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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

%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 k-th number with n-1 even-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 even-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 odd-indexed 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

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Last modified April 3 23:48 EDT 2020. Contains 333207 sequences. (Running on oeis4.)