

A143299


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


1



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
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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)=n1+Floor(n*tau), where tau=(1+sqrt(5))/2.
Let M(1) = 1, M(2) = 2 and for n >= 3, M(n) = M(n1).f(M(n2)) 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

Table of n, a(n) for n=1..105.


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)


CROSSREFS

Cf. A003622, A035513, A134561.
Sequence in context: A281856 A106696 A131839 * A239428 A257497 A266225
Adjacent sequences: A143296 A143297 A143298 * A143300 A143301 A143302


KEYWORD

nonn


AUTHOR

Clark Kimberling, Aug 05 2008


STATUS

approved



