20151110, 09:29  #155  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×3×7×109 Posts 
Quote:
I'd already given up on getting the right answer from you, when I saw from your later post that you think that "61#" means p_{61}# . So, for everyone's benefit, what N really was is, after all Code:
N=(283#1)*(2*283#1)*(3*283#1)*(4*283#1)*(5*283#1)*(6*283#1)*(7*283#1)*(8*283#1)*(9*283#1)*(10*283#1)*(11*283#1)*(12*283#1)*(13*283#1)*(14*283#1)*(15*283#1)*(16*283#1)*(17*283#1)*(18*283#1)*(19*283#1)*(20*283#1)*(21*283#1)*(22*283#1)*(23*283#1)*(24*283#1)*(25*283#1)*(26*283#1)*(27*283#1)*(28*283#1)*(29*283#1)*(30*283#1)*(31*283#1)*(32*283#1)*(33*283#1)*(34*283#1)*(35*283#1)*(36*283#1)*(37*283#1)*(38*283#1)*(39*283#1)*(40*283#1)*(41*283#1)*(42*283#1)*(43*283#1)+49*283# and 1088*N^3+1 is prime (because N was proven by Schickel) ...No, it isn't  this is just a coincidence. (and now Frank has another prime to prove, as he seems to enjoy doing them... ;) 

20151110, 12:10  #156  
Nov 2003
2^{6}×113 Posts 
Quote:
Quote:
the problem is equivalent to the cutting stock problem in O.R. (or the subset sum problem; take logs to turn it into a summation problem). These are NPC. However, I have seen no DISCUSSION. 

20151110, 16:31  #157  
"Rashid Naimi"
Oct 2015
Remote to Here/There
7×277 Posts 
Quote:
I can not do the runs myself anytime soon. 

20151110, 16:44  #158  
"Rashid Naimi"
Oct 2015
Remote to Here/There
7·277 Posts 
Quote:
I think I was absent the day the teacher talked about that. I have to cut down on my prime fighting hours so here is another try without fixing the code itself. Would this compute? (P_{61}#1 +42 x P_{61}#)!^{(}^{[B][B]P[/B][SUB]61[/SUB][B]#[/B])[/B]} + P_{61}# x 49 And The general format of the expression is: (P_{n}#1 +m x P_{n}#)!^{(}^{[B][B]P[/B][SUB]n[/SUB][B]#[/B])[/B]} + P_{n}# x k for positive integers n, m, and k where k <= P_{n} I would beg to differ on the coincidence claim though. I wouldn't expect that from a mathematician that you seem to be. Think about the statistics of it. Thank you for the reply. Last fiddled with by a1call on 20151110 at 17:29 

20151110, 17:31  #159 
"Curtis"
Feb 2005
Riverside, CA
1175_{16} Posts 
"think about the statistics of it"? Are you saying it *has* to be prime because of some property Serge didn't notice, or that it's merely *likely* to be prime due to its form? If the latter, this method of yours is utterly worthless; we have tons of forms that are "likely" to be prime but still need to be tested for primality.
If the former, you best get about proving how you KNOW it is prime without asking wolfram. 
20151110, 18:02  #160  
"Rashid Naimi"
Oct 2015
Remote to Here/There
7·277 Posts 
Quote:
Quote:


20151110, 20:50  #161 
"Curtis"
Feb 2005
Riverside, CA
41×109 Posts 
Well, then your method is utterly useless for finding large primes, since the form of your numbers doesn't allow for any of the special primality tests that run quickly. You should look into what the largest provenbyECPP prime is on the top 5000 website (primes.utm.edu), and also how long the proofs of those records took.
You might also enjoy quantifying just how much more likely your "astronomically more likely" claim is. If your method produces a list of numbers of, say, 10k digits of which 1% are expected to be prime, is that useful? Compare to what a sieve to 1e12 leaves for such a list; specifically, if I sieve a large pool of 10k digit numbers to 1e12, what percentage of remaining candidates should be prime? How do the candidatesperexpectedprime change at 100k digits for your method vs a sieve? It sounds like all you've done is think up an expression for which numbers are merely more likely to be prime than a sieved list of candidate numbers. While possibly interesting as a curiosity if it's vastly more likely, the lack of provability for candidates even half the size of 5000th place on the top 5000 renders it useless as a way of finding large primes. 
20151110, 21:03  #162 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
a rearrangement on each side turns this into (((m+1)*{{P_n}#}1)!) ^{{P_n}#} +{{P_n}#} * k = a*{{P_n}#} +k*{{P_n}#} doesn't it ? edit: my math even with edits doesn't show up so I am removing the Tex tags
Last fiddled with by science_man_88 on 20151110 at 21:08 
20151110, 21:07  #163  
"Rashid Naimi"
Oct 2015
Remote to Here/There
7×277 Posts 
Quote:
But please try to go through my numerous previous posts in this thread explaining reportedly this is not an attempt at finding a record breaking prime. I have already pointed out why I went through the exercise that I did not that long ago. My ultimate goal is to have a sum converge to less than the square of the largest factor of all the addends. I do not have that as of yet. But I think it is feasible to be formulated which would be the proof of primality for these types of primes. Through all the posts and insults, I have yet to see any valid argument as to why such a formulation can not exist. 

20151110, 22:39  #164 
Aug 2006
1011100110010_{2} Posts 
So as far as I can tell, the idea is to construct a big number B and a small number s and check B + s, B + 2s, B + 3s, ... until a prime is found.
Do you mean that you expect $(a\cdot p\#+b)^c+d$ to be less than $p^2$ for judicious choices of a, b, c, and d? 
20151110, 23:17  #165  
"Rashid Naimi"
Oct 2015
Remote to Here/There
7·277 Posts 
Quote:
The idea is to find a sum made of addends similar to my post 39 which are coprime (for Theorem 1, not so for Theorem 2). The truncated sum (no need to evaluate all digits) can be determined to be smaller than the (largestfactor)^2 (no need to determine the largest prime factor). please let me know if that makes sense. 

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