Continued Fractions of Harmonic Numbers

[David Wilson]

----- Original Message ----- 
From: "Leroy Quet" <qq-quet at mindspring.com>
To: <seqfan at ext.jussieu.fr>
Sent: Friday, May 27, 2005 3:12 PM
Subject: Continued Fractions of Harmonic Numbers


>I am wondering about the sequence where the nth term is the lowest
> positive m where the simple continued fraction of
> sum{k=1 to m} 1/k
> has exactly n terms.
>
> I get that the sequence starts 1, 2, 3, 6, 5,...
>
> Is this sequence in the EIS?

Is the fundamental sequence in the OEIS, namely, a(n) = number of terms in 
continued fraction of nth harmonic number?  If so, then your sequence can be 
described as the least inverse of a.

> Is there a term defined for every n? ie - is every number of terms
> represented among the continued fractions of harmonic numbers?
> Or are there some n's where no CF of any harmonic number has exactly n
> terms?

I am sure that as n grows, the above-described a(n) grows erratically 
without bound (although I would be surprised if you could come up with a 
decent lower bound).  In its erratic hopping around, it is not unthinkable 
that a(n) misses some small value m, then eventually n grows large enough 
that a(n) > m, so that a(n) = m has no solution.  This is fairly common 
behavior erratic sequences.

For example, consider the sequence

a(n) = SumOfDigits(3^n)/9

for n >= 2.  This sequence grows with approximately linear but erratic 
growth.  a(n) takes on every value m < 62, but apparently misses a(n) = m = 
62.  a(n) continues to grow until it becomes apparent (but provable at best 
with great difficulty) that a(n) will never again be as small as 62.  So we 
can confidently say that a^-1(62) does not exist.

> Another related sequence I wonder about:
>
> If there are a finite number of m's where, for every n, the mth harmonic
> number has exactly n terms in its CF (if there are no n's where an
> infinite number of harmonic numbers has exactly n terms in their CFs),
> then we can form the sequence of the *highest* integer m where the mth
> harmonic number has exactly n terms.
>
> Sequence starts: 1,...
>
> I do not know if this sequence has every term defined either.
>
> thanks,
> Leroy Quet