20101214, 22:12  #155 
Oct 2007
Manchester, UK
10100111101_{2} Posts 

20101214, 22:21  #156 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
this is more accurate to what i said. 
20101214, 22:55  #157 
Mar 2006
Germany
2×1,439 Posts 

20101215, 04:36  #158  
Aug 2006
2×29×103 Posts 
Quote:
Speaking of which I was just trying (unsuccessfully?) to get the developers to expose a particular function, or at least a specialization of it (essentially, a function to find a nontrivial factor of a number). Quote:
That's how I'd prefer to do it. That way I wouldn't have to check the argument at each step. Frankly, if you want to mess with that you should have to write your own code to handle it. (Who knows, maybe for that application there's a good way to deal with it.) 

20101215, 04:40  #159 
Aug 2006
13526_{8} Posts 
Depending on what you mean by "easy" and on how far apart the primes are, there might be.

20101215, 12:34  #160 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
I think we could use p^2=((pq)^2) mod q to eliminate composites there are ways we could use this like for example the difference between odd primes is always 2x for some x so ((2x)^2) mod q can be used which turns to to 4x^2 mod q. that used with the fact that all odd primes have squares 1 mod 2 and we cover all primes. Though this might be barking up the wrong tree.
Last fiddled with by science_man_88 on 20101215 at 12:35 
20101215, 14:27  #161  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20101215, 21:16  #162 
May 2004
New York City
2^{3}·23^{2} Posts 
If you're computing p^p_sum_mods to 10^12, how many digits
will the two highest successful values (new multiples of powers of ten) have, and will these numbers be equal? 
20101217, 19:38  #163 
May 2004
New York City
2^{3}·23^{2} Posts 
So what should we call this particular sequence?
I suspect it will turn up hidden numeric treasures. 
20101228, 01:12  #164  
May 2004
New York City
1000010001000_{2} Posts 
From above:
Quote:
Quote:
be evidence that all are  including the mersennes. Last fiddled with by davar55 on 20101228 at 01:21 

20101228, 01:35  #165  
Aug 2006
2·29·103 Posts 
Quote:
The hard part is the first of the three conditions; aside from that the conditional proof would be easy. But it wouldn't bring us any closer to an understanding of the infinitude or density of Mersenne primes. Frankly, I expect that showing that any reasonably natural exponential sequence (like Mersenne numbers) has infinitely many primes will be harder even than the prime tuple conjecture, which is mercifully linear. 

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