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

Author

Clark Kimberling, Aug 05 2008

Keywords

• nonn

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

Examples

			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)
		

Links

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

Crossrefs

Cf. A003622, A007895, A035513, A134561.

Programs

Mathematica

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

Formula

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