Nov 2016
2^{2}·3^{2}·5·13 Posts

A k is included in the conjecture if and only if this k has infinitely many prime candidates.
Thus, although these k's have a prime, they are excluded from the conjectures:
S8, k=27: Although (27*8^1+1)/gcd(27+1,81) is prime, but (27*8^n+1)/gcd(27+1,81) is prime only for n=1, because of the algebra factors, thus k=27 is excluded from S8.
S16, k=4: Although (4*16^1+1)/gcd(4+1,161) is prime, but (4*16^n+1)/gcd(4+1,161) is prime only for n=1, because of the algebra factors, thus k=4 is excluded from S16.
R4, k=1: Although (1*4^21)/gcd(11,41) is prime, but (1*4^n1)/gcd(11,41) is prime only for n=2, because of the algebra factors, thus k=1 is excluded from R4.
R4, k=4: Although (4*4^11)/gcd(41,41) is prime, but (4*4^n1)/gcd(41,41) is prime only for n=1, because of the algebra factors, thus k=4 is excluded from R4.
R8, k=1: Although (1*8^31)/gcd(11,81) is prime, but (1*8^n1)/gcd(11,81) is prime only for n=3, because of the algebra factors, thus k=1 is excluded from R8.
R8, k=8: Although (8*8^21)/gcd(81,81) is prime, but (8*8^n1)/gcd(81,81) is prime only for n=2, because of the algebra factors, thus k=8 is excluded from R8.
R8, k=64: Although (64*8^11)/gcd(641,81) is prime, but (64*8^n1)/gcd(641,81) is prime only for n=1, because of the algebra factors, thus k=64 is excluded from R8.
R16, k=1: Although (1*16^21)/gcd(11,161) is prime, but (1*16^n1)/gcd(11,161) is prime only for n=2, because of the algebra factors, thus k=1 is excluded from R16.
R16, k=16: Although (16*16^11)/gcd(161,161) is prime, but (16*16^n1)/gcd(161,161) is prime only for n=1, because of the algebra factors, thus k=16 is excluded from R16.
etc.
Last fiddled with by sweety439 on 20170611 at 21:35
