

A342066


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


0



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
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OFFSET

1,1


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 = (p1)*(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 p1 and p+1 is divisible by 3; and p1 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.


LINKS

Table of n, a(n) for n=1..32.


EXAMPLE

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.
Sequence in context: A237698 A345664 A052236 * A153378 A205230 A229539
Adjacent sequences: A342063 A342064 A342065 * A342067 A342068 A342069


KEYWORD

nonn


AUTHOR

Jon E. Schoenfield, Feb 27 2021


STATUS

approved



