20201010, 20:10  #1 
"Viliam Furík"
Jul 2018
Martin, Slovakia
5^{2}×13 Posts 
Repunit exponent mersenne numbers
I have TFed following repunit exponents to respective bit levels:
REP(19) > 89 bits (18 bits above k=1) REP(23) > 100 bits (~27 bits above k=1) REP(317) > 1070 bits (20 bits above k=1) For REP(23) I found a factor 5246666666666666666666614201, 92.083 bits, k = 236100. Has there been a search for factors of these numbers already? 
20201016, 10:33  #2 
Nov 2016
2^{2}×691 Posts 
A general form is R(R(p,a),b), where a>=2, b>=2 are integers (define R(p,b) = (b^p1)/(b1) = generalized repunit base b with length p)
These are the smallest prime R(p,b) with prime p>2 in all bases b<=1024: https://mersenneforum.org/showpost.p...&postcount=228, for perfect power bases, there are no possible primes except a very low prime p because of algebra factors, there are only 32 bases b<=1024 remain with no known repunit (probable) primes: {185, 269, 281, 380, 384, 385, 394, 396, 452, 465, 511, 574, 598, 601, 629, 631, 632, 636, 711, 713, 759, 771, 795, 861, 866, 881, 938, 948, 951, 956, 963, 1005, 1015}, all searched to p=100K The condition of R(R(p,a),b) is prime is R(p,a) is prime, and hence the bases a and b must not be perfect powers, however, for any integer pair (a,b), it is conjectured that only finitely many primes of the form R(R(p,a),b), since it is double exponent form, unlike R(p,b), which is conjectured to have infinitely many primes for all nonperfectpower bases b. 
20201016, 11:28  #3 
"Viliam Furík"
Jul 2018
Martin, Slovakia
145_{16} Posts 
Uhm, I'm sorry, but I think you don't understand what the thread is about. I am searching for factors of Mersenne numbers which exponent is a base10 repunit prime.
I have found that M(REP(23)) = M11111111111111111111111 has a factor. 
20201017, 17:38  #4 
"Viliam Furík"
Jul 2018
Martin, Slovakia
5^{2}·13 Posts 
Status update
REP(1031) > 3435 bits (13 bits above k=1)

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