mersenneforum.org Repunit exponent mersenne numbers
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 2020-10-10, 20:10 #1 Viliam Furik   Jul 2018 Martin, Slovakia 110111112 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?
 2020-10-16, 10:33 #2 sweety439     Nov 2016 1001001111102 Posts A general form is R(R(p,a),b), where a>=2, b>=2 are integers (define R(p,b) = (b^p-1)/(b-1) = 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 non-perfect-power bases b.
 2020-10-16, 11:28 #3 Viliam Furik   Jul 2018 Martin, Slovakia 223 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 base-10 repunit prime. I have found that M(REP(23)) = M11111111111111111111111 has a factor.
 2020-10-17, 17:38 #4 Viliam Furik   Jul 2018 Martin, Slovakia 223 Posts Status update REP(1031) -> 3435 bits (13 bits above k=1)

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