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A306590
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An irregular fractal sequence: underline all terms that follow a comma separating two digits of opposite parities. All underlined terms rebuild the starting sequence. (See the Comments section for more details.)
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1
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1, 3, 5, 7, 9, 10, 1, 11, 12, 3, 13, 14, 5, 15, 16, 7, 17, 18, 9, 19, 30, 10, 1, 31, 32, 11, 33, 34, 12, 3, 35, 36, 13, 37, 38, 14, 5, 39, 50, 15, 51, 52, 16, 7, 53, 54, 17, 55, 56, 18, 9, 57, 58, 19, 59, 70, 30, 10, 1, 71, 72, 31, 73, 74, 32, 11, 75, 76, 33, 77, 78, 34, 12, 3, 79, 90, 35, 91, 92, 36, 13, 93, 94, 37, 95, 96, 38, 14, 5
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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 the first digit of A and the last digit of the sequence have opposite parities. If this is not the case for A, we then extend the sequence with the smallest integer X not yet present in S whose first digit has the same parity as the last digit of the sequence.
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LINKS
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EXAMPLE
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After a(1) = 1 and a(2) = 3 we cannot have a(3) = 2 as this 2 would be the first underlined term, which is forbidden (the first underlined term of the sequence must be 1 as we want the underlined terms to rebuild the starting sequence). The same argument forbids a(3) = 4 (the digit 4 and the preceding digit 3, being of opposite parities, would force the term 4 to be the first one to be underlined).
This procedure forces the first 6 terms to be 1, 3, 5, 7, 9 and 10. But now something has changed: the last digit of the sequence is even. We must then duplicate the first term not yet duplicated, which is 1, as this digit 1 and the 0 of 10 have opposite parities. We thus have a(7) = 1, our genuine first underlined term. In the same spirit follow a(8) = 11 and a(9) = 12 which, again, ends the sequence with an even digit, allowing us to duplicate 3: a(10) = 3.
Etc.
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CROSSREFS
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Cf. A306580 where the same idea is involved.
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KEYWORD
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AUTHOR
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STATUS
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approved
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