%I #11 Aug 26 2022 07:28:39
%S 0,1,0,2,1,0,3,0,2,1,0,4,1,0,3,0,2,1,0,5,0,2,1,0,4,1,0,3,0,2,1,0,6,1,
%T 0,3,0,2,1,0,5,0,2,1,0,4,1,0,3,0,2,1,0,7,0,2,1,0,4,1,0,3,0,2,1,0,6,1,
%U 0,3,0,2,1,0,5,0,2,1,0,4,1,0,3,0,2,1,0
%N a(n) is the number of trailing 1's in the dual Zeckendorf representation of n (A104326).
%C The asymptotic density of the occurrences of k = 0, 1, 2, ... is 1/phi^(k+2), where phi = 1.618033... (A001622) is the golden ratio.
%C The asymptotic mean of this sequence is phi.
%H Amiram Eldar, <a href="/A356749/b356749.txt">Table of n, a(n) for n = 0..10000</a>
%e n a(n) A104326(n)
%e - ---- ----------
%e 0 0 0
%e 1 1 1
%e 2 0 10
%e 3 2 11
%e 4 1 101
%e 5 0 110
%e 6 3 111
%e 7 0 1010
%e 8 2 1011
%e 9 1 1101
%t fb[n_] := Module[{k = Ceiling[Log[GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; fr]; f[v_] := Module[{m = Length[v], k}, k = m; While[v[[k]] == 1, k--]; m - k]; a[n_] := Module[{v = fb[n]}, nv = Length[v]; i = 1; While[i <= nv - 2, If[v[[i ;; i + 2]] == {1, 0, 0}, v[[i ;; i + 2]] = {0, 1, 1}; If[i > 2, i -= 3]]; i++]; i = Position[v, _?(# > 0 &)]; If[i == {}, 0, f[v[[i[[1, 1]] ;; -1]]]]]; Array[a, 100, 0]
%Y Cf. A001622, A104326.
%Y Similar sequences: A003849, A035614, A276084, A278045.
%K nonn,base
%O 0,4
%A _Amiram Eldar_, Aug 25 2022