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%I #8 Mar 09 2020 15:30:04
%S 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,
%T 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,
%U 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
%N Number of terms in the Zeckendorf representation of every number in row n of the Wythoff array.
%C 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.
%C 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
%e Row 5 of the Wythoff array is (12, 20, 32, ...) and corresponding Zeckendorf representations all have 3 terms:
%e 12 = 1 + 3 + 8,
%e 20 = 2 + 5 + 13,
%e 32 = 3 + 8 + 21, etc.
%e From _Joerg Arndt_, May 14 2011: (Start)
%e The sequence as an irregular triangle:
%e 1, = M(1)
%e 1, = M(2)
%e 1, 2, = M(3) = M(2).f(M(1))
%e 1, 2, 2, = M(4) = M(3).f(M(2))
%e 1, 2, 2, 2, 3,
%e 1, 2, 2, 2, 3, 2, 3, 3,
%e 1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4,
%e 1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 2, 3, 3, 3, 4, 3, 4, 4,
%e 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, ...
%e (End)
%Y Cf. A003622, A035513, A134561.
%K nonn
%O 1,2
%A _Clark Kimberling_, Aug 05 2008
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