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Note that 504 = 7*8*9. For odd prime we have p^2 == 1 (mod 8). By Fermat's theorem and Euler's totient theorem we have p^6 == 1 (mod 7) for p != 7 and p^6 == 1 (mod 9) for p != 3. So 504 divides p^6 - 1 for p != 2, 3, 7.

There are no primes in this sequence except for a term not shown here, which is a(3) = (5^6 - 1)/504 = 31. Furthermore, omega(a(n)) = A001221(a(n)) >= 4 for n >= 7, and is exactly 4 for n = 7, 8, 9, 10, 13, 14, 16, 17, 23, ...

The set of prime factors of this sequence include all primes. First, a(7) is divisible by 2, and a(8) is divisible by 3 and 7. And also, for any prime q != 2, 3, 7, by Dirichlet's theorem on arithmetic progressions, there exists a prime p of the form k*q + d with d^6 == 1 (mod p), 0 < d < q. Since gcd(q,504) = 1, we have p^6 == d^6 == 1 (mod 504*q), so q is divisible by (p^6 - 1)/504.

Note that a(3)=31 for prime 5 is also an integer. - Michel Marcus, Jul 05 2018