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a(n) = (prime(n)^6 - 1)/504.
0

%I #74 Sep 08 2022 08:46:21

%S 3515,9577,47892,93345,293722,1180205,1760920,5090727,9424810,

%T 12542387,21387332,43976907,83691535,102222965,179480917,254167230,

%U 300266322,482316380,648691217,986073990,1652722232,2106190775,2369151382,2977639587,3327579585,4130856652

%N a(n) = (prime(n)^6 - 1)/504.

%C 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.

%C 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, ...

%C 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.

%C Note that a(3)=31 for prime 5 is also an integer. - _Michel Marcus_, Jul 05 2018

%e a(5) = (11^6 - 1)/504 = 3515, a(6) = (13^6 - 1)/504 = 9577, a(7) = (17^6 - 1)/504 = 47892, ...

%t Table[(Prime[n]^6 - 1) / 504, {n, 5, 40}] (* _Vincenzo Librandi_, Jul 13 2018 *)

%o (PARI) a(n)=(prime(n)^6 - 1)/504

%o (Magma) [(NthPrime(n)^6 - 1) div 504: n in [5..40]]; // _Vincenzo Librandi_, Jul 13 2018

%Y Cf. A024702, A089034.

%K easy,nonn

%O 5,1

%A _Jianing Song_, Jul 03 2018