OFFSET
1,2
COMMENTS
This sequence is motivated by the existence in A019278 of terms n such that s=sigma(n) is also a term of A019278. Those terms are a subsequence of this sequence.
The corresponding denominators are 1, 3, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 5, 15, 28, 127, 1, 1, 1, 127, 1, 39, 1, 1, 31, 1, 1, 682, 1, 9, 16, 1, 1, 310, 99, 1729, ...
Are there other terms, like 1 and 6 (see example)?
EXAMPLE
For n=1, sigma(1)=1, so 1 is obviously in the sequence.
For n=6, sigma(6)=12; sigma(sigma(6))/6 and sigma(sigma(12))/12 are both equal to 14/3, so they have same denominator 3; so 6 is in the sequence.
PROG
(PARI) isok(n) = {my(s = sigma(n), ss=sigma(s)); denominator(ss/n) == denominator(sigma(ss)/s); };
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Jul 23 2016
STATUS
approved