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A306808
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An irregular fractal sequence: underline a(n) iff the sum [a(n-1) + a(n)] is a palindrome; all underlined terms rebuild the starting sequence.
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2
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1, 9, 3, 1, 11, 2, 9, 4, 3, 1, 12, 5, 7, 6, 8, 10, 13, 14, 15, 16, 18, 17, 19, 20, 21, 22, 11, 23, 24, 25, 26, 27, 29, 28, 30, 31, 2, 9, 32, 33, 34, 35, 36, 37, 38, 40, 4, 3, 1, 39, 41, 42, 43, 12, 44, 45, 46, 47, 48, 49, 51, 52, 50, 5, 53, 54, 55, 57, 56, 58, 59, 7, 60, 6, 61, 62, 63, 64, 65, 67, 66, 68, 69, 8, 70, 72, 71, 73, 74
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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) = 9. S is extended by duplicating the first term A among the not yet duplicated terms of S, under the condition that the sum [a(n-1) + a(n)] is a palindrome. If this is not the case, we then extend S with the smallest integer X not yet present in S such that the sum [a(n-1) + a(n)] is not a palindrome. S is the lexicographically earliest 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) = 9
Can we duplicate a(1) to form a(3)? No, as a(2) + a(3) would be 10 and 10 is not a palindrome. We thus extend S with the smallest integer X not yet in S such that [a(2) + X] is not a palindrome. We get a(3) = 3.
Can we duplicate a(1) to form a(4)? Yes, as a(3) + a(4) = 4, which is a palindrome. We get a(4) = 1.
Can we duplicate a(2) to form a(5)? No, as a(4) + a(5) would be 10 and 10 is not a palindrome. We thus extend S with the smallest integer X not yet in S such that [a(4) + X] is not a palindrome; we get a(5) = 11.
Can we duplicate a(2) to form a(6)? No, as a(5) + a(6) would be 20 and 20 is not a palindrome. We thus extend S with the smallest integer X not yet in S such that [a(5) + X] is not a palindrome; we get a(6) = 2.
Can we duplicate a(2) to form a(7)? Yes, as [a(6) + a(7)] = 11, which is a palindrome. We get a(7) = 9.
Etc.
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CROSSREFS
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Cf. A306803 (which is obtained by replacing palindrome by prime in the definition).
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KEYWORD
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AUTHOR
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STATUS
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approved
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