Dividing Numerator or Denomninator Of Harmonic Numbers

[Edwin Clark]

On Tue, 13 Sep 2005, Leroy Quet wrote:

> Two days ago I submitted this sequence to the EIS.
> 
> >A000001 1,2,2,4,4,3,6,8,9,5,3,4,12,7,5,16
> >%N A000001 a(n) is lowest positive integer m such that n divides either 
> >the numerator or the denominator of the (reduced) H(m) = sum{k=1 to m} 1/k.
> >%e A000001 a(5) = 4 because H(4) = 25/12 is the first harmonic number with 
> >either its numerator or denominator divisible by 5.
> >a(6) = 3 because H(3) = 11/6 is the first harmonic number with either its 
> >numerator or denominator divisible by 6.
> >%Y A000001 A001008,A002805
> >%O A000001 1
> >%K A000001 ,more,nonn,
> 
> My question seems like it might have an obvious answer.
> (Yet when I posted it to sci.math, I received no replies as of today.)
> Does every positive integer divide the numerator or denominator
> of at least one harmonic number?
> 
> It seems as if a(n) might always be <= n.
> At least that is the way it is for all the terms above.
> 
Your conjecture that a(n) exists and is at most n holds at least up
to n = 1000.  
 
Moreover if we just ask for which n the denominator of H(n) is not 
divisible by n, computation shows that we obtain the following sequence:
(I haven't tried to prove it.) Note there are only 41 such values <= 1000.

ID Number: A074791
URL:       http://www.research.att.com/projects/OEIS?Anum=A074791
Sequence:  6,18,20,21,33,42,54,63,66,77,100,110,120,156,162,189,198,
           272,294,336,342,363,377,435,486,500,506,559,567,594,600,610,
           629,685,703,812,847,880,924,930,957,1067,1166,1210,1243,
           1247,1287,1320,1332,1458
Name:      Numbers n such that the numerator of sum(k=0,n-1,(k+1)/(n-k)) 
is not equal to N(n)-D(n) where N(n) is the numerator of the n-th 
harmonic number, and D(n) the denominator of the n-th harmonic number.