[seqfan] Re: A sequence and array
franktaw at netscape.net
franktaw at netscape.net
Sun Mar 7 22:11:24 CET 2010
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
.
...
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