computer trouble today

[N. J. A. Sloane]

SeqFans,

Tanya and I have privately discussed some of the possibly-recurrent sequences.
I think the result below may be interesting to people playing with
harmonic numbers.
It also gives rise to two new sequences:

Numbers n such that gcd(numerator(H(n)),numerator(H([n/2])))>1
(complete for n<=10^5):
7,506,1092,1755,3510,4896,52447

and the corresponding gcd's:
11,1093,1093,3511,3511,5557,104891

Curious observations:
1) All currently known terms of the second sequence are prime. What is
the smallest non-prime term (if any)?
2) The terms 1093 and 3511 (that appear several times in the second
sequence) are Wieferich primes. Is it a coincidence? Or maybe for
every Wieferich prime p, the numerators of all H(p-1), H((p-1)/2), and
H([(p-1)/4]) are divisible by p? Is the converse true?

Max

---------- Forwarded message ----------
From: Max Alekseyev <maxale at gmail.com>
Date: Mar 5, 2007 2:21 AM
Subject: Re: non recursions

On 3/4/07, Max Alekseyev <maxale at gmail.com> wrote:

> >     * A125581 a(n) = 11*a(n-1), a(0) = 77.
> >       Numbers n such that n does not divide the denominator of the n-th
> > harmonic number nor the denominator of the n-th alternating harmonic
> > number.
> >       Conjectured to be a geometric progression.
>
> Interesting conjecture, I will think more on it.

Alexander's original conjecture
%C A125581 Conjecture: a(n) = 7*11^n.
and equivalent Tanya's recurrent formula does NOT hold for A125581.

While A125581 indeed contains the geometric progression 7*11^n as a
subsequence, it also contains other geometric progressions such as:
506*1093^n, 1092*1093^n, 1755*3511^n, 3510*3511^n, and 4896*5557^n.
It may also contain some "isolated" terms (i.e., not participating in
the geometric progressions) but such terms are harder to find, and at
the moment I have no proof that they exist.

This is a sketch of my proof that geometric progression 7*11^n and the
others mentioned above belong to A125581.

Lemma 1. H'(n) = H(n) - H([n/2])

Lemma 2. For prime p and integer n>=p,
valuation(H(n),p) >= valuation(H([n/p]),p) - 1

Theorem. For an integer b>1 and a prime number p such that p divides
the numerators of both H(b) and H([b/2]), the geometric progression
b*p^n belongs to A125581.
Proof. It is enough to show that
valuation(H(b*p^n),p)>-n and valuation(H'(b*p^n),p)>-n.
By Lemma 2 we have
valuation(H(b*p^n),p) >= valuation(H(b),p) - n >= 1-n > -n.
 From this inequality Lemma 1 we have
valuation(H'(b*p^n),p) >= min{ valuation(H(b*p^n),p), valuation(H(b*p^n/2),p) }
>= min{ 1-n, valuation(H([b*p^n/2]),p) }
and we need to show that valuation(H([b*p^n/2]),p) >= 1-n.
Again by Lemma 2, we have
valuation(H([b*p^n/2]),p) >= valuation(H([b/2]),p) - n >= 1-n
that completes the proof.

It is easy to check that this Theorem holds for the aforementioned
geometric progressions.

Max