b is congruent to 1 modulo 3, then all Thabit numbers of the second kind base b are divisible by 3 (and greater than 3, since b ≥ 2), so there are no Thabit primes of the second kind base b.) The exponent of Thabit primes of the second kind cannot congruent to 1 mod 3 (except 1 itself), the exponent of Williams primes of the first kind cannot congruent to 4 mod 6, and the exponent of Williams primes of the second kind cannot congruent to 1 mod 6 (except 1 itself), since the corresponding polynomial to b is a reducible polynomial. (If n ≡ 1 mod 3, then (b+1)·bn + 1 is divisible by b2 + b + 1; if n ≡ 4 mod 6, then (b−1)·bn − 1 is divisible by b2 − b + 1; and if n ≡ 1 mod 6, then (b−1)·bn + 1 is divisible by b2 − b + 1) Otherwise, the corresponding polynomial to b is an irreducible polynomial, so if Bunyakovsky conjecture is true, then there are infinitely many bases b such that the corresponding number (for fixed exponent n satisfying the condition) is prime. ((b+1)·bn − 1 is irreducible for all nonnegative integer n, so if Bunyakovsky conjecture is true, then there are infinitely many bases b such that the corresponding number (for fixed exponent n) is prime) Least k ≥ 1 such that (n+1)·nk − 1 is prime are: (start with n = 2) 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 2, 1, 2, 1, 1, 4, 3, 1, 1, 1, 2, 7, 1, 2, 1, 2, 1, 2, 1, 1, 2, 4, 2, 1, 2, 2, 1, 1, 2, 1, 8, 3, 1, 1, 1, 2, 1, 2, 1, 5, 3, 1, 1, 1, 1, 3, 3, 1, 1, 5, 2, 1483, 1, 1, 1, 24, 1, 2, 1, 2, 6, 3, 3, 36, 1, 10, 8, 3, 7, 2, 2, 1, 2, 1, 1, 7, 1704, 1, 3, 9, 4, 1, 1, 2, 1, 2, 24, 25, 1, ... Least k ≥ 1 such that (n+1)·nk + 1 is prime are: (start with n = 2, 0 if no such k exists) 1, 1, 0, 1, 1, 0, 1, 2, 0, 2, 1, 0, 1, 1, 0, 1, 9, 0, 1, 1, 0, 2, 1, 0, 2, 1, 0, 5, 2, 0, 5,