--- In

primenumbers@yahoogroups.com, "Robert" <100620.2351@c...>

wrote:

>

> > >

> > > > But for expressions of the form x^2 + x + p there is no need

to

> > > > look for a longer one since it has been shown that p = 41

> > > > generates the longest one.

> > >

> > > Again, the same conjecture implies that arbitrarily long

> sequences

> > > of primes exist of the form x^2+x+p.

> > >

> > > What has been shown is that p=41 is the largest prime such that

> > > x^2+x+p is prime for all x, 0 <= x <= p-2.

> > >

> > > It has not been shown that x^2+x+p is never prime for 0 <= x

<=

> 40.

>

> Surely all one has to do is find a c in the equation x^2+x+c for

> which the following conditions apply:

>

> 2+c not divisible by 2, 3, 5

> and, 2+c meets all of:

>

> 1,5,6 mod 7

> 3,6,8,9,10 mod 11

> 1,3,4,5,6,7,10 mod 13

> 1,3,4,5,8,9,10,12 mod 17

> 4,5,8,11,12,13,14,16,18 mod 19

> 1,3,9,10,11,12,14,16,17,20,21 mod 23

> 3,5,6,7,9,10,12,13,14,16,21,22,26,27 mod 29

> 4,7,11,12,14,15,17,18,19,20,24,26,28,29,30 mod 31

> 1,6,7,8,10,11,12,13,15,16,17,22,24,25,28,32,35,36 mod 37

> 3,4,5,6,7,9,11,14,16,18,19,20,21,22,26,27,30,36,39,40 mod 41

> 1,5,6,8,10,11,14,17,19,22,23,24,26,27,28,29,30,34,36,37,38 mod 43

>

> c=41 is the first number to reach all of the conditions except the

> last, being 1mod7, 10mod11, 4mod13....but 0mod43

>

> Regards

>

> Robert Smith

>

> PS I may have gotten some of the register above incorrect, but

> someone will spot an error if I have made one. Thats what I like

> about you Primenumbers group.

>

> PPS 2+c = x^2+x+c with x=1

>

> PPPS this is the same logic as used in the determination of Payam

> numbers

#

Oops, I forgot to mention that the value of c, which contributes to

a run of >41 primes, must also clear similar mod n hurdles up to

sqrt c