A267371 Limiting sequence formed by starting with 01 and iteratively concatenating existing string with the shortest prefix that appears only once.

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

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: A270742 A164349 A094186 * A285205 A286654 A003849

Adjacent sequences: A267368 A267369 A267370 * A267372 A267373 A267374

Keyword

nonn

Author

Jeffrey Shallit, Jan 13 2016

Status

approved