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%I #80 Sep 28 2022 08:13:35
%S 1,3,1,1,3,3,3,1,1,3,1,1,3,1,1,3,3,3,1,1,3,3,3,1,1,3,3,3,1,1,3,1,1,3,
%T 1,1,3,3,3,1,1,3,1,1,3,1,1,3,3,3,1,1,3,1,1,3,1,1,3,3,3,1,1,3,3,3,1,1,
%U 3,3,3,1,1,3,1,1,3,1,1,3,3,3,1,1,3,3,3,1,1,3,3,3,1,1,3,1,1,3,1,1,3,3,3,1,1
%N a(1)=1, a(2)=3; all terms are either 1 or 3; each run of 3's is followed by a run of two 1's; and a(n) is the length of the n-th run of 3's.
%C It appears that the sequence can be calculated by any of the following three methods: (1) Start with 1 and repeatedly replace (simultaneously) all 1's with 1,3,1 and all 3's with 1,3,3,3,1. [Equivalently, trajectory of 1 under the morphism 1 -> 1,3,1; 3 -> 1,3,3,3,1. - _N. J. A. Sloane_, Nov 03 2019] (2) a(n)= A026490(2n). (3) Replace each 2 in A026465 (run lengths in Thue-Morse) with 3.
%C Length of n-th run of 1's in the Feigenbaum sequence A035263 = 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, .... - _Philippe Deléham_, Apr 18 2004
%C Another construction. Let S_0 = 1, and let S_n be obtained by applying the morphism 1 -> 3, 3 -> 113 to S_{n-1}. The sequence is the concatenation S_0, S_1, S_2, ... - D. R. Hofstadter, Oct 23 2014
%C a(n+1) is the number of times n appears in A003160. - _John Keith_, Dec 31 2020
%H Reinhard Zumkeller, <a href="/A080426/b080426.txt">Table of n, a(n) for n = 1..10000</a>
%H D. R. Hofstadter, <a href="/A075326/a075326_1.pdf">Anti-Fibonacci numbers</a>, Oct 23 2014.
%F a(1) = 1; for n>1, a(n) = A003156(n) - A003156(n-1). - _Philippe Deléham_, Apr 16 2004
%t Position[ Nest[ Flatten[# /. {0 -> {0, 2, 1}, 1 -> {0}, 2 -> {0}}]&, {0}, 8], 0] // Flatten // Differences // Prepend[#, 1]& (* _Jean-François Alcover_, Mar 14 2014, after _Philippe Deléham_ *)
%t nsteps=7;Flatten[SubstitutionSystem[{1->{3},3->{1,1,3}},{1},nsteps]] (* _Paolo Xausa_, Aug 12 2022, using D. R. Hofstadter's construction *)
%o (Haskell) -- following Deléham
%o import Data.List (group)
%o a080426 n = a080426_list !! n
%o a080426_list = map length $ filter ((== 1) . head) $ group a035263_list
%o -- _Reinhard Zumkeller_, Oct 27 2014
%o (PARI)
%o A080426(nmax) = my(a=[1], s=[[1, 3, 1], [], [1, 3, 3, 3, 1]]); while(length(a)<nmax, a=concat(vecextract(s,a))); a[1..nmax];
%o A080426(100) \\ _Paolo Xausa_, Sep 14 2022, using method (1) from comments
%o (Python)
%o def A080426(nmax):
%o a, s = "1", "".maketrans({"1":"131", "3":"13331"})
%o while len(a) < nmax: a = a.translate(s)
%o return list(map(int, a[:nmax]))
%o print(A080426(100)) # _Paolo Xausa_, Aug 30 2022, using method (1) from comments
%Y Cf. A026465, A026490, A035263, A003156, A328979, A003160.
%Y Arises in the analysis of A075326, A249031 and A249032.
%K nonn
%O 1,2
%A _John W. Layman_, Feb 18 2003
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