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Old 2017-02-23, 09:07   #9
sweety439
 
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Nov 2016

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If b and k are of these forms, then k is a Brier number (i.e. both Sierpinski number and Riesel number) to base b.

Code:
b               k
= 14 mod 15     = 4 or 11 mod 15
= 20 mod 21     = 8 or 13 mod 21
= 32 mod 33     = 10 or 23 mod 33
= 34 mod 35     = 6 or 29 mod 35
= 38 mod 39     = 14 or 25 mod 39
= 50 mod 51     = 16 or 35 mod 51
= 54 mod 55     = 21 or 34 mod 55
= 56 mod 57     = 20 or 37 mod 57
= 64 mod 65     = 14 or 51 mod 65
= 68 mod 69     = 22 or 47 mod 69
= 76 mod 77     = 34 or 43 mod 77
= 84 mod 85     = 16 or 69 mod 85
= 86 mod 87     = 28 or 59 mod 87
= 90 mod 91     = 27 or 64 mod 91
= 92 mod 93     = 32 or 61 mod 93
= 94 mod 95     = 39 or 56 mod 95
= 110 mod 111   = 38 or 73 mod 111
= 114 mod 115   = 24 or 91 mod 115
= 118 mod 119   = 50 or 69 mod 119
= 122 mod 123   = 40 or 83 mod 123
= 128 mod 129   = 44 or 85 mod 129
= 132 mod 133   = 20 or 113 mod 133
= 140 mod 141   = 46 or 95 mod 141
= 142 mod 143   = 12 or 131 mod 143
Generally, if there is a prime p divides both k+1 and b+1, and a prime q divides both k-1 and b+1, then k is both Sierpinski number (if gcd(k+1,b-1) = 1) and Riesel number (if gcd(k-1,b-1) = 1) to base b. (the covering set is both {p, q})

Thus, for the original Sierpinski/Riesel problems, if b+1 has at least two distinct odd prime factors, then it is easy to find a Sierpinski/Riesel k. Besides, for the reverse Sierpinski/Riesel problems, if neither k+1 nor k-1 is a power of 2 ("power of 2" includes 1), then it is easy to find a Sierpinski/Riesel base b.

Last fiddled with by sweety439 on 2017-02-23 at 09:15
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