Searching for generalized repunit PRP

[sweety439]

Is there a project to search the (probable) primes of the form (b^n-1)/(b-1) (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.

[sweety439]

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.

[sweety439]

Also, the link is the searching for primes of the form (a+1)^n-a^n: http://www.fermatquotient.com/PrimSerien/PrimPot.txt. More generally, is there a project to search for primes of the forms (a^n-b^n)/(a-b) with a>1, -a<b<a, prime n, and a and b are coprime? If b=1, this equals (a^n-1)/(a-1), if b=-1, this equals (a^n+1)/(a+1) (if n is odd), and if a-b=1, this equals a^n-(a-1)^n.

[paulunderwood]

Some are cofactors...

(a^p-1)/n

(a^p+1)/n

(a^p-b^p)/n

(a^p+b^p)/n

[carpetpool]

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^n-1)/(b-1) or (a+1)^n-a^n. As far as I know for both projects, All prime bases b < 10000 (for (b^n-1)/(b-1)) are tested to 20k digits. PRP factors found and tested.
FYI All bases a < 100 (for (a+1)^n-a^n) are tested to 20k digits. PRP factors found and tested.

Hope this helps. Check out here.

[sweety439]

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.

[sweety439]

These are files that I found the smallest odd prime p such that (b^p-1)/(b-1) or (b^p+1)/(b+1) is (probable) prime for 2<=b<=1025, and the smallest prime p such that (b+1)^p-b^p is (probable) prime for 1<=b<=1024. (not all (probable) primes are founded by me)

[sweety439]

Also, I have searched the smallest odd prime p such that (a^p-b^p)/(a-b) 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)

[Batalov]

'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^a-1 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!

[sweety439]

No, it is (a^p-b^p)/(a-b), not 2^a-1. This a is the base, not the exponent.

The file in that post lists the smallest odd prime p such that (a^p-b^p)/(a-b) 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).

[Batalov]

(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.)