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A165279
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Table read by antidiagonals: T(n, k) is the k-th number with n-1 odd-indexed Fibonacci numbers in its Zeckendorf representation.
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5
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1, 3, 2, 4, 5, 7, 8, 6, 15, 20, 9, 10, 18, 41, 54, 11, 13, 19, 49, 109, 143, 12, 14, 28, 52, 130, 287, 376, 21, 16, 36, 53, 138, 342, 753, 986, 22, 17, 39, 75, 141, 363, 897, 1973, 2583, 24, 23, 40, 96, 142, 371, 952, 2350, 5167, 6764, 25, 26, 44, 104, 198, 374
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OFFSET
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1,2
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COMMENTS
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For n>=0, row n is the monotonic sequence of positive integers m such that the number of odd-indexed Fibonacci numbers in the Zeckendorf representation of m is n.
We begin the indexing at 2; that is, 1=F(2), 2=F(3), 3=F(4), 5=F(5),...
Every positive integer occurs exactly once in the array, so that as a sequence it is a permutation of the positive integers.
For counts of even-indexed Fibonacci numbers, see A165278.
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LINKS
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EXAMPLE
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Northwest corner:
1....3....4....8....9...11...12...21...22...
2....5....6...10...13...14...16...17...23...
7...15...18...19...28...36...39...40...44...
20..41...49...52...53...75...96..104..107...
Examples:
12=8+3+1=F(6)+F(4)+F(2), zero odds, so 12 is in row 0.
28=21+5+2=F(8)+F(5)+F(3), two odds, so 28 is in row 2.
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MATHEMATICA
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f[n_] := Module[{i = Ceiling[Log[GoldenRatio, Sqrt[5]*n]], v = {}, m = n}, While[i > 1, If[Fibonacci[i] <= m, AppendTo[v, 1]; m -= Fibonacci[i], If[v != {}, AppendTo[v, 0]]]; i--]; Total[Reverse[v][[1 ;; -1 ;; 2]]]]; T = GatherBy[SortBy[ Range[10^4], f], f]; Table[Table[T[[n - k + 1, k]], {k, n, 1, -1}], {n, 1, Length[T]}] // Flatten (* Amiram Eldar, Feb 04 2020 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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