OFFSET
1,2
COMMENTS
A number n is called "abundant" if sigma(n) > 2n, and "highly abundant" if sigma(n) > sigma(m) for all m < n. With these definitions, it's possible for a number to be highly abundant but not abundant. (A similar situation occurs with 2 being prime and highly composite.)
Fischer shows that all highly abundant numbers greater than 20 are multiples of 6. Since 6 is perfect and multiples of perfect numbers are abundant, this list is finite and complete.
LINKS
Daniel Fischer, Prove that if Fn is highly abundant, then so is n, Mathematics Stack Exchange, Aug 13 2013
EXAMPLE
10 is in the sequence because sigma(10) > sigma(m) for m = 1 to 9, yet sigma(10) = 17 < 20.
PROG
(PARI) for(n=1, 1000, if((sum(i=1, n-1, sign(sigma(n)-sigma(i))) == n-1) && (sigma(n) <= 2*n), print1(n, ", "))) \\ Michel Marcus, Sep 21 2014
(PARI) is_A247334(n)={!for(i=2, n-1, sigma(n)>sigma(i)||return) && sigma(n)<=2*n} \\ M. F. Hasler, Oct 15 2014
CROSSREFS
KEYWORD
fini,full,nonn
AUTHOR
Andrew Rodland, Sep 13 2014
STATUS
approved