A342066 Primes p such that p^10 - 1 has 256 divisors.

1187, 4723, 33037, 66973, 72797, 87973, 100523, 197123, 219683, 229693, 276293, 278827, 440653, 448997, 482837, 562963, 601333, 621443, 670493, 742723, 846877, 892357, 1033427, 1149307, 1166027, 1245067, 1256747, 1614413, 1679773, 1865693, 1950323, 1970467

Offset: 1

Author

Jon E. Schoenfield, Feb 27 2021

Keywords

• nonn

Comments

Conjecture: sequence is infinite.

The only primes p such that p^10 - 1 has fewer than A309906(10)=256 divisors are 2, 3, 5, 7, 11, 13, and 43.

p^10 - 1 = (p-1)*(p+1)*(p^4 - p^3 + p^2 - p + 1)*(p^4 + p^3 + p^2 + p + 1). For every p > 11, one of these five factors is divisible by 11; one of p-1 and p+1 is divisible by 3; and p-1 and p+1 are consecutive even numbers, so one of them is divisible by 4 and their product is divisible by 8; thus, p^10 - 1 is divisible by 2^3 * 3 * 11.

For every term p with the exception of a(1)=1187, p^10 - 1 is of the form 2^3 * 3 * 11 * q * r * s * t, where q, r, s, and t are distinct primes > 11.

Examples

			For p = a(1) = 1187, p^10 - 1 = 2^3 * 3^3 * 11 * 593 * 1983522604541 * 1986867499321;
for p = a(2) = 4723, p^10 - 1 = 2^3 * 3 * 11 * 787 * 1181 * 45245048697451 * 497484826300381.
		

Crossrefs

Cf. A000005, A000040, A309906, A341670.