CF of Reciprocal Sum of A112373

[Paul D. Hanna]

Jeffrey (and Seqfans),
> > 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.
 
Excellent observation.  
Indeed, note the terms of A112373:
1,1,2,12,936,68408496,342022190843338960032,
584861200495456320274313200204390612579749188443599552, ...
where a(n+2) =(a(n+1)^3+a(n+1)^2)/a(n) with a(0)=1, a(1)=1.
 
It is remarkable to me that the reciprocal sum of these terms 
x = Sum_{n>=0} 1/A112373(n)
has the continued fraction expansion:
x = [2;1,1,2,2,6,12,78,936,73086,68408496,4999703411742,
342022190843338960032,1710009514450915230711940280907486,
584861200495456320274313200204390612579749188443599552,...]
 
such that the even partial quotients of the CF (when offset=0 at constant
term)
equal A112373 (for n>0): 
2,1,2,12,936,68408496,342022190843338960032,
584861200495456320274313200204390612579749188443599552,...
 
and the odd partial quotients equal A112373(n+1)/A112373(n):
1,2,6,78,73086,4999703411742,1710009514450915230711940280907486,...
 
> It would be nice to prove this.  Probably an easy induction, but
 
Yes, a proof would be nice; it may indeed have a simple reason behind it
all.
   
I will submit this CF and the constant to Neil when the time is right. 
I will include your nice recurrence and credit you with it. 
 
Thanks! 
       Paul
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