|
| |
|
|
A132075
|
|
An attempt to find a permutation of the positive integers with the property 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.
|
|
2
| |
|
|
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
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,2
|
|
|
COMMENTS
| The terms were found as follows. Start with 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 always selecting the first such truncated row that extends the terms already chosen. 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, (3) For every n > 0 there exists N > 0 such that the first n terms of A132075 are contained in every row of table A088643 from the N-th onwards.
|
|
|
CROSSREFS
| Cf. A088643.
Sequence in context: A122198 A122155 A106454 * A074846 A120225 A130685
Adjacent sequences: A132072 A132073 A132074 * A132076 A132077 A132078
|
|
|
KEYWORD
| easy,nonn
|
|
|
AUTHOR
| Paul Boddington (pbotherstuff(AT)yahoo.co.uk), Oct 30 2007, Mar 06 2010
|
| |
|
|
