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Old 2020-08-06, 06:43   #929
sweety439's Avatar
Nov 2016

2,347 Posts

Using the Riesel side as an example:

1. n must be >= 1 for all k

2. If (k*b^n-1)/gcd(kb-1,b-1) where n=1 is prime than k*b (i.e. MOB) will need a different prime because this prime would be (kb*b^0-1)/gcd(kb-1,b-1)

3. If (k*b^n-1)/gcd(kb-1,b-1) where n>1 is prime than k*b will have the same prime (in a slightly different form), i.e. (kb*b^(n-1)-1)/gcd(kb-1,b-1)

4. Assume that (k*b^1-1)/gcd(kb-1,b-1) is prime. (k*b^1-1)/gcd(kb-1,b-1) = (kb-1)/gcd(kb-1,b-1)

5. Conclusion: Per #2 and #4 the only time k*b needs a different prime than k is when (kb-1)/gcd(kb-1,b-1) is prime ((kb+1)/gcd(kb+1,b-1) for Sierp)
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