%I
%S 1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
%T 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
%U 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N a(0)=a(1)=a(2)=1, a(n)=0 for n>2.
%C With different signs this sequence is the convolutional inverse of the Fibonacci sequence: 1, 1, 1, 0, 0, ...  _Tanya Khovanova_, Jul 14 2007
%C Inverse binomial transform of A000124.  _R. J. Mathar_, Jun 13 2008
%C Partial sums give A158799. [_Jaroslav Krizek_, Dec 06 2009]
%H Andrei Asinowski, Cyril Banderier, Valerie Roitner, <a href="https://lipn.univparis13.fr/~banderier/Papers/several_patterns.pdf">Generating functions for lattice paths with several forbidden patterns</a>, (2019).
%F Given g.f. A(x), then B(a) = A(q) / q satisfies 0 = f(B(q), B(q^2)) where f(u, v) = v  u * (u  2).  _Michael Somos_, Oct 22 2013
%F Euler transform of length 3 sequence [ 1, 0, 1].  _Michael Somos_, Oct 22 2013
%F G.f. is third cyclotomic polynomial.
%F G.f.: (1  x^3) / (1  x).
%F Convolution inverse is A049347.  _Michael Somos_, Oct 22 2013
%e G.f. = 1 + x + x^2.
%e G.f. = 1/q + 1 + q.
%t a[ n_] := Boole[ n>=0 && n<=2]; (* _Michael Somos_, Oct 22 2013 *)
%o (PARI) {a(n) = n>=0 && n<=2}; /* _Michael Somos_, Oct 22 2013 */
%Y Cf. A049347.
%K easy,nonn
%O 0,1
%A _Paul Curtz_ and _Tanya Khovanova_, Jul 01 2007
