Previous name was : For N=Sum a(i).2^i, a(i)=0,1, set D(N)/D(2)=Sum i.a(i).2^(i-1); sequence gives n-th derivative of 16.
Actually, it appears that f(n) for n=0, 1, 2, ... gives 0, 0, 1, 1, 4, 4, 5, 5, 12, 12, 13, 13, 16, 16, 17, 17, 32, 32, 33, 33, 36, ... that is
A136013. -
Michel Marcus, Jul 18 2013 [This now follows from the comments in
A136013. -
Pontus von Brömssen, Sep 06 2020]