asymptotic formula for A006154

[A.N.W.Hone at kent.ac.uk]

Hi everybody - 

I was interested in this question. One can easily get the leading order part of 
the asymptotics  from Cauchy's root test, so that a(n)/n!~ (1/R)^n * (subexponential part), 
where R=log(1+sqrt(2)) is the radius of convergence of the series. The subexponential 
part is here just a constant 1/(R*cosh(R))=sqrt(2)/2/log(1+sqrt(2)) which indeed gives 

a(n) ~ sqrt(2)/2*n!/log(1+sqrt(2))^(n+1). 

To find this part, one can certainly use Cauchy's integral formula for the coefficients, but 
it requires a bit more work. I used a result (based on this formula) from an article by Philippe Flajolet, Symbolic Enumerative Combinatorics and Complex Asymptotic Analysis, Algorithms Seminar 2000-2001, F. Chyzak (ed.), INRIA, (2002), pp. 161-170 [see Theorem 5 and Table 2, noting that 1/(1-sinh(x)) just has a simple pole at x=R]. Rather than using this general result, 
for this function it might be possible to make a simple choice of contour and get it more directly. 

Anyway, perhaps this reference will be very useful for other seqfans who want to work out the 
asymptotics of a sequence from its g.f./e.g.f. It is available from the INRIA website. 

All the best, 
Andy 
 




----- Original Message -----
From: Max Alekseyev <maxale at gmail.com>
Date: Sunday, February 18, 2007 7:33 pm
Subject: Re: asymptotic formula for A006154
To: ralf at ark.in-berlin.de, Simon Plouffe <simon.plouffe at gmail.com>, seqfan at ext.jussieu.fr

> On 2/17/07, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
> > > So the entry would be
> > >
> > > %F A006154 a(n) ~ sqrt(2)/2*(n-1)!/log(1+sqrt(2))^n.
> >
> > Does someone have or know of a proof?
> 
> It should easy follow from the expression of Laurent series
> coefficients as contour integrals, and Cauchy integral formula:
> http://mathworld.wolfram.com/LaurentSeries.html
> http://mathworld.wolfram.com/CauchyIntegralFormula.html
> 
> In particular, log(1+sqrt(2)) is a pole for function 1/(1-sinh(x)).
> 
> Max
> 
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