generatingfunctions

[Ralf Stephan]

[On the original question by Clark Kimberling about patterns in the
transform 
  - Let S the set of numbers in the integer sequence generated by the ogf
    (or other gf, the poster did not reply to an eMail asking for this)
	A(z), what is the ogf generating the inverse series, i.e., the set
	{integers} without S?]
	
> >>Mitch Harris:
> >>> >Possibly someone knows of references to special cases?
> >>>
> >>> I think this is hard in general: take r(z) = 1/(1-2z), so r(n) = 2^n. I
> >>> don't know of a 'good' (rational function or even special function)
> >>> representation of the ogf
> >>>
> >>>   \sum z^n [n=2^k]
> >>
> >>[log2(n)]-[log2(n-1)], n>1, would be cheating?
>... 
> but what is the gf of floor(log n)? there's the recurrence, but no closed
> form for it (that I know of).

I only see infinite sums here. 'Closed form' usually means none
of these are present? It seems uninteresting to consider
D'(z) = 1/(1-x) - sum(k>0, x^2^k) which generates
{1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1...}, or do you get anywhere from
there?

The alternatives I see are on the other hand developing log2 into an
infinite sum. Using partial sums of this, then taking lcm of denominators,
and fiddle with removing of floor, finally having at most an infinite
sum of rational ogfs, perhaps of certain well-known polynomials?

On the third hand, there are other gfs wide over my head.


ralf