Skip to main content

Return to Revisions

3 of 3

Commonmark migration

edited Jun 17, 2020 at 9:04

edited Jun 17, 2020 at 9:04

1

Oasis, 4 bytes

Code:

nj+U

Explanation:

Extended version:

nj+00
 0 = a(0)
 0 = a(1)
a(n) =
n # Push n
 j # Get the largest divisor under n
 + # Add to a(n - 1)

answered Jun 24, 2017 at 11:47

44.7k
6
83
248