# [seqfan] Re: sequence, array, some questions

Robert Israel israel at math.ubc.ca
Sun Mar 7 20:29:33 CET 2010

```
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
> .
>
> It is easy to prove that a(n)=1 if and only if n is a prime.  Here's a
> corner of the array whose ith row is all n such that a(n)=i, so that row
> 1 is the primes:
>
> 2....3....5....7....11....13....
> 4....9....25...49...121...169...
> 6....8....15...27...35....77...
> 10...14...16...21...22....26...
> 12...32...45...243...
> 18...20...28...63...
> 44...50...52...68...
>
> Which of the the following statements are true?
>
> (1)  Row i includes all (prime)^i.

True.
If n = p^i where p is prime, and k = p^j q where q is not divisible by p,
then n+k is coprime to q.  The only way for n+k to divide k is that n+k is
a power of p.  If i <> j this is impossible, so we must have i=j and
q = p^m - 1 for some m, with 1 <= m <= i.  Thus a(p^i) = i.

Cheers,
Robert Israel

> (2)  Every positive integer >1 is in the array.
>
> (3)  The numbers in column 1 are even.
>
> Clark Kimberling
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

```