A092053: Value of Continued Fraction [1;1/2,1/3,1/4,...,1/n,...]

[Joseph Biberstine]

	It seems that by defining h[k] as the "numerator" (note in our case it
won't be restricted to integer outputs) of the k-th convergent to x in
the traditional fashion (h[-2]=0, h[-1]=1, h[n]=h[n-1]*a[n] + h[n-2] and
a[n] in the n-th partial quotient of x, in our case h[n-1]/(n+1) +
h[n-2]), we find:

	h[2*k-1] = h[2*k] = A1803[k-1]/2^A5187[k]

	Unfortunately I cannot find any such apparent patterns in h[]'s brother
function, k[].  Hope this provides any insight.

-JRB

Paul D. Hanna wrote:
> Seqfans, 
> Can anyone evaluate the continued fraction:
>    x = [1;1/2,1/3,1/4,...,1/n,...].
>  
> Convergence is very slow: 
> at 500000 partial quotients, x = 1.751943215111853159301... ;
> at 600000 partial quotients, x = 1.751942411573727042118...
>  
> The convergents of the CF are interesting, and are related to Pi/2.
>  
> See A092053: 
> "Denominators of the convergents of the continued fraction expansion
> [1;1/2,1/3,1/4,...,1/n,...]."
> COMMENT: 
> "Numerators of convergents are A001902 (successive denominators 
> of Wallis's product approximation to Pi/2). 
> Sum of numerators and denominators equals powers of 2: 
> A001902(n) + a(n) = 2^A092054(n)."
>  
> Thanks, 
>    Paul