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A325783
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Reading the first row of this array, or the first column, or the successive antidiagonals is the same as reading this sequence.
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2
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1, 2, 2, 2, 3, 2, 2, 4, 5, 2, 3, 6, 7, 8, 3, 2, 9, 10, 11, 12, 2, 2, 13, 14, 15, 16, 17, 2, 4, 18, 19, 20, 21, 22, 23, 4, 5, 24, 25, 26, 27, 28, 29, 30, 5, 2, 31, 32, 33, 34, 35, 36, 37, 38, 2, 3, 39, 40, 41, 42, 43, 44, 45, 46, 47, 3, 6, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 6, 7, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 7, 8, 69
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OFFSET
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1,2
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COMMENTS
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The array is always extended by its antidiagonals with the smallest term not yet present that doesn't lead to a contradiction. The sequence is thus the lexicographically earliest of its kind.
This regular pattern appears: . . . . 3 . . 4 5 . . 6 7 8 . . 9 10 11 12 . . 13 14 15 16 17 . . 18 19 20 21 22 23 . . This is the first time that these terms appear in the sequence. So it is possible to calculate the terms of this pattern. - Bernard Schott, Jun 03 2019
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LINKS
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FORMULA
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a(n*(n+1)/2) = a(n*(n-1)/2+1) = a(n). - Rémy Sigrist, May 21 2019
T(n+1,k+1) = A000027(n,k) + 2 if both sequences are read as square arrays. - Charlie Neder, Jun 03 2019
For 2 <= q <= k:
a(k*(k+1)/2 + 2) = (k-2)*(k-1)/2 + 3.
a(k*(k+1)/2 + q) = (k-2)*(k-1)/2 + q + 1.
a(k*(k+1)/2 + k) = a(k*(k+3)/2) = (k-2)*(k-1)/2 + k + 1 = (k^2-k+4)/2. (End)
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EXAMPLE
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Array:
1 2 2 2 3 2 2 4 5 2 3 ...
2 3 4 6 9 13 18 24 31 39 48 ...
2 5 7 10 14 19 25 32 40 49 59 ...
2 8 11 15 20 26 33 41 50 60 71 ...
3 12 16 21 27 34 42 51 61 72 84 ...
2 17 22 28 35 43 52 62 73 85 98 ...
2 23 29 36 44 53 63 74 86 99 113 ...
4 30 37 45 54 61 75 87 100 112 129 ...
5 38 46 55 62 76 88 101 113 130 146 ...
2 47 56 63 77 89 102 114 131 147 164 ...
3 57 64 78 90 101 115 132 148 165 183 ...
...
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CROSSREFS
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Cf. A325784 and A325785 where the same idea is developped, but restricted to, respectively, the first row and the first column of the arrays presented.
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KEYWORD
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AUTHOR
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STATUS
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approved
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