A000120; another recurrence?

[FJ]

--- On Wed 05/24, Joshua Zucker < joshua.zucker at gmail.com > wrote:
I think that all your notation below boils down to "shifting the terms by 2^k adds 1 to all of them", which seems pretty clear to me!

 --- On Wed 05/24, Marc LeBrun < mlb at fxpt.com > wrote:
Here's a suggestion you might try: Shifting corresponds to multiplying the GF or those polynomials by powers of x, and concatenating just requires shifting them enough and then adding them. Your construction should be equivalent to a set of identities involving the GF that depend on its "self similarity" under these operations. Then you might perhaps prove it valid in the simplest case and extend it by induction. Hope that helps, good luck!

 --- On Wed 05/24, N. J. A. Sloane < njas at research.att.com > wrote:
Francois, isn't your result pretty obvious - and provable by induction, using that fact that the next 2^m binary numbers are obtained by putting a 1 in front of the first 2^m binary numbers?

Thank you all for your different perspectives.  Not being a professional mathematician, I found them very educational.  I guess it does look pretty obvious when you boil away all that notation, doesn't it?  And the GF approach to shifting is not something I would have thought of.  Thanks!

-Francois.

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