A325783 Reading the first row of this array, or the first column, or the successive antidiagonals is the same as reading this sequence.

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

Offset: 1

Author

Eric Angelini, May 21 2019

Keywords

• nonn
• tabl

Comments

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

Examples

			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 ...
  ...
		

Crossrefs

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.

Cf. A000124, A000217, A000027.

Formula

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

From Bernard Schott, Jun 03 2019: (Start)

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)