Karlheinz Brandenburg

Uncwilly,
Did you read about what I wanted to write about?
It is sure a good idea but it isn't my direction

[QUOTE=blob100;206628]Retina, this is totally serious...[/QUOTE]Okay, sorry about that. How about "Austin Powers" instead. He is my personal nemesis/hero. [size=1][color=#ffffff]And my long lost brother also.[/color][/size]

If it is your hero, write them a letter (you can't participate in the "name your hero" competition).

By reading "Solved And Unsloved Problems in Number Theory", I found this conjecture: there are ifinitely many primes q of the form 2p+1 where p a prime.
And after some playing with this conjecture, I found a stronger one:
conjecture: there are infinitely many primes [tex] q=2^{n+1}Q+2^{n+1}-1[/tex] such that [tex]m=2^{n}Q+2^{n}-1[/tex] is also a prime, where Q is not neccesarily a prime number.

[QUOTE=blob100;206656]By reading "Solved And Unsloved Problems in Number Theory", I found this conjecture: there are ifinitely many primes q of the form 2p+1 where p a prime.
And after some playing with this conjecture, I found a stronger one:
conjecture: there are infinitely many primes [tex] q=2^{n+1}Q+2^{n+1}-1[/tex] such that [tex]m=2^{n}Q+2^{n}-1[/tex] is also a prime, where Q is not neccesarily a prime number.[/QUOTE]

STOP MAKING CONJECTURES. You don't know enough mathematics
to be able to say anything new or meaningful.

This new conjecture is no stronger than the one you quoted.

Neither is new. Both are subsumed by conjectures that really ARE
stronger. Look up Schinzel's Conjecture and the Bateman-Horn Conjecture.

And your notation is lousy. Your "conjecture" is better stated as:

s1 := 2^n R - 1
and
s2 := 2^(n+1) R - 1

are both prime i.o. for some R \in Z depending on n and for all n.
[there are other ways of stating it as well, i,e. s1 and 2s1 + 1 are prime i.o. ] By presenting it as a trinary form as you do, you disguise the
fact that s2 = 2s1 + 1.

It is a simple sub-case of Schinzel's Conjecture.

I didn't conjecture... I just made a new way to say the original conjecture, I found in the book.
Thats all...

[QUOTE=R.D. Silverman;206661]STOP MAKING CONJECTURES. You don't know enough mathematics
to be able to say anything new or meaningful.

This new conjecture is no stronger than the one you quoted.

Neither is new. Both are subsumed by conjectures that really ARE
stronger. Look up Schinzel's Conjecture and the Bateman-Horn Conjecture.

And your notation is lousy. Your "conjecture" is better stated as:

s1 := 2^n R - 1
and
s2 := 2^(n+1) R - 1

are both prime i.o. for some R \in Z depending on n and for all n.
[there are other ways of stating it as well, i,e. s1 and 2s1 + 1 are prime i.o. ] By presenting it as a trinary form as you do, you disguise the
fact that s2 = 2s1 + 1.

It is a simple sub-case of Schinzel's Conjecture.[/QUOTE]

Actually, allow me to give you the following exercize:

Assume that p and 2p+1 are both prime infinitely often.

Use this assumption to prove your "new" conjecture.

Yes, I know that "my" conjecture can be proven by the first conjecture.
The point is that I found it as the same conjecture...
As I wrote, I was playing with the conjecture and found my variation as the same conjecture, just stronger. Thats why this conjecture is trivially proven by the first one and the first one is trivially proven by mine.

[QUOTE=blob100;206670]Yes, I know that "my" conjecture can be proven by the first conjecture.
The point is that I found it as the same conjecture...
As I wrote, I was playing with the conjecture and found my variation as the same conjecture, just stronger.


[/QUOTE]

No. It is weaker. I will leave determining why as an exercize for the
student.

[QUOTE]

Thats why this conjecture is trivially proven by the first one and the first one is trivially proven by mine.[/QUOTE]

The latter statement is not true. Again, I leave it as an exercize.
(hint: think "quantifiers")

Stop MAKING these conjectures, and start ASKING QUESTIONS.

Oh, and let's see your proof.

[QUOTE=R.D. Silverman;206690]No. It is weaker.[/QUOTE]

Is it? It seems to be exactly the same to me.