Dividing Numerator or Denomninator Of Harmonic Numbers

[Franklin T. Adams-Watters]

This conjecture is almost certainly correct.

It is easier to ignore the numerators, and just look at the denominators.  Thus we sharpen the conjecture to:

For any n, there exists m <= n such that n divides the denominator of H_m.

It is trivial to show that this is correct when n is a prime power; then m=n.

Note next that divisibility of Den(H_n) by p^e can change only at a multiple of p^e, and that on at least one side of any such multiple, the denominators will be divisible by p^e.  That is, if

a/b + 1/(k*p^e) = c/d,

then if b is not divisible by p^e, d must be.

It follows that if n = p^e * q^f, with (without loss of generality) q^f > p^e, then either Den(H_(q^f)) is divisible by p^e, or the next larger multiple of p^e will have its denominator divisible by p^e; and since that must me smaller than 2*q^f, that denominator will still be divisible by p^e.  Hence the conjecture holds for any number with at most 2 distinct prime divisors.

I don't see right now how to go to larger numbers of prime divisors.

We can show that there is at least a solution for every n.  We need the lemma that the values of n for which p^e does not divide Den(H_n) have density zero.  The case p=2 is trivial; if n>=2^k, then 2^k | Den(H_n).
For any other prime, at least (p-1)/(2p) of the values between any k*p^e and (k+1)*p^e have the requisite condition.  But this also applies for p^(e+1), p^(e+2), etc.  So the overall probability of non-divisibility by p^e is at least (p+1)/(2p) * (p+1)/(2p) * ... = 0.

This being the case, the density of numbers where Den(H_m) is not divisible by n for arbitrary n must also be zero; hence the complement is certainly non-empty.  Q.E.D.

Leroy Quet <qq-quet at mindspring.com> wrote:
>... 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.
>
>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.
-- 
Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645


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