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%I #8 Feb 04 2020 08:12:34
%S 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,
%T 130,287,376,21,16,36,53,138,342,753,986,22,17,39,75,141,363,897,1973,
%U 2583,24,23,40,96,142,371,952,2350,5167,6764,25,26,44,104,198,374
%N Table read by antidiagonals: T(n, k) is the k-th number with n-1 odd-indexed Fibonacci numbers in its Zeckendorf representation.
%C 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.
%C We begin the indexing at 2; that is, 1=F(2), 2=F(3), 3=F(4), 5=F(5),...
%C Every positive integer occurs exactly once in the array, so that as a sequence it is a permutation of the positive integers.
%C For counts of even-indexed Fibonacci numbers, see A165278.
%C Essentially, (row 0)=A054204, (column 1)=A035508.
%e Northwest corner:
%e 1....3....4....8....9...11...12...21...22...
%e 2....5....6...10...13...14...16...17...23...
%e 7...15...18...19...28...36...39...40...44...
%e 20..41...49...52...53...75...96..104..107...
%e Examples:
%e 12=8+3+1=F(6)+F(4)+F(2), zero odds, so 12 is in row 0.
%e 28=21+5+2=F(8)+F(5)+F(3), two odds, so 28 is in row 2.
%t 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 *)
%Y Cf. A165276, A165277, A165278.
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Sep 13 2009
%E More terms from _Amiram Eldar_, Feb 04 2020