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A293843
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Split the infinite binary word A030302, from left to right, into the largest possible cubefree chunks; a(n) = length of n-th cubefree chunk.
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3
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5, 7, 4, 4, 10, 3, 9, 5, 2, 3, 5, 12, 9, 10, 2, 3, 7, 9, 2, 5, 4, 2, 4, 2, 4, 2, 4, 6, 6, 3, 8, 15, 6, 12, 5, 2, 14, 2, 6, 2, 4, 6, 3, 11, 13, 8, 2, 2, 4, 3, 5, 7, 4, 2, 6, 5, 2, 5, 2, 2, 3, 2, 5, 2, 5, 7, 4, 3, 7, 7, 7, 3, 7, 15, 7, 19, 7, 2, 5, 7, 11, 5, 5
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OFFSET
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1,1
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COMMENTS
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Using A030190 instead of A030302 leads to the same sequence except for the first term (that would equal 6).
The word A030302 contains infinitely many consecutive triples of 0's (corresponding for example to the last binary digits of multiples of 8), hence A030302 has no infinite cubefree suffix, and this sequence is well defined for any n > 0.
a(n) >= 2 for any n > 0.
This sequence is unbounded: for any n > 0:
- hence we can choose a number in A286262, say c, with 2*n digits in binary,
- and the subword of A030302 corresponding to c will participate in no more than two cubefree chunks,
- and one of those chunks will have at least length n, QED.
a(n) n
---- --
5 1
7 2
10 5
12 12
15 32
19 76
21 212
25 412
35 418
36 2305
39 5118
47 5516
59 49014
63 104902
67 261530
71 478638
75 1016483
79 2148745
83 4532050
87 9534639
91 20011894
95 41896466
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LINKS
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EXAMPLE
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The following table shows the first terms of the sequence, alongside the corresponding cubefree chunks:
n a(n) n-th chunk
-- ---- ----------
1 5 11011
2 7 1001011
3 4 1011
4 4 1100
5 10 0100110101
6 3 011
7 9 110011011
8 5 11011
9 2 11
10 3 100
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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