If $n$ is a positive integer, we say that n is abundant if $\sigma(n)>2n$.
How can I find the six smallest abundant positive integers?
It should be: 12, 18, 20, 24, 30, 36.
Asked
Active
Viewed 115 times
0
-
Right, they are $12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 108, 112, 114, 120, 126, 132, 138, 140, 144, 150, 156, 160, 162, 168, 174, 176, 180, 186, 192, 196, 198, 200, 204, 208, 210, 216, 220, 222, 224, 228, 234, 240, 246, 252, 258, 260, 264, 270, \ldots$, see OEIS. – Dietrich Burde Nov 21 '17 at 10:57
-
But how can I find them easily without knowing that row? – WinstonCherf Nov 21 '17 at 11:02
-
Just by computing the function $\sigma(n)$. There is a formula - did you try to search already? – Dietrich Burde Nov 21 '17 at 11:03
-
Yes, I know the formula. So I just have to write down all the functions σ(n) for n=1 till n=36? There is no easier way? – WinstonCherf Nov 21 '17 at 11:04
-
Yes, this would be most efficient. We quickly see whether or not $\sigma(n)>2n$ for any $n\le 36$. Of course, we may skip the prime numbers etc. So we start with $\sigma(4)=7$, which is not greater $8$, then $\sigma(6)=12$, which is close, but no, etc. – Dietrich Burde Nov 21 '17 at 11:05