> >

> > > 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