Number of divisors of num and denom of harmonic #s

[Max Alekseyev]

On 5/29/07, Leroy Quet <qq-quet at mindspring.com> wrote:
> Just curious.
>
> What are the first 50-100 or so terms of the sequence where a(n) =
>
> A129567(n) (mod A129568(n))?
>
> (A129568 is the number of divisors of the numerator of the nth harmonic
> number H(n). {H(n) = sum{k=1 to n} 1/k.}
> And A129567 is the number of divisors of the denominator of the nth
> harmonic number H(n).)
>
> A129568 has been extended, but A129567 currently only has a few terms.

The first 100 terms of A129567:

1, 2, 4, 6, 12, 6, 12, 16, 48, 48, 96, 96, 192, 192, 192, 240, 480,
320, 640, 320, 160, 160, 320, 640, 1920, 1920, 3840, 3840, 7680, 7680,
15360, 18432, 9216, 9216, 9216, 9216, 18432, 18432, 18432, 18432,
36864, 18432, 36864, 73728, 73728, 73728, 147456, 147456, 442368,
442368, 442368, 442368, 884736, 663552, 663552, 663552, 663552,
663552, 1327104, 1327104, 2654208, 2654208, 1769472, 2064384, 2064384,
1032192, 2064384, 2064384, 4128768, 4128768, 8257536, 12386304,
24772608, 24772608, 24772608, 24772608, 12386304, 12386304, 24772608,
24772608, 41287680, 41287680, 82575360, 82575360, 82575360, 82575360,
82575360, 165150720, 330301440, 330301440, 330301440, 330301440,
330301440, 330301440, 330301440, 330301440, 660602880, 660602880,
660602880, 220200960

> I would bet that A129568(n) divides A129567(n) for many n's.

The only n for which A129568(n) does NOT divide A129567(n) below 169 are:
18 and 22.

Max



Max Alekseyev wrote:
>...
>The only n for which A129568(n) does NOT divide A129567(n) below 169 are:
>18 and 22.

Oh, darn. I was hoping that EVERY term of A129568 divided its 
corresponding term in A129567.
:)

Well, I knew that was a long shot. Still, I wonder if there is a theorem 
in there somewhere. Probably not.

Maybe someone should submit the terms of the sequence which lists the 
positive integers n where d(numerator(H(n))) does not divide 
d(denominator(H(n))), where, again, H(n) = sum{k=1 to n}1/k, and d(m) is 
the number of positive divisors of m, of course.


Let's see: we have 18,22,...

Thanks,
Leroy Quet