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20210624, 02:29  #1 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·7·263 Posts 
Interesting properties about Mersenne(related) exponents and Wagstaff(related) exponents
Mersenne number (Mp, 2^p1) and Wagstaff number (Wp, (2^p+1)/3):
3 (2^21, both Mp and Wp primes) 5 (2^2+1, both Mp and Wp primes) 7 (2^31 or 4^1+3, both Mp and Wp primes) 11 (a Sophie Germain prime == 3 mod 4 (and >3), thus Mp cannot be prime (divisible by 2*p+1), but Wp is still prime) (Mp is not prime, but Wp is still prime, this is because 11 is itself a Wagstaff prime) 13 (4^23, both Mp and Wp primes) 17 (2^4+1, both Mp and Wp primes) 19 (4^2+3, both Mp and Wp primes) 23 (a Sophie Germain prime == 3 mod 4 (and >3), thus Mp cannot be prime (divisible by 2*p+1), but Wp is still prime) 29 (a Sophie Germain prime == 1 mod 4 (and >5), thus Wp cannot be prime (divisible by 2*p+1)) 31 (2^51, both Mp and Wp primes)  2^5  43 (Mp is not prime, but Wp is still prime, this is because 43 is itself a Wagstaff prime) 61 (2^63, both Mp and Wp primes)  2^6  67 (2^6+3, neither Mp not Wp is prime, this is because 67 is irregular prime, this number is the prime which Mersenne makes fault) [between 67 and 127, there are two additional numbers makes Mp prime (89 and 107) and two additional numbers makes Wp prime (79 and 101)] 127 (2^71, both Mp and Wp primes)  2^7  [between 2^7 and 2^8, there is no primes makes Mp prime, but there are three primes makes Wp primes, all of them are regular primes]  2^8  257 (2^8+1, neither Mp not Wp is prime, this is because 257 is irregular prime, this number is the prime which Mersenne makes fault) Originally, Mersenne may think that 2^p  1 is prime if and only if p = 2^k ± 1 or p = 4^k ± 3 for some natural number k, now this become the New Mersenne Conjecture, i.e. (p = 2^k ± 1 or p = 4^k ± 3) + (Mersenne number 2^p1 is prime) + (Wagstaff number (2^p+1)/3 is prime) is never 2 Double Mersenne number (MMp) and double Wagstaff number (WWp): MMp and WWp are primes for p < 2^3 (except WW2, which is not integer) but not for 2^3 < p < 2^5 MMp is prime for p = 2, 3, 5, 7, but not for p = 13, 17, 19, 31 WWp is prime for p = 3, 5, 7, but not for p = 11, 13, 17, 19, 23, 31 MM127 may be prime, since 127 = M7, note that for M5 = 31, MM31 is not prime, but for M3 = 7, MM7 is prime WW43 may be prime, since 43 = W7, note that for W5 = 11, WW11 is not prime, but for W3 = 3, WW3 is prime I conjectured that MMp and WWp are composite for p>7, except p = M7 (=127) and p = W7 (=43) note that for p = M5 (=31) and p = W5 (=11), both numbers are composite for Mersenne and Wagstaff numbers with exponent Fermat primes (i.e. MFn and WFn), they are primes for 3, 5, 17, but composite for 257 and 65537 Mersennerelated number (2^(p^(r+1))1)/(2^(p^r)1) and Wagstaffrelated number (2^(p^(r+1))+1)/(2^(p^r)+1): the Wagstaffrelated number has trivial factor 3 for p=3, thus we take out this factor when p=3 the r=0 case is the original Mersenne/Wagstaff number, for r=1: the two numbers are both primes for p=3 and p=7, but for p=5, they are both composites (semiprimes, 601*1801 and 251*4051, respectively), original I conjectured that for r=1, there are no other primes besides p=3 and p=7, but unfortunately, the Mersennerelated number for p=59 is also prime, however, no other primes found for the Wagstaffrelated number, and we can conjecture that the number of the Mersennerelated primes and the Wagstaffrelated primes for r=1 (and for r>1) is finite. for r>1 and p>2 (the p=2 case for Mersennerelated number is exactly the Fermat numbers, and the p=2 case for Wagstaffrelated number are not integers), the only known primes is for r=2 and p=3, and I conjectured that there are no other primes of this type. Unique primes in base 2: the formula is Phi(n,2)/gcd((Phi(n,2),n), for the case that gcd((Phi(n,2),n)>1, we let f(p) = ord_p(2) = A014664(primepi(p)) for odd prime p, all Phi(p^r*f(p),2) are divisible by p, and thus we consider the number Phi(p^r*f(p),2)/p for r=1, this number is prime for p = 3, 5, 7, 19, 43, 127, all these primes are related to Mersenne primes or Wagstaff primes, and hence related to New Mersenne Conjecture, besides, all these primes are themselves unique primes in base 2, this number is prime for p < 2^3, and three additional number (19, 43, 127), 19 is a prime make both Mersenne prime and Wagstaff prime (note that this number is not prime for p = 13 or 17), 43 is Wagstaff prime W7 (note that this number is not prime for p = W5 = 11), 127 is Mersenne prime M7 (note that this number is not prime for p = M5 = 31), original I conjectured that for r=1, there are no other primes, but unfortunately, the number for p=719 is a probable prime, note that 719 = prime(128) = prime(2^7) for r=2, this number is prime only for p = 3 and 7 (like Mersennerelated number (2^(p^(r+1))1)/(2^(p^r)1) and Wagstaffrelated number (2^(p^(r+1))+1)/(2^(p^r)+1), which for r=1, is also only prime for p = 3 and 7) and conjectured no others, for r>2, the only known primes is for r=3 and p=3 (like Mersennerelated number (2^(p^(r+1))1)/(2^(p^r)1) and Wagstaffrelated number (2^(p^(r+1))+1)/(2^(p^r)+1), which for r>1 and p>2, is also only prime for r=2 and p=3), and I conjectured that there are no other primes of this type. Besides, for the odd numbers n such that Phi(n,2) and Phi(2*n,2) are both primes, the known such n are 3, 5, 7, 13, 15, 17, 19, 31, 49, 61, 85, 127, 345 (and additional numbers 9, 21, 27 if we take out the factor gcd(Phi(n,2,n))), they are the primes in New Mersenne Conjecture (3, 5, 7, 13, 17, 19, 31, 61, 127) plus the square of a prime (49) and three additional numbers (15, 85, 345), and I conjectured that there are no other such n. For n = 11, 23, 35 (and almost all n == 11 mod 12 with 2*n+1 prime), Phi(n,2) is not prime and divisible by 2*n+1, and for n = 11, 23, 35, Phi(n,2)/(2*n+1) is prime (Phi(n,2)/(2*n+1) is prime for n = 11, 20, 23, 35, 39, 48, 83, 96, 131, 231, 303, 375, 384, 519, 771, 848, 1400, 1983, 2280, 2640, 2715, 3359, 6144, 7736, 7911, 11079, 13224, 16664, 24263, 36168, 130439, ... (it can be proved that all such even n except 20 are divisible by 8, and note that 3*2^r are such n for r = 4, 5, 7, 11), and it is unknown whether there are infinitely many such n), and for n == 4 mod 8, Phi(n,2) cannot be prime because of the Aurifeuillian factorization (Phi(nL,2) and Phi(nM,2), reference: https://oeis.org/A250197/a250197_2.txt), and we want to know whether these numbers are primes: Code:
formula n=11 n=23 n=35 Phi(2*n,2) prime prime composite Phi(4*n L,2) and Phi(4*n M,2) both primes both composites both primes Phi(8*n,2) composite prime prime Phi(2^n1,2), Phi(2^n+1,2), Phi(2*(2^n1),2) are primes for n<=7 (except Phi(63,2)), but Phi(2*(2^n+1),2) is prime only for n<=4, like that MMp, WMp, WWp are primes for primes p<=7, but MWp is prime only for primes p<=3 Conjectures: * For every n not == 4 mod 8, there are infinitely many primes p not dividing n such that Phi(n*p,2) are primes (the special case for n = 1 and n = 2 are that there are infinitely many Mersenne primes and infinitely many Wagstaff primes), however, no such primes p are known for n = 27, 35, 41, 47, ..., see text file for the status (for n = 27, p = 2 makes prime if we take out the factor gcd(Phi(n,2,n))) * For every odd n, there are only finitely many r such that Phi(n*2^r,2) is prime (the special case for n = 1 is that there are only finitely many Fermat primes) * For every m, there are infinitely many n such that Phi(n,2)/(m*n+1) is prime * (Related to New Mersenne Conjecture) for odd n, (n = 2^k ± 1 or p = 4^k ± 3) + (Phi(n,2) is prime) + (Phi(2*n,2) is prime) is 2 only for n = 1, 9, 33, 49, 61, 63, 65, 85, 129, 345 * (n = 2^k ± 1 or p = 4^k ± 3) + (Phi(n,2) is prime) + (Phi(2*n,2) is prime) is 3 only for 3, 5, 7, 13, 15, 17, 19, 31, 127 * Phi(2^n1,2) is prime only for n = 2, 3, 4, 5, 7 * Phi(2^n+1,2) is prime only for n = 1, 2, 3, 4, 5, 6, 7 * Phi(2*(2^n1),2) is prime only for n = 1, 2, 3, 4, 5, 6, 7 * Phi(2*(2^n+1),2) is prime only for n = 1, 2, 4 (also n = 3 if we take out the factor gcd(Phi(n,2,n))) Last fiddled with by sweety439 on 20210624 at 06:45 
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