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Old 2017-06-11, 21:33   #315
sweety439
 
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Nov 2016

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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+1,8-1) is prime, but (27*8^n+1)/gcd(27+1,8-1) 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+1,16-1) is prime, but (4*16^n+1)/gcd(4+1,16-1) is prime only for n=1, because of the algebra factors, thus k=4 is excluded from S16.

R4, k=1: Although (1*4^2-1)/gcd(1-1,4-1) is prime, but (1*4^n-1)/gcd(1-1,4-1) is prime only for n=2, because of the algebra factors, thus k=1 is excluded from R4.

R4, k=4: Although (4*4^1-1)/gcd(4-1,4-1) is prime, but (4*4^n-1)/gcd(4-1,4-1) is prime only for n=1, because of the algebra factors, thus k=4 is excluded from R4.

R8, k=1: Although (1*8^3-1)/gcd(1-1,8-1) is prime, but (1*8^n-1)/gcd(1-1,8-1) is prime only for n=3, because of the algebra factors, thus k=1 is excluded from R8.

R8, k=8: Although (8*8^2-1)/gcd(8-1,8-1) is prime, but (8*8^n-1)/gcd(8-1,8-1) is prime only for n=2, because of the algebra factors, thus k=8 is excluded from R8.

R8, k=64: Although (64*8^1-1)/gcd(64-1,8-1) is prime, but (64*8^n-1)/gcd(64-1,8-1) is prime only for n=1, because of the algebra factors, thus k=64 is excluded from R8.

R16, k=1: Although (1*16^2-1)/gcd(1-1,16-1) is prime, but (1*16^n-1)/gcd(1-1,16-1) is prime only for n=2, because of the algebra factors, thus k=1 is excluded from R16.

R16, k=16: Although (16*16^1-1)/gcd(16-1,16-1) is prime, but (16*16^n-1)/gcd(16-1,16-1) 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 2017-06-11 at 21:35
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