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 2021-04-26, 04:39 #1 rudy235     Jun 2015 Vallejo, CA/. 5×197 Posts Frequency of REPUNIT primes Frequency of REPUNITS Generalized repunits are $${a^n-1} \over {a-1}$$ 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 egamma *log2(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?
 2021-04-26, 06:17 #2 slandrum   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. (4p-1)/3 = (2p-1)(2p+1)/3 (8p-1)/7 = (2p-1)(4p+2p+1)/7 Last fiddled with by slandrum on 2021-04-26 at 06:43
2021-04-26, 10:49   #3
rudy235

Jun 2015
Vallejo, CA/.

98510 Posts

Slight correction:
Quote:
 egamma *log2(y) Which produces an estimate of 52 (rounding down) Mersenne numbers under 82,258,9933 which is very close to the actual number of 51

 2021-04-26, 13:11 #4 Dr Sardonicus     Feb 2017 Nowhere 107018 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 qth power for any prime q, taking p prime and substituting x = c into the cyclotomic polynomial identity $\frac{x^{np}-1}{x^{n}-1}\;=\;\frac{\prod_{d\mid np}\Phi_{d}(x)}{\prod_{\delta\mid n}\Phi_{\delta}(x)}$ automatically produces more than one proper factor unless n = p^k, a power of the prime p. In that case, we have $\frac{(x^{p^{k+1}} - 1)}{(x^{p^{k}} - 1)}\; =\; \Phi_{p^{k+1}}(x)$ 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).
2021-04-26, 13:34   #5
rudy235

Jun 2015
Vallejo, CA/.

5×197 Posts

Quote:
 Originally Posted by Dr Sardonicus Elaborating on the earlier post explaining the paucity of prime repunits to bases which are perfect powers: [SNIP] ...automatically produces more than one proper factor unless n = p^k, a power of the prime p. In that case, we have $\frac{(x^{p^{k+1}} - 1)}{(x^{p^{k}} - 1)}\; =\; \Phi_{p^{k+1}}(x)$ 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).
I fully understand the reasons behind the paucity of prime repunits to bases that are perfect powers.

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

2021-04-26, 14:00   #6
Dr Sardonicus

Feb 2017
Nowhere

454510 Posts

Quote:
 Originally Posted by rudy235 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
I'm not certain, but I'm guessing you've got plenty of company.

 2021-04-26, 23:17 #7 Batalov     "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
2021-04-27, 00:29   #8
rudy235

Jun 2015
Vallejo, CA/.

11110110012 Posts

Quote:
 Originally Posted by Batalov It is easy. It is ... $$k * e^\gamma * log_2 {y}$$, where k is a koefficient
Ok Serge. Thanks
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 2021-04-27 at 00:30 Reason: Forgot “3”

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