Frequency of REPUNIT primes

rudy235 · 2021-04-26, 04:39

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

e^gamma *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?

slandrum · 2021-04-26, 06:17

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

rudy235 · 2021-04-26, 10:49

Slight correction:

Quote:

e^gamma *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

Dr Sardonicus · 2021-04-26, 13:11

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

$\frac{x^{np}-1}{x^{n}-1}\;=\;\frac{\prod_{d\mid np}\Phi_{d}(x)}{\prod_{\delta\mid n}\Phi_{\delta}(x)}$

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).

rudy235 · 2021-04-26, 13:34

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

Dr Sardonicus · 2021-04-26, 14:00

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.

Batalov · 2021-04-26, 23:17

It is easy. It is ... $k * e^\gamma * log_2 {y}$,
where k is a koefficient

rudy235 · 2021-04-27, 00:29

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

sweety439 · 2021-05-28, 02:17

Currently there are 32 bases 2<=b<=1024 without known generalized repunit (probable) primes > R2: (exclude the bases with no possible generalized repunit primes > R2)

{185, 269, 281, 380, 384, 385, 394, 396, 452, 465, 511, 574, 598, 601, 631, 632, 636, 711, 713, 759, 771, 795, 861, 866, 881, 938, 948, 951, 956, 963, 1005, 1015}

See the text file for the status for bases 2<=b<=1024

References for list of generalized repunit (probable) primes sorted by base:

Bases 2 to 160 and some larger bases of the form m^r+1

Bases 2 to 999 (broken link: from wayback machine cached copy)

2021-04-26, 04:39	#1
rudy235 Jun 2015 Vallejo, CA/. 1111100010₂ 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 e^gamma *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 11·13 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 2021-04-26 at 06:43

2021-04-26, 23:17	#7
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 10010100000100₂ Posts	It is easy. It is ... \(k * e^\gamma * log_2 {y}\), where k is a koefficient

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2021-04-26, 13:11	#4
Dr Sardonicus Feb 2017 Nowhere 11043₈ 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 $\frac{x^{np}-1}{x^{n}-1}\;=\;\frac{\prod_{d\mid np}\Phi_{d}(x)}{\prod_{\delta\mid n}\Phi_{\delta}(x)}$ 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).