--- In

primenumbers@yahoogroups.com,

"ALBERTO" <arfzelaya@...> wrote:

> Let p any prime of the form 6k+-1

> and n any integer number

...

> if an integer x is not equal to pn+-k

> it won't be equal to 6ab+-a+-b

Let's try to parse that carefully.

(I failed to do so, previously, sorry.)

Proposition: Let x be a positive integer for which there exists

at least one triplet (a,b,s) with integers a >= b > 0

and a sign s = +/-1 such that x = 6*a*b + s*(a+b).

Then there exists at least one such triplet for which

at least one of 6*a + s and 6*b + s is prime.

Proof: 6*x + 1 = (6*a + s)*(6*b + s) has at least one prime divisor.

Example 1: With x = 1627604, there are 5 triplets,

yielding only 1 prime, since 6*x + 1 = 5^10.

Example 2: With x = 6197024, there are 63 triplets,

yielding 7 primes, since 6*x + 1 = 23#/6.

Question for Alberto: What has this to do with H+L ?

David (with apologies for earlier stupidity)