- I started with this question: It is possible to find a number "k"

that: 6k+1 returns a prime, in a way that we are sure of it

primality, with no need of testing it?

I'm speaking of "k" that are non zero positives integers.

This is how I started and follow my research:

1)

I studied the numbers of the form: 6k+1 and 6k+1

I separated these numbers into three categories:

a) 6k+1 or 6k-1 that are primes

b) 6k+1 that aren't primes

c) 6k-1 that aren't primes.

In the categories "b" and "c", I wonder why those "k" don't return a

prime. Is there any rule for them? After I looked at them for a

while I realized this:

CAT B: 6k+1

Let P be a prime greater than 3

If: k = (P^2-1)/6 + P.j then 6k+1 is composite, where "j" is

every integer greater or equal to zero.

Then I obtained this:

Let k be a number of this form:

i) k= (P^2-1)/6 a where P is the nth prime number,

and 0<a<(P^2-1)/6

If:

k (Q^2-1)/6 is not a multiple of Q; then 6k+1 is prime with

this considerations: Q is every prime that is lesser to P and

greater than 3, so we need to do the test for the validity

of "k" "m" times.

I also noticed this:

If we took a k, P and Q with the conditions stated in i) then we

have that:

If: k - (Q^2-1)/6 is multiple of Q, being Q every prime that is

lesser to P and greater than 3 then 6k+1 is multiple of every Q that

divides: k - (Q^2-1)/6

CAT C: 6k-1

Let P be a prime greater than 3, and Q the next prime after P with

this considerations: Q-P is not a multiple of six.

Then we have:

If: k = (P.Q + 1)/6 + P.j then 6k-1 is composite, where "j" is

every integer greater or equal to zero. - develator81 dice:

> I started with this question: It is possible to find a number "k"

Indeed you can. k=1, k=2 and k=3 are good examples!

> that: 6k+1 returns a prime, in a way that we are sure of it

> primality, with no need of testing it?

Bernardo Boncompagni

________________________________

"Wars not make one great"

Yoda

When's who

A timeline of history makers

http://whenswho.redgolpe.com

Factorization of

special form numbers

http://factors.redgolpe.com

________________________________ - Hi,

Please refer to my communication of Oct.17,2004 to Yahoo prime number group on the subject of "Prime Structure" where I have discussed the occurrence of

6*n -1 and 6*n + 1 type primes and the governable factors for the appearance of such primes in prime sequence. In the same communication I have discussed the impact of such prime representation on the many intriguing prime properties like presence or absence of twin primes, prime formulae for prime generation, computation of prime pairs satisfying Goldbach Conjecture etc. If you need further information or wish to discuss the topic plese feel free to email me.

L.J.balasundaram

develator81 <develator81@...> wrote: I started with this question: It is possible to find a number "k"

that: 6k+1 returns a prime, in a way that we are sure of it

primality, with no need of testing it?

I'm speaking of "k" that are non zero positives integers.

This is how I started and follow my research:

1)

I studied the numbers of the form: 6k+1 and 6k+1

I separated these numbers into three categories:

a) 6k+1 or 6k-1 that are primes

b) 6k+1 that aren't primes

c) 6k-1 that aren't primes.

In the categories "b" and "c", I wonder why those "k" don't return a

prime. Is there any rule for them? After I looked at them for a

while I realized this:

CAT B: 6k+1

Let P be a prime greater than 3

If: k = (P^2-1)/6 + P.j then 6k+1 is composite, where "j" is

every integer greater or equal to zero.

Then I obtained this:

Let k be a number of this form:

i) k= (P^2-1)/6 a where P is the nth prime number,

and 0<a<(P^2-1)/6

If:

k (Q^2-1)/6 is not a multiple of Q; then 6k+1 is prime with

this considerations: Q is every prime that is lesser to P and

greater than 3, so we need to do the test for the validity

of "k" "m" times.

I also noticed this:

If we took a k, P and Q with the conditions stated in i) then we

have that:

If: k - (Q^2-1)/6 is multiple of Q, being Q every prime that is

lesser to P and greater than 3 then 6k+1 is multiple of every Q that

divides: k - (Q^2-1)/6

CAT C: 6k-1

Let P be a prime greater than 3, and Q the next prime after P with

this considerations: Q-P is not a multiple of six.

Then we have:

If: k = (P.Q + 1)/6 + P.j then 6k-1 is composite, where "j" is

every integer greater or equal to zero.

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