20161213, 18:42  #34 
Nov 2016
8B6_{16} Posts 
Usually, for all integers a, b with a>1, a<b<a, and a and b are coprime, then there exists odd prime p such that (a^pb^p)/(ab) is prime. However, because of algebra factorization, all numbers of these forms (with odd prime p) are not prime:
(4^p1^p)/(41) (9^p1^p)/(91) (9^p4^p)/(94) (16^p1^p)/(161) (16^p9^p)/(169) (25^p1^p)/(251) (25^p4^p)/(254) (25^p9^p)/(259) (25^p16^p)/(2516) (36^p1^p)/(361) (36^p25^p)/(3625) ... (8^p1^p)/(81) (except p=3) (8^p+1^p)/(8+1) (27^p1^p)/(271) (except p=3) (27^p+1^p)/(27+1) (27^p8^p)/(278) (except p=3) (27^p+8^p)/(27+8) (except p=3) (64^p1^p)/(641) (64^p+1^p)/(64+1) (64^p27^p)/(6427) (except p=3) (64^p+27^p)/(64+27) (125^p1^p)/(1251) (125^p+1^p)/(125+1) (125^p8^p)/(1258) (125^p+8^p)/(125+8) (125^p27^p)/(12527) (125^p+27^p)/(125+27) (except p=3) (125^p64^p)/(12564) (125^p+64^p)/(125+64) (216^p1^p)/(2161) (216^p+1^p)/(216+1) (except p=3) (216^p125^p)/(216125) (216^p+125^p)/(216+125) (except p=3) ... (32^p1^p)/(321) (32^p+1^p)/(32+1) (128^p1^p)/(1281) (except p=7) (128^p+1^p)/(128+1) (except p=7) ... (4^p+1^p)/(4+1) (except p=3) (64^p+1^p)/(64+1) (324^p+1^p)/(324+1) (1024^p+1^p)/(1024+1) (1024^p+81^p)/(1024+81) (1024^p+625^p)/(1024+625) (2500^p+1^p)/(2500+1) (2500^p+81^p)/(2500+81) (2500^p+2401^p)/(2500+2401) (5184^p+1^p)/(5184+1) (5184^p+625^p)/(5184+625) (5184^p+2401^p)/(5184+2401) ... (81^p+4^p)/(81+4) (81^p+64^p)/(81+64) (625^p+4^p)/(625+4) (625^p+64^p)/(625+64) (625^p+324^p)/(625+324) (2401^p+4^p)/(2401+4) (2401^p+64^p)/(2401+64) (2401^p+324^p)/(2401+324) (2401^p+1024^p)/(2401+1024) ... No numbers of these form can be prime. In general, if there exists prime r such that a and b are both rth powers, then no primes can be in the form (a^pb^p)/(ab) except (a^rb^r)/(ab), if there exists two (or more) different primes r, s such that a and b are both rth powers and sth powers, then no primes can be in the form (a^pb^p)/(ab), and if a*b is of the form 4k^4 with integer k, then no primes can be in the form (a^pb^p)/(ab) except the special case (4^3(1)^3)/(4(1)). Last fiddled with by sweety439 on 20161213 at 18:49 
20161213, 21:39  #35 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·2,281 Posts 
If you are retracing the path along the steps of Mike Oakes, JeanLouis Charton, Robert Price, Lelio Paula, Richard Fischer and others  and boldly go where many men went 10+ years ago, for your own edification  that's totally fine.
Just don't call it new. The only reason that you are not seeing anyone on the path is ...because they are miles ahead of you! There is no need to dump fat tables of tiny primes (or composites). Quality, not quantity counts. All small cases are known. Recreating them with a trivial script is a matter of seconds. The only "new" direction to go is up. 
20161214, 12:51  #36  
Nov 2016
4266_{8} Posts 
Quote:
Last fiddled with by sweety439 on 20161214 at 12:51 

20161215, 14:15  #37  
Nov 2016
2·5·223 Posts 
Quote:
Last fiddled with by sweety439 on 20161215 at 14:15 

20161215, 19:01  #38 
Nov 2016
2·5·223 Posts 
The generalized repunit (probable) primes in bases 2<=b<=1025 are in this text file, but there are still 56 nonperfect power bases b without known generalized repunit (probable) prime. All of them are checked to at least p=7000.
Last fiddled with by sweety439 on 20161215 at 19:03 
20161215, 21:35  #39  
"Sam"
Nov 2016
465_{8} Posts 
Quote:
48^5854347^58543 (18^256671)/17 

20161216, 17:57  #40 
Nov 2016
4266_{8} Posts 

20161217, 08:35  #41 
"Sam"
Nov 2016
3·103 Posts 
Also check out https://en.wikipedia.org/wiki/Mersenne_prime for lists of the original Mersenne primes, and all other types of generalizations, such as those of the forms (a^pb^p)/(ab), (a^p+b^p)/(a+b), etc. Also check how closely the primes p (which make (a^pb^p)/(ab) and or (a^p+b^p)/(a+b) prime) are distributed. I will give some of the basic conditions to help you understand more about these forms:
For numbers of the form (a^pb^p)/(ab) integers a, b. (If it is (a^p+b^p)/(ab), b is negative and p is odd) There are infinitely many primes of the form (a^pb^p)/(ab) if (a, b) are coprime, not perfect rth powers, 4ab is not x^4, and a > 1, a < b < a. For all primes p, there are infinitely many (a, b) pairs such that (a^pb^p)/(ab) is prime, followed by this, for any fixed integers a, b, or (ab), and a prime p, there are also infinitely many primes of the form (a^pb^p)/(ab). For each pair of positive integers (a, b) with a > 1, there are a unique set of generalizations for that base which are (a^pb^p)/(ab) (a^p+b^p)/(a+b) ((a^2)^p+(b^2)^p)/(a^2+b^2) ... ((a^(2^n))^p+(b^(2^n))^p)/(a^(2^n)+b^(2^n)) For each prime p and fixed (a, b) values which the conditions listed above hold, there should be infinitely many n values (as above) such that ((a^(2^n))^p+(b^(2^n))^p)/(a^(2^n)+b^(2^n)) is prime. All these are conjectures, as of now, and not proven to be true. 
20161226, 17:48  #42  
Nov 2016
4266_{8} Posts 
Quote:
No (probable) prime found. Now, all bases b<=600 are checked to at least p=10000. Last fiddled with by sweety439 on 20161226 at 18:29 

20161226, 18:28  #43 
Nov 2016
2×5×223 Posts 
I saw factordb and found that all bases b<=1025 are checked to at least 8000, some bases 701<=b<=1025 are even checked to 10000 (theses bases are 752, 861, 866, 872, 881, 932, 956), no primes found with 7001<=p<=8000 except (907^73311)/906, and no primes found with 701<=b<=1025 and 8001<=p<=10000.
Reseving the base 701<=b<=1025 which were not checked to 10000 to 8500 (including bases 711, 713, 731, 759, 771, 795, 820, 938, 948, 951, 963, 996, 1005, 1015), using factordb. 
20161226, 23:26  #44  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,281 Posts 
Quote:
(598^21)/597 is prime. ...Then, of course, (570^129071)/569 is a PRP. I found it in 2 minutes after 1 minute of sieving. Big fat tables are useful only if they contain at least the easiest of all possible results that would potentially make up for the time lost reading. 

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