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 2017-02-23, 09:07 #9 sweety439   Nov 2016 22×3×5×47 Posts 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