OFFSET
1,2
COMMENTS
All colossally abundant (CA) numbers are also superabundant (SA). (Proof. If n is CA and k < n, then sigma(n)/n = n^{epsilon}*sigma(n)/n^{1+epsilon} >= n^{epsilon}*sigma(k)/k^{1+epsilon} > k^{epsilon}*sigma(k)/k^{1+epsilon} = sigma(k)/k, and so n is SA.)
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Jun 07 2011
STATUS
approved