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A132075
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A conjectured permutation of the positive integers such that for every n, a(n) is the largest number among a(1), a(2), ..., a(n) that when added to a(n+1) gives a prime.
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3
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1, 2, 3, 4, 7, 6, 5, 14, 9, 10, 13, 16, 15, 8, 11, 20, 17, 12, 19, 24, 23, 18, 25, 22, 21, 26, 27, 34, 33, 28, 31, 30, 29, 32, 35, 36, 37, 46, 43, 40, 39, 44, 45, 38, 41, 42, 47, 50, 59, 54, 55, 58, 51, 62, 65, 48, 61, 52, 57, 56, 53, 60, 49, 64, 63, 68, 69, 70, 67, 72, 77, 80, 71, 66, 73, 78, 79, 84, 83, 90, 89, 74, 75, 76, 81, 82, 85, 88, 93, 86, 95, 104, 107, 92
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OFFSET
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1,2
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COMMENTS
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The terms are defined as follows. Start by choosing the initial terms: 1, 2, 3. Then write the rows of table A088643 backwards but always leave off the last three quarters of the terms. This gives: [], [], [], [1], [1], [1], [1], [1, 2], [1, 2], [1, 2], [1, 4], [1, 4, 3,], [1, 4, 3] etc. Then build the sequence up by repeatedly choosing the first such truncated row that extends the terms already chosen. [Edited by Peter Munn, Aug 19 2021]
It is not until the 26th truncated row - [1, 2, 3, 4, 7, 6] - that the initial list is extended at all. It is unclear whether this process can be continued indefinitely, although I have verified by computer that it generates a sequence of at least 2000 terms. Conjecturally: (1) the sequence is infinite, (2) it is the unique sequence containing infinitely many complete rows of table A088643, and (3) for every n > 0 there exists N > 0 such that the first n terms of this sequence are contained in every row of table A088643 from the N-th onwards.
Maybe the idea could be expressed more concisely by defining this sequence as the limit of the reversed rows of A088643? - M. F. Hasler, Aug 04 2021
It seems we do not know of an existence proof for the limit of the reversed rows of A088643. - Peter Munn, Aug 19 2021
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LINKS
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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