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A281185 a(0)=0, a(1)=1, a(2)=0; thereafter, a(2n) = a(n) + a(n+1) for n >= 2, a(2n+1) = a(n) for n >= 1.

%I A281185 #36 Jan 28 2022 01:34:33
%S A281185 0,1,0,1,1,0,2,1,1,1,2,0,3,2,2,1,2,1,3,1,2,2,3,0,5,3,4,2,3,2,3,1,3,2,
%T A281185 4,1,4,3,3,1,4,2,5,2,3,3,5,0,8,5,7,3,6,4,5,2,5,3,5,2,4,3,4,1,5,3,6,2,
%U A281185 5,4,5,1,7,4,6,3,4,3,5,1,6,4,7,2,7,5,5,2,6,3,8,3,5,5,8,0,13,8,12,5
%N A281185 a(0)=0, a(1)=1, a(2)=0; thereafter, a(2n) = a(n) + a(n+1) for n >= 2, a(2n+1) = a(n) for n >= 1.
%C A281185 A "bow" sequence. The bow sequences are a family of recursive sequences defined to have the flipped recursion from the Stern sequence A002487 (called bow for the opposite end of the boat from the stern). The bow sequences require two initial conditions: a(1)=alpha, a(2)=beta. We also define a(0)=0, although it does not enter into the recursion.
%C A281185 The bow sequences then follow the recursion a(2n) = a(n) + a(n+1) for n at least 2, and a(2n+1) = a(n). This particular bow sequence has initial conditions a(1)=0, a(2)=1 and (along with the sequence A106345 with initial conditions a(1)=1, a(2)=0) is of particular importance when studying the general bow sequences.
%H A281185 Rémy Sigrist, <a href="/A281185/b281185.txt">Table of n, a(n) for n = 0..25000</a>
%H A281185 M. Dennison, <a href="http://hdl.handle.net/2142/16821">A Sequence Related to the Stern Sequence</a>, Ph.D. dissertation, University of Illinois at Urbana-Champaign, 2010.
%H A281185 Melissa Dennison, <a href="https://www.emis.de/journals/JIS/VOL22/Dennison/dennis3.html">On Properties of the General Bow Sequence</a>, J. Int. Seq., Vol. 22 (2019), Article 19.2.7.
%e A281185 a(3) = a(1) = 1, a(4) = a(2) + a(3) = 0 + 1 = 1, a(5) = a(2) = 0.
%p A281185 f:=proc(n) option remember;
%p A281185 if n=0 then 0
%p A281185 elif n=1 then 1
%p A281185 elif n=2 then 0
%p A281185 else
%p A281185    if n mod 2 = 0 then f(n/2)+f(1+n/2) else f((n-1)/2) fi;
%p A281185 fi;
%p A281185 end;
%p A281185 [seq(f(n),n=0..150)]; # _N. J. A. Sloane_, Apr 26 2017
%t A281185 b[0]=0; b[1]=1; b[2]=0; b[n_?EvenQ]:=b[n]=b[n/2]+b[n/2+1]; b[n_?OddQ]:=b[n]=b[(n-1)/2]
%Y A281185 Cf. A002487, A106345.
%K A281185 nonn,look,easy
%O A281185 0,7
%A A281185 _Melissa Dennison_, Apr 12 2017
%E A281185 Edited by _N. J. A. Sloane_, Apr 26 2017