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A080426
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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.
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10
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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, 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, 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
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OFFSET
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1,2
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COMMENTS
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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.
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
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
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LINKS
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FORMULA
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MATHEMATICA
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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 *)
nsteps=7; Flatten[SubstitutionSystem[{1->{3}, 3->{1, 1, 3}}, {1}, nsteps]] (* Paolo Xausa, Aug 12 2022, using D. R. Hofstadter's construction *)
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PROG
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(Haskell) -- following Deléham
import Data.List (group)
a080426 n = a080426_list !! n
a080426_list = map length $ filter ((== 1) . head) $ group a035263_list
(PARI)
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];
(Python)
a, s = "1", "".maketrans({"1":"131", "3":"13331"})
while len(a) < nmax: a = a.translate(s)
return list(map(int, a[:nmax]))
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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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