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A356749
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a(n) is the number of trailing 1's in the dual Zeckendorf representation of n (A104326).
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4
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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, 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, 0, 3, 0, 2, 1, 0, 5, 0, 2, 1, 0, 4, 1, 0, 3, 0, 2, 1, 0
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OFFSET
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0,4
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COMMENTS
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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.
The asymptotic mean of this sequence is phi.
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LINKS
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EXAMPLE
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- ---- ----------
0 0 0
1 1 1
2 0 10
3 2 11
4 1 101
5 0 110
6 3 111
7 0 1010
8 2 1011
9 1 1101
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MATHEMATICA
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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]
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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