Talk:Highly abundant number

Five and Seven
Are 5 and 7 highly abundant? I never heard of highly abundant numbers before reading this, so maybe I don't understand, but how as I understand this $$\sigma$$(4)=3 and $$\sigma$$(prime)=1.-- Randall Bart &lt;wiki@randallbart.com&gt; 05:08, 21 October 2006 (UTC)

I'm fairly certain there are no highly abundant primes above 3, since for any prime p, $$\sigma$$(p) = p+1, and since p-1 (for any prime greater than 3) is an even integer, it will always have divisors: 1,2, some integer k,itself, and maybe more. so since (p-1)+1+2 > p+1, there shouldn't be any more primes in this sequence. --habitue 06:42, 8 February 2007 (UTC)

There is a proof here from Daniel Fischer that every highly abundant number greater than 3, 20, 630 is divisible by 2, 6, 12 respectively. Notice also that because every multiple of a perfect or abundant number is abundant, 20 is the last possible non-abundant highly abundant number, making the complete list of non-abundant highly abundant numbers: 1, 2, 3, 4, 6, 8, 10, 16. I've conjectured that for every integer there is exists a last highly abundant number not divisible by that integer. Jaycob Coleman (talk) 13:46, 17 October 2013 (UTC)