A000120; another recurrence?

[Marc LeBrun]

 >=Francois
 > L_k seems to be the first 2^k terms of sequence A000120.

(A000120(n) = 2[n]1 in rebase notation; I suppose I should submit an 
advertisement).

 > The question is, how can I go about proving this?

You might try an approach using the generating function,

   G.f.: 1/(1-x) * Sum(k>=0, x^(2^k)/(1+x^(2^k))).
   - Ralf Stephan (ralf(AT)ark.in-berlin.de), Apr 19 2003

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!