- View SourceYou have not proved a thing; you merely launched a new hypothesis.

You are not proving anywhere that, for a fixed factorial, it exists a twin prime that

added to the factorial will always give another twin pair. You only give some evidence.

Indeed, a very very little evidence for it. Moreover, you don't need to prove that the sum

is a twin, if you can simply prove that it exists a twin for every factorial, then you

would have proven the TP conjecture! But you can't prove it with your idea, unless you

build a chain of numbers that are always twins (you have to prove it) and you prove that

adding that twins to factorials give new twins. That's it, to make explicit the twins that

your chain will use.

Nevertheless, your idea about that relation between two twin primes seems beautiful to me.

Will all the twin primes be related to a factorial and another twin prime? If it is true,

is it trivial? If it is not trivial, is there a simple formula to regulate this fact? How

many primes will you add to the same factorial before you change to the next factorial?

Because factorials grow a lot faster than twin primes do (at least I think so)

Or will it be a consecuence of the small law of numbers? Can you go further with your

checking? Go up with the twins to, say, 10^8, if you can, and then return with the results

if you think they are relevant.

Jose Brox

----- Original Message -----

From: "Milton Brown" <miltbrown@...>

To: "primenumbers" <primenumbers@yahoogroups.com>

Sent: Thursday, May 05, 2005 10:54 PM

Subject: [PrimeNumbers] Infinite Number of Twin Primes

Here is a demonstration that there are an infinite number of Twin Primes,

and a method of finding them.

Consider the following twin primes:

(5,7)

(11,13) (17, 19)

(29,31) (41,43)

(149,151) (179,181) ...

Now put the appropriate factorial ahead of them

(2!+3,2!+5)

(3!+5,3!+7) (3!+11,3!+13) note (3!+17,3!+19=25) does not woork

(4!+5,4!+7) (4!+17,4!+19)

(5!+29,5!+31) (5!+59,5!+61)

This is the pattern: Start at the next factorial and add prior pairs of twin primes,

until you obtain a pair of twin primes for that factorial.

The equivalent theorem is "There is always a pair of twin primes between a factorial

and its successor factorial."

Then Ininite Twin Primes are proved since 2!, 3!, ..., n!, ... are infinite in number.

The familiar formula:

n!+2, n!+3, ..., n!+n

shows that these numbers are all composite,

but for this value of n we start testing the previous twin primes at

n!+(n+1) and (n-1)! > n+1 for n>4. This is easily proved by induction on n:

(n-1)! > n+1

n! = n(n-1)! > n^2+n > n+2 since

n^2 > 2.

Milton L. Brown

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The Prime Pages : http://www.primepages.org/

Yahoo! Groups Links - View SourceI found that

4!+857, 4!+859 adds (881,883)

allowing the primes generated by the n factorial to go "backwards" to be checked with

former factorials.

But then I got stuck. No twin primes with (881,883) with n! from 3 to 7.

Jose Brox

PS Milton still has an escape, if he has a way to determine a twin that added with the

factorial will give another twin... but this is silly, cause that method would prove TP

conjecture itself!

----- Original Message -----

From: "Décio Luiz Gazzoni Filho" <decio@...>

To: <primenumbers@yahoogroups.com>

Sent: Friday, May 06, 2005 1:13 AM

Subject: Re: [PrimeNumbers] Infinite Number of Twin Primes

On Thursday 05 May 2005 19:49, you wrote:

> At 04:54 PM 5/5/2005, Milton Brown wrote:

> >This is the pattern: Start at the next factorial and add prior pairs of

> >twin primes,

> >until you obtain a pair of twin primes for that factorial.

>

> How do you know that will work (that you will always obtain a pair of twins

> for that factorial)?

The answer is, he doesn't, because it won't work. This one didn't even need a

computer search; it fell to a search by hand. I would appreciate a third

check (I've already double-checked locally).

Start from (3,5).

2!+3, 2!+5 adds (5,7). Values found until now: (3,5), (5,7).

3!+5, 3!+7 adds (11,13). 3!+11, 3!+13 adds (17,19).

Values found until now: (3,5), (5,7), (11,13), (17,19).

4!+5, 4!+7 adds (29,31). 4!+17, 4!+19 adds (41,43).

Values found until now: (3,5), (5,7), (11,13), (17,19), (29,31), (41,43).

5!+17, 5!+19 adds (137,139). 5!+29, 5!+31 adds (149,151). 5!+149, 5!+151 adds

(269,271).

Values found until now: (3,5), (5,7), (11,13), (17,19), (29,31), (41,43),

(137,139), (149,151), (269,271).

6!+137, 6!+139 adds (857,859).

Values found until now: (3,5), (5,7), (11,13), (17,19), (29,31), (41,43),

(137,139), (149,151), (269,271), (857,859).

Finally, 7! doesn't produce any values. And thus Milton's conjecture is

demolished, as about anything that he brainfarts on this list.

Décio

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