A306174 a(n) = (prime(n)^6 - 1)/504.

3515, 9577, 47892, 93345, 293722, 1180205, 1760920, 5090727, 9424810, 12542387, 21387332, 43976907, 83691535, 102222965, 179480917, 254167230, 300266322, 482316380, 648691217, 986073990, 1652722232, 2106190775, 2369151382, 2977639587, 3327579585, 4130856652

Offset

5,1

Comments

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

Links

Table of n, a(n) for n=5..30.

Example

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

Mathematica

Table[(Prime[n]^6 - 1) / 504, {n, 5, 40}] (* Vincenzo Librandi, Jul 13 2018 *)

Prog

(PARI) a(n)=(prime(n)^6 - 1)/504
(Magma) [(NthPrime(n)^6 - 1) div 504: n in [5..40]]; // Vincenzo Librandi, Jul 13 2018

Crossrefs

Cf. A024702, A089034.

Sequence in context: A268352 A178871 A317162 * A205071 A263302 A205278

Adjacent sequences: A306171 A306172 A306173 * A306175 A306176 A306177

Keyword

easy,nonn

Author

Jianing Song, Jul 03 2018

Status

approved