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A165278
Table read by antidiagonals: T(n, k) is the k-th number with n-1 even-indexed Fibonacci numbers in its Zeckendorf representation.
5
2, 5, 1, 7, 3, 4, 13, 6, 9, 12, 15, 8, 11, 25, 33, 18, 10, 17, 30, 67, 88, 20, 14, 22, 32, 80, 177, 232, 34, 16, 24, 46, 85, 211, 465, 609, 36, 19, 27, 59, 87, 224, 554, 1219, 1596, 39, 21, 29, 64, 122, 229, 588, 1452, 3193, 4180, 41, 23, 31, 66, 156, 231, 601
OFFSET
1,1
COMMENTS
For n>=0, row n is the monotonic sequence of positive integers m such that the number of even-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 odd-indexed Fibonacci numbers, see A165279.
Essentially, (row 0)=A062879, (column 1)=A027941, (column 2)=A069403.
EXAMPLE
Northwest corner:
2....5....7...13...15...18...20...34...36...
1....3....6....8...10...14...16...19...20...
4....9...11...17...22...24...27...29...31...
12..25...30...32...46...59...64...66...72...
Examples:
20=13+5+2=F(7)+F(5)+F(3), zero evens, so 20 is in row 0.
19=13+5+1=F(7)+F(5)+F(2), one even, so 19 is in row 1.
22=21+1=F(8)+F(2), two evens, so 22 is in row 2.
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Sep 13 2009
EXTENSIONS
More terms from Amiram Eldar, Feb 04 2020
STATUS
approved