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发表于 2025-5-1 04:56
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由二次三项式a*x^2-b*x+p生成的多个连续素数之最小p
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1 2 a(1) = 2 (a prime), x^2 + 2 gives a prime for x = 1.
2 3 a(2) = 3 (a prime), 2*x^2 + 3 gives distinct primes for x = 1 to 2.
3 5
4 5 a(4) = 5 (a prime), 2*x^2 + 5 gives distinct primes for x = 1 to 4.
5 7
6 7 a(6) = 7 (a prime), 4*x^2 + 7 gives distinct primes for x = 1 to 6.
7 11
8 11
9 11
10 11 a(10) = 11 (a prime), 2*x^2 + 11 gives distinct primes for x = 1 to 10.
11 13
12 13 a(12) = 13 (a prime), 6*x^2 + 13 gives distinct primes for x = 1 to 12.
13 17
14 17
15 17
16 17 a(16) = 17 (a prime), 6*x^2 + 17 gives distinct primes for x = 1 to 16.
17 19
18 19 a(18) = 19 (a prime), 10*x^2 + 19 gives distinct primes for x = 1 to 18.
19 23
20 23
21 23
22 23 a(22) = 23 (a prime), 3*x^2 - 3*x + 23 gives distinct primes for x = 1 to 22.
23 29
24 29
25 29
26 29
27 29
28 29 a(28) = 29 (a prime), 2*x^2 + 29 gives distinct primes for x = 1 to 28.
29 31 a(29) = 31 (a prime), 2*x^2 - 4*x + 31 gives distinct primes for x = 1 to 29.
30 41
31 41
32 41
33 41
34 41
35 41
36 41
37 41
38 41
39 41
40 41 a(40) = 41 (a prime), x^2 - x + 41 gives distinct primes for x = 1 to 40.
41 647 a(41) = 647 (a prime), abs(36*x^2 - 594*x + 647) gives distinct primes for x = 1 to 41.
42 1277 a(42) = 1277 (a prime), abs(36*x^2 - 666*x + 1277) gives distinct primes for x = 1 to 42.
43 1979 a(43) = 1979 (a prime), abs(36*x^2 - 738*x + 1979) gives distinct primes for x = 1 to 43.
44 2753 a(44) = 2753 (a prime), abs(36*x^2 - 810*x + 2753) gives distinct primes for x = 1 to 44.
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