[seqfan] Re: sequence, array, some questions
T. D. Noe
noe at sspectra.com
Sun Mar 7 21:14:10 CET 2010
At 11:43 AM -0800 3/7/10, Robert Israel wrote:
>On Sat, 6 Mar 2010, Kimberling, Clark wrote:
>
>> 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
>> .
>
>There are several mistakes here. For example, a(6) should be 4, not, 3,
>as 6+k divides 6*k for k = 3, 6, 12 and 30. If I'm not mistaken, the
>first 30 terms should be
>
>0, 1, 1, 2, 1, 4, 1, 3, 2, 4, 1, 7, 1, 4, 4, 4, 1, 7, 1, 7, 4, 4, 1, 10,
>2, 4, 3, 7, 1, 13
>
>A Maple program to calculate a(n) is
>
>A:= n -> nops(remove(hastype,[isolve((n+k)*x=n*k)],nonposint));
That would make it A063647 Number of ways to write 1/n as the difference of
exactly two unit fractions.
Tony
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