A000120; another recurrence?

N. J. A. Sloane njas at research.att.com
Thu May 25 02:03:43 CEST 2006


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?
Neil





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