Waldek Hebisch <[EMAIL PROTECTED]> writes:

> Martin Rubey wrote:

> > The hard part is to get the asymptotic at infinity :-)  As far as I know,
> > functions of the form
> > 
> > (a+b*n)^n*q(n)
> > 
> > (a,b are fractions, q a rational function) does not necessarily satisfy a
> > difference equation.
> > 
> 
> Hmm, if f(n) = (a+b*n)^nW_1(n)/W_2(n) and W_2(n+1) and W_2(n) are non-zero,
> then we have
> 
> W_2(n+1)*W_1(n)*f(n+1) = (a+n*n)*W_2(n)*W_1(n+1)*f(n)         (**)

No, I don't think so:

  (a+b*(n+1))^(n+1) / (a+b*n)^n is not rational.

Martin


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