CF of Reciprocal Sum of A112373

[Mitch Harris]

> From: Jeffrey Shallit 
> > From: "Paul D Hanna" 
> > 
> > Seqfans,
> >        This may or may not be surprising, but I find it 
> interesting neve=
> > rtheless.
> > The sum of the reciprocal terms of Andrew Hone's sequence A112373:
> >  x Sum_{n>0} 1/A112373(n) =
> > 
> >      2.584401724019776724812076147...
> > where A112373 is defined by: =
> > 
> > a(n+2) = (a(n+1)^3+a(n+1)^2)/a(n) with a(0)=1, a(1)=1 =
> > 
> >  =
> > 
> > is a constant with an interesting Continued Fraction:
> >  =
> > 
> > x = [2; 1, 1, 2, 2, 6, 12, 78, 936, 73086, 68408496, 4999703411742,
> >  342022190843338960032, 1710009514450915230711940280907486, 
> 584861200495=
> > 456320274313200204390612579749188443599552,...]
> >  =
> > 
> > I wonder if the terms of the above CF has any recurrence pattern?
> 
> Yes:  the even-indexed terms (such as 12 = 2*6) appear to be 
> the product of the previous two terms.
> 
> The odd-indexed terms (such as 78=6*12 + 6) appear to be the product
> of the previous two terms, plus the term two behind.
> 
> It would be nice to prove this.  Probably an easy induction, but
> I don't have time right now.  Very beautiful.

Paul's conjecture, that the sequence b(n) defined by

   b(2n) = a(n)
   b(2n+1) = a(n+1)/a(n),

matches your observation, that

   1) b(2n) = b(2n-1)b(2n-2)
   2) b(2n+1) = (b(2n)+1)b(2n-1),

is not hard to prove (using the definition of a(n)):

   1) b(2n-1)b(2n-2) = a(n)/a(n-1) * a(n-1) = a(n) = b(2n).
   2) (b(2n)+1)b(2n-1) = (a(n)+1)a(n)/a(n-1)
                       = (a(n)^2+a(n))a(n)/(a(n-1)a(n))
                       = a(n+1)/a(n) = b(2n+1)

(and base cases by inspection)

As to the CF conjecture, i.e. that b(n) is the CF, from computation
of the CF of the partial sums (that is \sum_{k=0..n} 1/a(n)), the last
element of the CF always seems to be a(n), but the next to last isn't
always a(n-1), so a straightforward induction probably won't work 
exactly right. Any ideas on how to attack this otherwise (or get around 
this "probably")?

Mitch