[[seqfan] Few INVERT transforms requested. A000045 --> A000129, how ?]

[Christian G. Bower]

Like John Laymen, I'm not using the Maple procedure, but rather my own
program

Antti Karttunen <Antti.Karttunen at iki.fi> wrote:

> 
> Dear Maple-users!
> 
> If any of you have Maple and Neil's transform-package easily at hand
> (see http://www.research.att.com/~njas/sequences/transforms.txt )
> could you compute for me the following transforms:
> 
> INVERT([1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]);
> 
> Here the sequence in brackets is A019590, and the result SHOULD BE A000045,
> the Fibonacci Numbers, which is obvious when one thinks about
> the combinatorial interpretation of INVERT, as explained in:
> http://www.research.att.com/~njas/doc/eigen.ps
> or:
> http://www.research.att.com/~njas/doc/eigen.pdf
> 
> (but I'm not sure about by my thoughts until I see the experimental data!)
> 
> and also this:
> 
>
INVERT([1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946]);
>  (i.e. our Dear  Fibonaccis)
> 
> and with different offsets:
> 
>
INVERT([0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946]);

I get:
A000129 Pell Numbers

> 
>
INVERT([1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946]);

I get:
A007484 E(2,7)
> 
> Should some of these result A000129?
> And if you can prove it with g.f.ology or with some combinatorial identity,
> then even better.

Don't have time to do proofs right now. Since Pell numbers are defined by
simple recurrence, should be easy to convert to a g.f. and thus show
equivalence to the transformations.

Christian