A214981 -id:A214981 - cp's OEIS frontend

A214979 A179180 - A214977.

0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 2, 1, 1, 1, 0, 0, 1, 2, 3, 2, 2, 1, 0, 1, 0, 0, 0, 1, 2, 3, 4, 6, 4, 3, 3, 2, 2, 1, 2, 2, 1, 1, 1, 0, 0, 1, 2, 3, 4, 6, 6, 7, 9, 7, 6, 5, 5, 5, 4, 4, 4, 2, 1, 2, 2, 3, 2, 2, 1, 0, 1, 0, 0, 0, 1, 2, 3, 4, 6, 6, 7, 9, 9, 10, 11, 13, 15, 13, 12, 11, 9, 8, 8, 8, 8

Offset: 1

Clark Kimberling, Oct 22 2012

nonn

Let U(n) and V(n) be the number of terms in the Lucas representations and Zeckendorf (Fibonacci) representations, respectively, of all the numbers 1,2,...,n. Then a(n) = V(n) - U(n). Conjecture: a(n) >= 0 for all n, and a(n) = 0 for infinitely many n.

			(See A214977 and A179180.)

Clark Kimberling, Table of n, a(n) for n = 1..10000
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.

Cf. A214977, A179180, A214980, A214981, A000032, A000045.

z = 200;
s = Reverse[Sort[Table[LucasL[n - 1], {n, 1, 70}]]];
t1 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2,1]], # > 0 &]] &, Range[z]];
u[n_] := Sum[t1[[k]], {k, 1, n}]
u1 = Table[u[n], {n, 1, z}] (* A214977 *)
s = Reverse[Table[Fibonacci[n + 1], {n, 1, 70}]];
t2 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2,1]], # > 0 &]] &, Range[z]];
v[n_] := Sum[t2[[k]], {k, 1, n}]
v1 = Table[v[n], {n, 1, z}]  (* A179180 *)
w=v1-u1 (* A214979 *)
Flatten[Position[w, 0]]  (* A214980 *)
(* Peter J. C. Moses, Oct 18 2012 *)

A214977 Number of terms in Lucas representations of 1,2,...,n.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 11, 13, 15, 16, 18, 20, 22, 24, 27, 30, 31, 33, 35, 37, 39, 42, 45, 47, 50, 53, 56, 57, 59, 61, 63, 65, 68, 71, 73, 76, 79, 82, 84, 87, 90, 93, 96, 100, 104, 105, 107, 109, 111, 113, 116, 119, 121, 124, 127, 130, 132, 135, 138, 141, 144

Offset: 1

Clark Kimberling, Oct 22 2012

nonn

See the conjecture at A214979.

			n..Lucas(n)..# terms...A214977(n)
1..1.........1.........1
2..2.........1.........2
3..3.........1.........3
4..4.........1.........4
5..4+1.......2.........6
6..4+2.......2.........8
7..7.........1.........9
8..7+1.......2.........11
9..7+2.......2.........13

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A214978, A214979, A214980, A214981, A000032.

z = 200; s = Reverse[Sort[Table[LucasL[n - 1], {n, 1, 70}]]]; t1 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2,1]], # > 0 &]] &, Range[z]]; u[n_] := Sum[t1[[k]], {k, 1, n}]; u1 = Table[u[n], {n, 1, z}]
(* Peter J. C. Moses, Oct 18 2012 *)

A214980 Positions of zeros in A214979.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 9, 10, 16, 17, 24, 26, 27, 28, 45, 46, 71, 73, 74, 75, 121, 122, 194, 196, 197, 198, 320, 321, 516, 518, 519, 520, 841, 842, 1359, 1361, 1362, 1363, 2205, 2206, 3566, 3568, 3569, 3570, 5776, 5777, 9344, 9346, 9347, 9348, 15125

Offset: 1

Clark Kimberling, Oct 22 2012

nonn

Let U(n) and V(n) be the number of terms in the Lucas representations and Zeckendorf (Fibonacci) representations, respectively, of all the numbers 1,2,...,n. Then A214980 is the sequence of zeros of the sequence A214979(n) = V(n) - U(n). It is conjectured at A214979 that A214980 is infinite.

Clark Kimberling, Table of n, a(n) for n = 1..58

Cf. A214977, A214978, A214979, A214981, A000032, A000045.

Mathematica
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(See the program at A214979.)
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A214979 A179180 - A214977.

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A214977 Number of terms in Lucas representations of 1,2,...,n.

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A214980 Positions of zeros in A214979.

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