mersenneforum.org Sums of all Squares
 Register FAQ Search Today's Posts Mark Forums Read

2010-12-14, 22:12   #155
lavalamp

Oct 2007
London, UK

52116 Posts

I still have no idea what you meant.

Quote:
 Originally Posted by science_man_88 I forgot to mention p^2-q^2 can be made into 2* (the sum of the numbers (q-p)) -(p+q)
Please use TeX, that's what it's for:
$p^{2} - q^{2} = 2*\left(\sum_{i=q}^{p}i\right) - p - q$

 2010-12-14, 22:21 #156 science_man_88     "Forget I exist" Jul 2009 Dumbassville 836910 Posts $p^{2} - q^{2} = 2*\left(\sum_{i=q}^{p}i\right) -(p+q)$ this is more accurate to what i said.
2010-12-14, 22:55   #157
kar_bon

Mar 2006
Germany

22·23·31 Posts

Quote:
 Originally Posted by science_man_88 this is more accurate to what i said.
It's the same:

-(p+q) = (-1)*(p+q) = ((-1)*p + (-1)*q) = ((-p) + (-q)) = -p -q

2010-12-15, 04:36   #158
CRGreathouse

Aug 2006

2·2,969 Posts

Quote:
 Originally Posted by lavalamp This program is for use (ideally) by people who already know what they're doing, I don't see why these specific loops would need to be hidden. On a personal note, I would prefer knowing what code is going to do when I run it, rather than be unsure because I don't happen to know the current primelimit.
On the other hand, you don't know what happens when you run isprime() or factor(). In fact most functions have this kind of hidden stuff going on and don't expose it.

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:
 Originally Posted by lavalamp It would be possible to get the best of both worlds though, have a generic loop that can call whichever loop seems most appropriate, and the numerous specific loops for people who already know what they want, it only takes a few lines in the documentation.
That's a possibility. It's sort of the approach used by factor: there's an extra function factorint() where you can customize it to some degree with the optional flag argument.

Quote:
 Originally Posted by lavalamp If there's no way (or it is at least very hard) to make the forbigprime loop efficient AND able to jump around with the index, then it seems reasonable to deny the ability to change it, for that loop only.
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.)

2010-12-15, 04:40   #159
CRGreathouse

Aug 2006

2·2,969 Posts

Quote:
 Originally Posted by science_man_88 if theres a easy formula for calculating the distance between primes that doesn't use a random n in 6n I could see a way to extend it the way it is, but using a shift in vectors.
Depending on what you mean by "easy" and on how far apart the primes are, there might be.

2010-12-15, 12:34   #160
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by CRGreathouse Depending on what you mean by "easy" and on how far apart the primes are, there might be.
I think we could use p^2=((p-q)^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 2010-12-15 at 12:35

2010-12-15, 14:27   #161
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

202618 Posts

Quote:
 Originally Posted by science_man_88 I think we could use p^2=((p-q)^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.
composites.

 2010-12-15, 21:16 #162 davar55     May 2004 New York City 2·2,099 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?
 2010-12-17, 19:38 #163 davar55     May 2004 New York City 2×2,099 Posts So what should we call this particular sequence? I suspect it will turn up hidden numeric treasures.
2010-12-28, 01:12   #164
davar55

May 2004
New York City

2·2,099 Posts

From above:

Quote:
 Originally Posted by davar55 While this is current, any other mathematicians here want to comment on whether this sequence is provably infinite? Perhaps by a density heuristic argument?
Quote:
 Originally Posted by CRGreathouse It's heuristically infinite with the n-th term about 2.3n * 10^n. I doubt it can be proven infinite with present technology, but a conditional proof on the k-tuple conjecture seems plausible (though it would give bounds wildly out of proportion with the true size of the terms).
If we prove that enough sequences' infinitude (like this one) depends on that k-tuple conjecture, then proving even one of them infinite would
be evidence that all are - including the mersennes.

Last fiddled with by davar55 on 2010-12-28 at 01:21

2010-12-28, 01:35   #165
CRGreathouse

Aug 2006

173216 Posts

Quote:
 Originally Posted by davar55 If we prove that enough sequences' infinitude (like this one) depends on that k-tuple conjecture, then proving even one of them infinite would be evidence that all are - including the mersennes.
I'm skeptical. The proof I had in mind wouldn't generalize to anything like Mersenne numbers. It would just set up lots of numbers such that (because of congruence conditions) omitted interior numbers could not be prime and such that the numbers were candidates for the prime tuple conjecture, and such that the sum of the first 1, 2, ..., N numbers to their own powers would cover all congruence classes mod 10^n.

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.

 Similar Threads Thread Thread Starter Forum Replies Last Post a1call Miscellaneous Math 42 2017-02-03 01:29 Nick Number Theory Discussion Group 0 2016-12-11 11:30 3.14159 Miscellaneous Math 12 2010-07-21 11:47 CRGreathouse Math 6 2009-11-06 19:20 m_f_h Puzzles 45 2007-06-15 17:46

All times are UTC. The time now is 15:50.

Tue Nov 24 15:50:56 UTC 2020 up 75 days, 13:01, 4 users, load averages: 1.63, 1.73, 1.79