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A143299
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Number of terms in the Zeckendorf representation of every number in row n of the Wythoff array.
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1
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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
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1,2
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COMMENTS
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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
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LINKS
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EXAMPLE
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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.
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)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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