Dividing Numerator or Denomninator Of Harmonic Numbers

[David Harden]

>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?
>
>thanks,
>Leroy Quet

First of all, it is easy to see that no even positive integer will
ever divide the numerator of a harmonic number in lowest terms, since the highest power of 2 in the set {1,...,n} divides only one integer in that set-- namely, itself.
However, it may be a very hard problem to determine whether or not every odd positive integer divides the numerator of a harmonic number. I will not address this problem here, because it seems to be too hard.

Due to the above-noted property of even numbers, any hope for proving that every positive integer divides the numerator or denominator of a harmonic number would seem to lie in proving that statement for denominators.

Here is a proof:

(n=1 is trivial, so it is omitted)
First of all, we handle the case where n is a power of a prime. In this case, the least common multiple of the integers from 1 to n-1 is not a multiple of n and therefore it is easy to show that the nth harmonic number has a denominator which is a multiple of n.

Now suppose n is not a power of a prime.

Let the prime factorization of n be n=(p_1^e_1)*...*(p_k^e_k).
It is easy to see that it is possible to choose a sequence {E_i}(1 <= i <= k) of positive integers such that E_i >= e_i for all i and, for all i from 1 to k-1, p_i^E_i/(p_(i+1)^E_(i+1)) > 3.
Note that, between any two consecutive multiples of p_i^E_i, there are at least 3 multiples of p_(i+1)^E_(i+1) due to the way the E_i's were chosen.
For N between p_1^E_1 and 2*(p_1^E_1), the denominator of the Nth harmonic number will be a multiple of p_1^E_1.
Between these two numbers, there are at least 3 multiples of p_2^E_2. Call 3 consecutive ones among these m_1, m_2 and m_3.
Maybe the m1th harmonic denominator is a multiple of p_2^E_2. Maybe it isn't.
If it isn't, then the power to which p_2 appears in the factorization of the m_1th harmonic denominator is < E_2. This makes it easy to prove that, in this case, the power to which p_2 appears in the factorization of the m_2th harmonic denominator is E_2.
So, in either case, choosing N to be an integer on (an appropriate) one of the intervals [m_1,m_2), [m_2,m_3) will allow the Nth harmonic denominator to be a multiple of p_1^E_1 and p_2^E_2. And that interval also has at least 3 multiples of of p_3^E_3 (if k > 2)... 
Continuing in this fashion, one may wind up with a set of p_k^E_k values of N such that the Nth harmonic denominator is a multiple of (p_1^E_1)*...*(p_k^E_k) and therefore a multiple of n (since E_i >= e_i for all i). This concludes the proof.

Enjoy knowing that all of the terms of your sequence are defined, and thanks for the problem!

---- David