Asymptote of A003242

[Alec Mihailovs]

From: <hv at crypt.org>

> By my calculation, A003242(n) should tend to k^n, where k is given by:
>  1 = sum_i=1^inf{ 1/(k^i + 1) }

I'm not sure that it is helpful, but I noticed the following (easy to prove) 
identity,

sum(1/(k^i + 1), i=1..infinity)=-sum((-1)^i/(k^i - 1), i=1..infinity)

Alec Mihailovs 




The Knopmacher/Prodinger paper cited in the the sequence mentions the
aymptotics giving the formula:

c(n) ~ A*rho^(-n) = (0.456387)*(1.750243)^n

It does not give a closed form for these values, but it relates them to
generating functions.

rho is the dominant pole of sigma(z)
A=1/(rho*sigma'(rho))

The link to the paper is out of date. Use:
http://web.wits.ac.za/Academic/Science/Maths/Staff/Knopfmacher/Publications.htm

It's the last paper in the 1992-1998 section.

Christian

------ Original Message ------

> By my calculation, A003242(n) should tend to k^n, where k is given by:
>   1 = sum_i=1^inf{ 1/(k^i + 1) }
> 
> By numerical methods I find k ~= 1.750241291718309, can anyone suggest
> how to solve that equation to get a closed form?
> 
...