%I
%S 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,
%T 177,232,34,16,24,46,85,211,465,609,36,19,27,59,87,224,554,1219,1596,
%U 39,21,29,64,122,229,588,1452,3193,4180,41,23,31,66,156,231,601
%N Table read by antidiagonals: T(n, k) is the kth number with n1 evenindexed 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 evenindexed 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 oddindexed Fibonacci numbers, see A165279.
%C Essentially, (row 0)=A062879, (column 1)=A027941, (column 2)=A069403.
%e Northwest corner:
%e 2....5....7...13...15...18...20...34...36...
%e 1....3....6....8...10...14...16...19...20...
%e 4....9...11...17...22...24...27...29...31...
%e 12..25...30...32...46...59...64...66...72...
%e Examples:
%e 20=13+5+2=F(7)+F(5)+F(3), zero evens, so 20 is in row 0.
%e 19=13+5+1=F(7)+F(5)+F(2), one even, so 19 is in row 1.
%e 22=21+1=F(8)+F(2), two evens, so 22 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, A165279.
%K nonn,tabl
%O 1,1
%A _Clark Kimberling_, Sep 13 2009
%E More terms from _Amiram Eldar_, Feb 04 2020
