[seqfan] Re: sequence, array, some questions

Kimberling, Clark ck6 at evansville.edu
Sun Mar 7 23:34:32 CET 2010


Thanks, Robert, for the proof of (1).  As a couple of people have noted,
the sequence didn't fit the description - because I neglected to say
that k>=n. In that case, the original sequence 0,1,1,2,1,3,1,3,2,4,...
and the array (all copied below) seem okay.

If the restriction that k>=n is removed, the array starts thus:

2....3....5....7....11....13.... (primes)
4....9....25...49...121...169... (prime squares) 
8....27...125.....               (cubes) 
6....10...14...15...16....21...  
32...243...
64...
12...18...20...28...44...45...50...52...    
   
For this array, questions (2) and (3) look easier.

Clark Kimberling



-----Original Message-----
From: seqfan-bounces at list.seqfan.eu
[mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Robert Israel
Sent: Sunday, March 07, 2010 1:30 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: sequence, array, some questions



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/
>


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