Continued Fractions of Harmonic Numbers

[zak seidov]

Possible values of 
length of c.f. in question are
(listing only tvalues <100)

1,2,3,4,5,6,7,8,9,10,11,15,16,17,18,19,20,21,23,24,25,26,29,31,32,33,34,35,36,39,40,41,42,43,44,46,47,48,49,50,54,57,59,60,61,63,64,65,66,67,68,69,70,71,72,73,76,78,79,80,81,82,84,87,88,89,90,93,94,95,96,99

Hence there are many missings:
12,13,14,22,27,28,30, etc..
right?

Zak




--- Joshua Zucker <joshua.zucker at gmail.com> wrote:
> On 5/27/05, Leroy Quet <qq-quet at mindspring.com>
> wrote:
> > 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,...
> 
> I get 1,2,3,6,5,7,8,10,14,9,18 but a(12) may not
> exist.
> Is my algorithm working OK?  My spot checks suggest
> that the program is good.
>  
> > Is this sequence in the EIS?
> 
> I don't find it.
> 
> > 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 find many numbers that don't seem to be present,
> and judging by the
> rate of increase I doubt it'll come back and pick
> things up later. 
> There's no 12, 13, 14, 22, 27, 28, ...
> among the first thousand harmonic series partial
> sums, anyway.
> 
> > 
> > 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.
> 
> Again, the rate of growth suggests to me that the
> sequence never comes
> back down to the small numbers, so I conjecture the
> obvious stuff,
> 1,4,3,6,5,7,11,... (also not in EIS)
> 
> Same conjecture, undefined for 12,13,14,22,27,28...
> (also doesn't seem
> to be in EIS).
> 
> If someone wants me to get the program to spit out
> more terms of all
> these sequences, let me know.
> 
> --Joshua Zucker
> 
> 


		
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