2020-10-17, 06:55 | #1 |
Nov 2016
2^{2}·691 Posts |
Some notable smallest repunit/nexus prime
Let
a(b) = Smallest prime p such that the generalized repunit number (b^p-1)/(b-1) is prime b(p) = Smallest base b such that the generalized repunit number (b^p-1)/(b-1) is prime c(b) = Smallest prime p such that the nexus number (b+1)^p-b^p is prime d(p) = Smallest base b such that the nexus number (b+1)^p-b^p is prime a'(b) = Smallest prime p such that the generalized nega-repunit number (b^p+1)/(b+1) is prime b'(p) = Smallest base b such that the generalized nega-repunit number (b^p+1)/(b+1) is prime c'(b) = Smallest prime p such that the nega-nexus number ((b+1)^p+b^p)/(2b+1) is prime d'(p) = Smallest base b such that the nega-nexus number ((b+1)^p+b^p)/(2b+1) is prime Then b(109) = d(109) = b'(109) = d'(109) = 12 b(317) = d(317) = 10 b(37) = 61 and d(37) = 39 are both larger than near numbers (i.e. b(p) and d(p), where p is prime and <= 61) b(37) = b'(41) = 61, but b'(37) and b(41) are small (between 2^3 and 2^4): b'(37) = 16, b(41) = 14 b(67) = b'(71) = 46, but b'(67) and b(71) are very small (less than 2^3): b'(67) = 5, b(71) = 3 b(p) = 2 and d(p) = 1 for primes p such that 2^p-1 is prime (of course): 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, ..., the first two primes not including here are 11 and 23, because they are Sophie Germain primes == 3 mod 4, thus 2^p-1 is divisible by 2p+1 and cannot be prime, also, for such primes p, (b^p-1)/(b-1) for 3-smooth bases b are also divisible by 2p+1, and also cannot be prime we have d(11) = d(23) = 5 (for such primes p, (b+1)^p-b^p is divisible by 2p+1 for b = 1, 2, 8, thus d(p) >= 4) The Mersenne exponents are 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, ..., and the Wagstaff exponents are 3, 5, 7, 11, 13, 17, 19, 23, 31, 43, 61, 79, 101, 127, the differences are: * Mersenne has 2 (of course Wagstaff cannot have 2, since the number is not integer) * Wagstaff has the two numbers in the previous section: 11 & 23 * Wagstaff has an additional Wagstaff prime W7 = 43 (also W5 = 11), note that both Mersenne and Wagstaff have the Mersenne primes M5 = 31 and M7 = 127) * Both Mersenne and Wagstaff has two additional numbers between 2^6 and 2^7: Mersenne has 89 & 107, Wagstaff has 79 & 101 b'(79) = b'(101) = 2, and b(79) = b(101) = 22 if we do not allow p = 2, then a(18) = 25667 and c(18) = 1607 are both very large, also for a(96) = 3343 and c(96) = 1303 for the prime 239, b(239) = 223 and a(223) = 239, also d(239) = 367 and c(367) = 239, note that 223 and 367 are also primes b(179), b'(181), d(181), d'(179) are large, but b'(179), b(181), d'(181), d(179) are small (b^p-1)/(b-1) is divisible by p if p = b-1, such numbers are primes when divided by p for b = 3, 4, 6, 18, 4358, ... (b^p+1)/(b+1) is divisible by p if p = b+1, such numbers are primes when divided by p for b = 4, 18, ... ((b+1)^p+b^p)/(2b+1) is divisible by p if p = 2b+1, such numbers are primes when divisible by p for b = 2, 3, 6, 18, ... Last fiddled with by sweety439 on 2020-10-17 at 07:02 |
2020-10-17, 09:38 | #2 |
Nov 2016
2^{2}·691 Posts |
M5 = 31, M7 = 127, W5 = 11, W7 = 43
It is known that MM5, MM7, WW5, WW7 are primes However, MMM5 and WWW5 are both known to be composite Two open problems: Is MMM7 prime? Is WWW7 prime? I guess no, MM7 is the largest double-Mersenne prime and WW7 is the largest double-Wagstaff prime Also, 127 is the largest number which satisfy all three conditions of New Mersenne Conjecture. By this conjecture, WM5 and WM7 are primes, however, MW5 and MW7 are not primes (if they are primes, then they will disprove this conjecture :)) (since W5 = 11 and W7 = 43 are not of the form 2^n+-1 or 4^n+-3) These six numbers are also composite: WMM5, MWM5, WWM5, MMW5, WMW5, MWW5 However, no known status for these six numbers for 7 instead of 5 By the way, M7 and W7 are the largest known primes such that Phi(p*znorder(p,2),2)/p is prime (the other known such primes are 3, 5, 7, 19, they also satisfy that Mp and W7 are primes, however, W7 does not satisfy that) For Phi(p^2*znorder(p,2),2)/p, the only known such primes are 3 and 7, and no known primes for Phi(p^3*znorder(p,2),2)/p and above |
2020-10-25, 09:39 | #3 |
Nov 2016
2^{2}·691 Posts |
* If p is larger, then Rp(b) is harder to be prime, the smallest base b such that Rp(b) is prime is about gamma*p (gamma is the Euler constant 0.577215664901...)
In general .... (let Rp(b) = (b^p-1)/(b-1)) * If 2p+1 is prime (i.e. p is Sophie Germain prime), then Rp(b) is harder to be prime, since half of the bases b should be removed as Rp(b) is divisible by 2p+1 * If 4p+1 is prime, then a quarter of the bases b should be removed * If 6p+1 is prime, then a sixth of the bases b should be removed * If 8p+1 is prime, then an eighth of the bases b should be removed etc. Also, it seems that if p is irregular prime (37, 59, 67, 101, 103, 131, 149, 157, 233, 257, 263, 271, 283, 293, 307, 311, 347, 353, 379, 389, 401, 409, 421, 433, 461, 463, 467, 491, 523, 541, 547, 557, 577, 587, 593, ...), then Rp(b) (also Rp(b)/(2*p+1), Rp(b)/(4*p+1), Rp(b)/(6*p+1), Rp(b)/(8*p+1), etc. if Rp(b) is divisible by 2(4,6,8,etc.)*p+1) is harder to be prime (than the regular primes nearby to p), e.g. for p=37, the smallest base b such that Rp(b) is prime is 61, which is much larger than the primes nearby to it (29 -> 6, 31 -> 2, 41 -> 14, 43 -> 15), and for p = 67, the smallest base b such that Rp(b) is prime is 46, which is also larger than the primes nearby to it (53 -> 24 (this is because 2*53+1 is prime, then 24>19), 59 (also irregular) -> 19, 61 -> 2, 71 -> 3, 73 -> 11) Note that Mersenne thinks R67(2) and R257(2) are primes, but they are composite, and 67 and 257 are irregular primes |
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