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A306801
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An irregular fractal sequence: underline a(n) iff [a(n-1) + a(n)] is divisible by 3; all underlined terms rebuild the starting sequence.
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1
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1, 3, 2, 1, 4, 6, 3, 5, 8, 9, 7, 2, 1, 10, 12, 11, 4, 13, 15, 6, 3, 14, 17, 18, 16, 5, 20, 21, 19, 8, 23, 24, 9, 22, 25, 27, 26, 7, 2, 1, 28, 30, 29, 10, 31, 33, 12, 32, 35, 36, 34, 11, 4, 37, 39, 38, 13, 40, 42, 15, 6, 3, 41, 44, 45, 43, 14, 47, 48, 46, 17, 50, 51, 18, 49, 52, 54, 53, 16, 5, 56, 57, 55, 20, 59, 60, 21, 58, 61, 63
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OFFSET
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1,2
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COMMENTS
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The sequence S starts with a(1) = 1 and a(2) = 3. S is extended by duplicating the first term A among the not yet duplicated terms, under the condition that [A + the last term Z of the sequence] is divisible by 3. If this is not the case, we then extend S with the smallest integer X not yet present in S such that [X + the last term Z of the sequence] is not divisible by 3. This is the lexicographically first sequence with this property.
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LINKS
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EXAMPLE
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S starts with a(1) = 1 and a(2) = 3.
Can we duplicate a(1) to form a(3)? No, as a(2) + a(3) would be 4 and 4 is not divisible by 3; we thus extend S with the smallest integer X not yet in S such that [X + a(2)] is not divisible by 3. We get X = 2 and thus a(3) = 2.
Can we duplicate a(1) to form a(4)? Yes, as now [a(1) + a(3)] is divisible by 3; we get thus a(4) = 1.
Can we duplicate a(2) to form a(5)? No, as a(4) + a(2) would be 4 and 4 is not divisible by 3; we thus extend S with the smallest integer X not yet in S such that [X + a(4)] is not divisible by 3. We get X = 4 and thus a(5) = 4.
Can we duplicate a(2) to form a(6)? No, as a(5) + a(2) would be 7 and 7 is not divisible by 3; we thus extend S with the smallest integer X not yet in S such that [X + a(5)] is not divisible by 3. We get X = 6 and thus a(6) = 6.
Can we duplicate a(2) to form a(7)? Yes, as now [a(2) + a(6)] is divisible by 3; we get thus a(7) = 3.
Can we duplicate a(3) to form a(8)? No, as a(7) + a(3) would be 5 and 5 is not divisible by 3; we thus extend S with the smallest integer X not yet in S such that [X + a(6)] is not divisible by 3. We get X = 6 and thus a(8) = 5.
Etc.
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CROSSREFS
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Cf. A122196 (which is obtained by replacing 3 by 2 in the definition of this sequence).
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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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