236062339335365181118768\

264647548098518281973426206257327260311399656484832821211

f(100000) is computed in 2 seconds and has 43293 digits.

It is far harder to find the smallest constants to avoid

all small factors in x^2+x+A.

http://www.primepuzzles.net/conjectures/conj_017.htm lists that.

The record is A=2457080965043150051 which gives 281 as smallest factor.

--

Jens Kruse Andersen