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 A267371 Limiting sequence formed by starting with 01 and iteratively concatenating existing string with the shortest prefix that appears only once. 6
 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1 LINKS Michael S. Branicky, Table of n, a(n) for n = 1..10000 EXAMPLE Starting with 01, the shortest prefix occurring only once is 0, so the next string is (01)(0) = 010. After that it becomes (010)(01) = 01001, etc. MATHEMATICA preLo[{np_, s_}] := NestWhile[#+1&, np, StringCount[s, StringTake[s , #], Overlaps->True]>1&] a267371[n_] := Map[Boole[#=="1"]&, Characters[Last[Nest[{preLo[#], StringJoin[#[[2]], StringTake[#[[2]], preLo[#]]]}&, {1, "01"}, n]]]]a267371[13]] (* Hartmut F. W. Hoft, Mar 22 2024 *) PROG (Python) from itertools import count, islice def agen(): # generator of terms astr, k, mink = "01", 2, 1 while True: yield from map(int, astr[:k]) for k in range(1, len(astr)+1): if astr[1:].count(astr[:k]) == 0: break mink = max(mink, k) astr += astr[:k] print(list(islice(agen(), 133))) # Michael S. Branicky, Mar 23 2024 CROSSREFS Cf. A267372, A267373, A267374. Sequence in context: A373338 A164349 A094186 * A285205 A286654 A003849 Adjacent sequences: A267368 A267369 A267370 * A267372 A267373 A267374 KEYWORD nonn AUTHOR Jeffrey Shallit, Jan 13 2016 STATUS approved

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Last modified July 23 10:38 EDT 2024. Contains 374547 sequences. (Running on oeis4.)