[seqfan] Re: A sequence and array

[franktaw at netscape.net]

Since n+k always divides n^2+n*k, the relation n+k divides n*k is 
equivalent to n+k divides n^2.  Thus a(n) = (tau(n^2)-1)/2:

0, 1, 1, 2, 1, 4, 1, 3, 2, 4, 1, 7, 1, 4, 4, 4, 1, 7, 1, 7

This is A063647.  You can find a great deal more information there.

(Here using tau(n) for the number of divisors of n.)

Franklin T. Adams-Watters

-----Original Message-----
From: Kimberling, Clark <ck6 at evansville.edu>

Seqfans,

Let a(n) be the number of positive integers k such that n+k divides n*k.
(Equivalently, 1/n + 1/k = 1/m for some integer m.)  The first 30 terms
of a(n) are 0,1,1,2,1,3,1,3,2,4,1,5,1,4,3,4,1,6,1,6,4,4,1,8,2,4,3,6,1,10
.
...