Continued-Fraction-Generated Sequence Also Equals...

[Leroy Quet]

[also posted to sci.math]

I wrote:
> Consider the continued fraction:
> 
> [1; 1,1, 1/2,1/2, 1/3,1/3, 1/4,1/4,...,ceiling((m-2)/2)]
> 
> = c(m), for m >= 3,  
> 
> where the total number of (rational) CF-terms is (m-1).
> 
> Let c(1) = 1, and c(2) = 2, so we have:
> 
> {c(j)} -> 1, 2, 2, 3/2, 7/4, 19/12, 61/36,... 
> 
> 
> Now, let us take 'another' sequence {c'(j)},
> where:
> 
> c'(1) = 1,  c'(2) = 2.
> 
> And c'(1+m) =
> 
> (sum{j=0 to floor((m-1)/2)} c'(m- 2j)) /ceiling(m/2).
> 
> In other words, 
> c'(1+m) is the average of every-other previous c'(),
> the average of {c'(m),c'(m-2),c'(m-4),...,c'(1 or 2)}.
> 
> 
> 
> So, you guessed it,
> 
> 
> each c(k) = c'(k).
> 
> Also,
> 
> c(1+m) = c(m)/ceiling(m/2) + c(m-1)(1 -1/ceiling(m/2)),
> 


I forgot to mention that the above is true ONLY for m >= 4.

I also will rewrite the above recursion as to be in-terms of only 
integers.

Let n(m) = c(m)*(floor((m-1)/2))!*(ceiling((m-1)/2))!.

So, then, for m >= 3,

n(m+2) = n(1+m) + n(m) *floor(m/2) *ceiling(m/2),

with n(3) = 2, n(4) = 3.

n(m)   (first term is n(3))
..2, 3, 7, 19, 61, ...

(Not in the EIS, actually.)

Anyone out there making progress on a closed-form for {c(m)} or {n(m)},
or on the limit of the c's?


> which should (easily?) lead to a closed-form for c(m).
> 
> But I have not the time now to explore further...
> 
> By the way, what is the closed-form for
> 
> limit{m-> oo} c(m),
> 
> anyway?
> 

thanks,
Leroy Quet