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A352745 a(n) is the number of Lyndon factors of the Fibonacci string of length n.

%I A352745 #16 Aug 06 2024 11:33:25
%S A352745 1,1,1,2,2,3,4,4,6,5,8,6,10,7,12,8,14,9,16,10,18,11,20,12,22,13,24,14,
%T A352745 26,15,28,16,30,17,32,18,34,19,36,20,38,21,40,22,42,23,44
%N A352745 a(n) is the number of Lyndon factors of the Fibonacci string of length n.
%C A352745 The Fibonacci string of length n is defined Fibonacci(n) = cat(Fibonacci(n - 1), Fibonacci(n - 2)) for 1 < n and the initial conditions Fibonacci(0) = "1" and Fibonacci(1) = "0", where 'cat' is the operation of concatenating strings. The length of Fibonacci(n) is A352744(1, n). The sequence starts: "1", "0", "01", "010", "01001", "01001010", ...
%C A352745 Apart from the first four terms seems to be identical with A117248.
%H A352745 Guy Melançon, <a href="https://citeseerx.ist.psu.edu/pdf/5b369eca3b04c04ae429b9de318e246d2e9b9fbc">Lyndon factorization of infinite words</a>, STACS 96 (Grenoble, 1996), 147-154, Lecture Notes in Comput. Sci., 1046, Springer, Berlin, 1996.
%H A352745 Wikipedia, <a href="https://en.wikipedia.org/wiki/Lyndon_word">Lyndon word</a>
%e A352745 The Lyndon factorization of the Fibonacci strings of length n = 0..9.
%e A352745 [0] ["1"]
%e A352745 [1] ["0"]
%e A352745 [2] ["01"]
%e A352745 [3] ["01", "0"]
%e A352745 [4] ["01", "001"]
%e A352745 [5] ["01", "00101", "0"]
%e A352745 [6] ["01", "00101", "001", "001"]
%e A352745 [7] ["01", "00101", "0010010100101", "0"]
%e A352745 [8] ["01", "00101", "0010010100101", "00100101", "001", "001"]
%e A352745 [9] ["01", "00101", "0010010100101", "0010010100100101001010010010100101", "0"]
%p A352745 with(StringTools): A352745 := n -> nops(LyndonFactors(Fibonacci(n))):
%p A352745 seq(A352745(n), n = 0..12);
%Y A352745 Cf. A000045, A014707, A074650, A117248, A211100, A352746, A352744.
%K A352745 nonn,more
%O A352745 0,4
%A A352745 _Peter Luschny_, Apr 06 2022