Continued-Fraction-Generated Sequence Also Equals...

[Leroy Quet]

[also posted to sci.math]

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 (positive 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)),

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