## HARDY LITTLEWOOD CONJECTURE ANALISYS. FIRST PROPOSITION.

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• Let p any prime of the form 6k+-1 and n any integer number and k the k value corresponding to each prime of the form 6k+-1 (for instance k values corresponding
Message 1 of 3 , Jan 5, 2011
Let p any prime of the form 6k+-1
and n any integer number
and k the k value corresponding to each prime of the form 6k+-1
(for instance k values corresponding to prime 5 is 1 because 5=(6*1)-1
and k value for 109 is 18 because 109=6*18+1)
Therefore the algebraic expression
pn+-k is equivalent to 6ab+-a+-b
In other words if an integer x is not equat to pn+-k it won't be equal to 6ab+-a+-b
I wish to know if in your opinion this proposition is right or wrong.
Kind regards,
Alberto Zelaya
• ... That was utter nonsense, since 17 = 6*3 - 1 is prime and 31 = 17*2 - 3. Abject apologies for relying on dud code, without performing a mental sanity check.
Message 2 of 3 , Jan 6, 2011

> There is no prime
> p = 6*k +/- 1 such that
> 31 = p*n +/- k with the signs matching.

That was utter nonsense, since
17 = 6*3 - 1 is prime and
31 = 17*2 - 3.

Abject apologies for relying on dud code,
without performing a mental sanity check.

David
• ... 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
Message 3 of 3 , Jan 6, 2011
"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)
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