20161206, 14:52  #1 
Nov 2016
2496_{10} Posts 
Searching for generalized repunit PRP
Is there a project to search the (probable) primes of the form (b^n1)/(b1) (generalized repunit base b) or (b^n+1)/(b+1) (generalized repunit base b)? There are links to this, http://www.fermatquotient.com/PrimSerien/GenRepu.txt (base b for 2<=b<=152)
http://www.fermatquotient.com/PrimSerien/GenRepuP.txt (base b for 2<=b<=152) These are text files I searched the repunit PRP for b up to +257 and n up to 10000. 
20161206, 14:58  #2 
Nov 2016
2^{6}·3·13 Posts 
Some bases have algebra factors.
These bases are numbers of the form k^r with integer k != 1, 0, 1 and integer r > 1, and numbers of the form 4k^4 with integer k > 0. The list of all such bases are ..., 1024, 1000, 729, 512, 343, 243, 216, 125, 64, 32, 27, 8, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, ... and ..., 5184, 2500, 1024, 324, 64, 4. 
20161206, 15:03  #3 
Nov 2016
2^{6}·3·13 Posts 
Also, the link is the searching for primes of the form (a+1)^na^n: http://www.fermatquotient.com/PrimSerien/PrimPot.txt. More generally, is there a project to search for primes of the forms (a^nb^n)/(ab) with a>1, a<b<a, prime n, and a and b are coprime? If b=1, this equals (a^n1)/(a1), if b=1, this equals (a^n+1)/(a+1) (if n is odd), and if ab=1, this equals a^n(a1)^n.

20161206, 15:38  #4 
Sep 2002
Database er0rr
3^{2}·389 Posts 

20161207, 01:35  #5 
"Sam"
Nov 2016
2×3×53 Posts 
I don't think there is a project (just yet) but I am starting my own (I don't know if you'd find it interesting or not) for find PRP factors of numbers of the form (b^n1)/(b1) or (a+1)^na^n. As far as I know for both projects, All prime bases b < 10000 (for (b^n1)/(b1)) are tested to 20k digits. PRP factors found and tested.
FYI All bases a < 100 (for (a+1)^na^n) are tested to 20k digits. PRP factors found and tested. Hope this helps. Check out here. Last fiddled with by carpetpool on 20161207 at 02:08 
20161207, 11:26  #6 
Nov 2016
2^{6}×3×13 Posts 
Before, I used factordb.com to find all repunit PRPs with 152<=b<=257 and 257<=b<=152 and n<=10000. Since in http://www.fermatquotient.com/, it was already searched for 2<=b<=151 and 151<=b<=2.

20161207, 11:35  #7 
Nov 2016
2^{6}·3·13 Posts 
These are files that I found the smallest odd prime p such that (b^p1)/(b1) or (b^p+1)/(b+1) is (probable) prime for 2<=b<=1025, and the smallest prime p such that (b+1)^pb^p is (probable) prime for 1<=b<=1024. (not all (probable) primes are founded by me)

20161207, 11:46  #8 
Nov 2016
2^{6}×3×13 Posts 
Also, I have searched the smallest odd prime p such that (a^pb^p)/(ab) is prime for 1<a<=50, a<b<a, and a and b are coprime. (for (a^p+b^p)/(a+b), note that since p is odd, it equals (a^p(b)^p)/(a(b)))
All of the "NA" terms were searched to at least p=10000. (0 if no possible prime) Last fiddled with by sweety439 on 20161207 at 11:48 
20161207, 16:03  #9 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10001111000111_{2} Posts 
'Those who don't know history are doomed to repeat it.' (George Santayana)
Would you please stop reposting in every possible forum thread trivial results that are known for a century? Sadly you forgot to post in http://mersenneforum.org/forumdisplay.php?f=24 that for 1<a<=50, 2^a1 is prime for 2, 3, 5, 7, 13, 17, 19, 31. Yeah, yeah, with a little effort you could also post that this is OEIS sequence A000043. Thanks a lot! 
20161207, 18:12  #10 
Nov 2016
4700_{8} Posts 
No, it is (a^pb^p)/(ab), not 2^a1. This a is the base, not the exponent.
The file in that post lists the smallest odd prime p such that (a^pb^p)/(ab) is prime. (for 1<a<=50, a<b<a, and a and b are coprime) These are (a, b) that I found no primes (to searched to at least p=10000): (32, 5), (43, 7), (44, 43), (46, 11), (46, 31), (47, 33), (49, 46), (49, 46), (50, 43), (50, 37). Last fiddled with by sweety439 on 20161207 at 18:15 
20161210, 00:46  #11 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×43×71 Posts 
(49^16747+46^16747)/95 is a PRP ... Too small for PRPtop  the limit was raised to 30000 digits.
(46^45281+11^45281)/57 is a PRP Btw, I wonder if you know how to sieve them properly. Or is it "up to the first 10000 primes"? (Hint: most of the small primes will not divide any of these candidates.) Last fiddled with by Batalov on 20161210 at 22:40 
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