OFFSET
1,1
COMMENTS
For any number k with two or more distinct prime divisors, the sum of divisors of k^n is composite, so each term is of the form p^j where p is prime and j >= 1, i.e., all terms are prime powers (A246655). Additionally, sigma(k^n) = sigma(p^(j*n)) = (p^(j*n + 1) - 1)/(p - 1) is composite when j*n + 1 is composite, so a(n) must be of the form p^j where j*n + 1 is prime.
LINKS
Jon E. Schoenfield, Table of n, a(n) for n = 1..200
EXAMPLE
a(1) = 2 because sigma(1^1) = sigma(1) = 1 (not prime), but sigma(2^1) = sigma(2) = 1 + 2 = 3 (prime).
a(3) = 4 because sigma(1^3) = 1 (not prime), sigma(2^3) = 1 + 2 + 4 + 8 = 15 (composite), sigma(3^3) = 1 + 3 + 9 + 27 = 40 (composite), but sigma(4^3) = sigma(2^6) = 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127 (prime).
a(19) = 1174562876521148458974062689 = 17^22 because sigma((17^22)^19) is prime and sigma(k^19) is not prime for any smaller value of k.
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Mar 11 2017
STATUS
approved