A143299 Number of terms in the Zeckendorf representation of every number in row n of the Wythoff array.

1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 2, 3, 3, 3, 4, 3, 4, 4, 2, 3, 3, 3, 4, 3, 4, 4, 3, 4, 4, 4, 5, 2, 3, 3, 3, 4, 3, 4, 4, 3, 4, 4, 4, 5, 3, 4, 4, 4, 5, 4, 5, 5, 2, 3, 3, 3, 4, 3, 4, 4, 3, 4, 4, 4, 5, 3, 4, 4, 4, 5, 4, 5, 5, 3, 4, 4, 4, 5, 4, 5, 5, 4, 5, 5, 5, 6, 2, 3, 3, 3, 4, 3, 4, 4, 3, 4, 4, 4, 5, 3, 4, 4

Offset

1,2

Comments

Every number in a row of the Wythoff array has the same number of Zeckendorf summands as the first number in the row; hence A035513(n) is the number of Zeckendorf summands of A003622(n)=n-1+Floor(n*tau), where tau=(1+sqrt(5))/2.

Let M(1) = 1, M(2) = 2 and for n >= 3, M(n) = M(n-1).f(M(n-2)) where f() increments by one and the dot stands for concatenation, then this sequence is 0.M(1).M(2).M(3).M(4)... , see the example. - Joerg Arndt, May 14 2011

Links

Paolo Xausa, Table of n, a(n) for n = 1..10946

Formula

a(n) = A007895(n-1) + 1. - Paolo Xausa, Jun 17 2024

Example

Row 5 of the Wythoff array is (12, 20, 32, ...) and corresponding Zeckendorf representations all have 3 terms:
12 = 1 + 3 + 8,
20 = 2 + 5 + 13,
32 = 3 + 8 + 21, etc.
From Joerg Arndt, May 14 2011: (Start)
The sequence as an irregular triangle:
1,        = M(1)
1,        = M(2)
1, 2,     = M(3) = M(2).f(M(1))
1, 2, 2,  = M(4) = M(3).f(M(2))
1, 2, 2, 2, 3,
1, 2, 2, 2, 3, 2, 3, 3,
1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4,
1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 2, 3, 3, 3, 4, 3, 4, 4,
1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 2, 3, 3, 3, 4, 3, 4, 4, 2, 3, 3, 3, ...
(End)

Mathematica

Flatten[Nest[{Flatten[#], #[[1]] + 1} &, {1, 2}, 9]] (* Paolo Xausa, Jun 17 2024 *)

Crossrefs

Cf. A003622, A007895, A035513, A134561.

Sequence in context: A106696 A131839 A373887 * A239428 A257497 A266225

Adjacent sequences: A143296 A143297 A143298 * A143300 A143301 A143302

Keyword

nonn

Author

Clark Kimberling, Aug 05 2008

Status

approved