OFFSET
2,1
COMMENTS
a(1) corresponds to odd multiperfect numbers greater than 1, if they exist.
Does every integer greater than 1 have a solution?
65 is a solution for every positive n congruent to 2 mod 4.
LINKS
Alexander Violette, Table of n, a(n) for n = 2..200
Wikipedia, Divisor function
EXAMPLE
65 is the smallest odd number k > 1 such that sigma_2(k)/k is an integer, hence a(2)=65.
PROG
(PARI) a(n) = my(k=3); while (sigma(k, n) % k, k+=2); k; \\ Michel Marcus, May 30 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Alexander Violette, May 13 2026
STATUS
approved
