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

%I A143299 #17 Jun 17 2024 15:46:17
%S A143299 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 A143299 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 A143299 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 A143299 Number of terms in the Zeckendorf representation of every number in row n of the Wythoff array.
%C A143299 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 A143299 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
%H A143299 Paolo Xausa, <a href="/A143299/b143299.txt">Table of n, a(n) for n = 1..10946</a>
%F A143299 a(n) = A007895(n-1) + 1. - _Paolo Xausa_, Jun 17 2024
%e A143299 Row 5 of the Wythoff array is (12, 20, 32, ...) and corresponding Zeckendorf representations all have 3 terms:
%e A143299 12 = 1 + 3 + 8,
%e A143299 20 = 2 + 5 + 13,
%e A143299 32 = 3 + 8 + 21, etc.
%e A143299 From _Joerg Arndt_, May 14 2011: (Start)
%e A143299 The sequence as an irregular triangle:
%e A143299 1,        = M(1)
%e A143299 1,        = M(2)
%e A143299 1, 2,     = M(3) = M(2).f(M(1))
%e A143299 1, 2, 2,  = M(4) = M(3).f(M(2))
%e A143299 1, 2, 2, 2, 3,
%e A143299 1, 2, 2, 2, 3, 2, 3, 3,
%e A143299 1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4,
%e A143299 1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 2, 3, 3, 3, 4, 3, 4, 4,
%e A143299 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 A143299 (End)
%t A143299 Flatten[Nest[{Flatten[#], #[[1]] + 1} &, {1, 2}, 9]] (* _Paolo Xausa_, Jun 17 2024 *)
%Y A143299 Cf. A003622, A007895, A035513, A134561.
%K A143299 nonn
%O A143299 1,2
%A A143299 _Clark Kimberling_, Aug 05 2008