20210426, 04:39  #1 
Jun 2015
Vallejo, CA/.
5×197 Posts 
Frequency of REPUNIT primes
Frequency of REPUNITS
Generalized repunits are \({a^n1} \over {a1}\) Thus if a =2 we have the Mersenne Numbers The Mersenne Exponent primes are 2, 3, 5, 7, 13, 17, 19, 31, 61, 89..... http://oeis.org/A000043 If a =3 we have the following Exponent primes 3, 7, 13, 71, 103, 541, 1091, 1367, 1627, 4177, 9011, 9551, 36913, 43063, 49681, 57917, 483611, 877843, 2215303 http://oeis.org/A028491 If a =4 we have only this prime Exponent 2 which produces "5" if a =5 we have the following Exponent primes 3, 7, 11, 13, 47, 127, 149, 181, 619, 929, 3407 https://oeis.org/A004061 If a =6 we have these Exponent primes 2, 3, 7, 29, 71, 127, 271, 509, 1049, 6389, 6883, 10613, 19889, 79987, 608099, 1365019 http://oeis.org/A028491 We can ontinue to a=7,10,11,12... 8 and 9 are very disappointing 8 has 1 and 9 has none. https://oeis.org/A004063. for 7 http://oeis.org/A004023 for 10 http://oeis.org/A005808 for 11 https://oeis.org/A004064 for 12 So, at this point, we have a theory about the frequency of Mersenne Exponents called the Lenstra–Pomerance–Wagstaff conjecture which states that the number of Mersenne primes with exponent p less than y is asymptotically e^{gamma} *log_{2}(y) Which produces an estimate of 45 Mersenne numbers under 82,258,9933 which is quite close to the actual number of 51. GIVEN THAT FOR BASE 10 (Actual REPUNITS) the number of exponents n that produce a Prime Repunit under 5,79477 is only 9 at present, is there a generalization of the Lenstra–Pomerance–Wagstaff conjecture that could be adapted to other Generalised Repunits? or at least Specifically to the (Base 10) Repunits? 
20210426, 06:17  #2 
Jan 2021
California
2·5·7 Posts 
It's not really surprising or disappointing that the bases that are powers have 0 or 1 repunit primes.
(4^{p}1)/3 = (2^{p}1)(2^{p}+1)/3 (8^{p}1)/7 = (2^{p}1)(4^{p}+2^{p}+1)/7 Last fiddled with by slandrum on 20210426 at 06:43 
20210426, 10:49  #3  
Jun 2015
Vallejo, CA/.
985_{10} Posts 
Slight correction:
Quote:


20210426, 13:11  #4 
Feb 2017
Nowhere
10701_{8} Posts 
Elaborating on the earlier post explaining the paucity of prime repunits to bases which are perfect powers:
If n > 1, taking the base a = c^n, c > 1 not a q^{th} power for any prime q, taking p prime and substituting x = c into the cyclotomic polynomial identity automatically produces more than one proper factor unless n = p^k, a power of the prime p. In that case, we have Thus with a = 8 = 2^3, the only candidate repunit prime is (8^3  1)/(8  1) = 73, and with a = 2^9, the only candidate is (2^27  1)/(2^9  1). 
20210426, 13:34  #5  
Jun 2015
Vallejo, CA/.
5×197 Posts 
Quote:
What I would love to understand is how to extend the Lenstra–Pomerance–Wagstaff conjecture to those bases which are not perfect powers. like 10, 5, 6 and 12 

20210426, 14:00  #6 
Feb 2017
Nowhere
4545_{10} Posts 

20210426, 23:17  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9,433 Posts 
It is easy. It is ... \(k * e^\gamma * log_2 {y}\),
where k is a koefficient 
20210427, 00:29  #8  
Jun 2015
Vallejo, CA/.
1111011001_{2} Posts 
Quote:
So for 2 k= 1 What are then the values of k for 3, 4, 5, 6, 7, 10 Last fiddled with by rudy235 on 20210427 at 00:30 Reason: Forgot “3” 

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