20100225, 14:14  #221 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2407_{16} Posts 
Karlheinz Brandenburg

20100225, 14:17  #222 
Jan 2010
379 Posts 
Uncwilly,
Did you read about what I wanted to write about? It is sure a good idea but it isn't my direction Last fiddled with by blob100 on 20100225 at 14:20 
20100225, 14:54  #223 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{7}·47 Posts 
Okay, sorry about that. How about "Austin Powers" instead. He is my personal nemesis/hero. And my long lost brother also.
Last fiddled with by retina on 20100225 at 14:57 
20100225, 15:21  #224 
Jan 2010
379 Posts 
If it is your hero, write them a letter (you can't participate in the "name your hero" competition).

20100225, 17:54  #225 
Jan 2010
379 Posts 
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 such that is also a prime, where Q is not neccesarily a prime number. Last fiddled with by blob100 on 20100225 at 17:55 
20100225, 18:56  #226  
Nov 2003
2^{2}×5×373 Posts 
Quote:
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 BatemanHorn 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 subcase of Schinzel's Conjecture. 

20100225, 19:01  #227 
Jan 2010
101111011_{2} Posts 
I didn't conjecture... I just made a new way to say the original conjecture, I found in the book.
Thats all... 
20100225, 19:02  #228  
Nov 2003
2^{2}×5×373 Posts 
Quote:
Assume that p and 2p+1 are both prime infinitely often. Use this assumption to prove your "new" conjecture. 

20100225, 19:44  #229 
Jan 2010
379 Posts 
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. 
20100225, 22:21  #230  
Nov 2003
2^{2}·5·373 Posts 
Quote:
student. Quote:
(hint: think "quantifiers") Stop MAKING these conjectures, and start ASKING QUESTIONS. Oh, and let's see your proof. 

20100226, 06:06  #231 
Aug 2006
2·11·271 Posts 

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