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%I #25 Jun 07 2019 11:08:40
%S 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,
%T 19,20,21,22,23,4,5,24,25,26,27,28,29,30,5,2,31,32,33,34,35,36,37,38,
%U 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
%N Reading the first row of this array, or the first column, or the successive antidiagonals is the same as reading this sequence.
%C 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.
%C 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
%F a(n*(n+1)/2) = a(n*(n-1)/2+1) = a(n). - _Rémy Sigrist_, May 21 2019
%F T(n+1,k+1) = A000027(n,k) + 2 if both sequences are read as square arrays. - _Charlie Neder_, Jun 03 2019
%F From _Bernard Schott_, Jun 03 2019: (Start)
%F For 2 <= q <= k:
%F a(k*(k+1)/2 + 2) = (k-2)*(k-1)/2 + 3.
%F a(k*(k+1)/2 + q) = (k-2)*(k-1)/2 + q + 1.
%F a(k*(k+1)/2 + k) = a(k*(k+3)/2) = (k-2)*(k-1)/2 + k + 1 = (k^2-k+4)/2. (End)
%e Array:
%e 1 2 2 2 3 2 2 4 5 2 3 ...
%e 2 3 4 6 9 13 18 24 31 39 48 ...
%e 2 5 7 10 14 19 25 32 40 49 59 ...
%e 2 8 11 15 20 26 33 41 50 60 71 ...
%e 3 12 16 21 27 34 42 51 61 72 84 ...
%e 2 17 22 28 35 43 52 62 73 85 98 ...
%e 2 23 29 36 44 53 63 74 86 99 113 ...
%e 4 30 37 45 54 61 75 87 100 112 129 ...
%e 5 38 46 55 62 76 88 101 113 130 146 ...
%e 2 47 56 63 77 89 102 114 131 147 164 ...
%e 3 57 64 78 90 101 115 132 148 165 183 ...
%e ...
%Y 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.
%Y Cf. A000124, A000217, A000027.
%K nonn,tabl
%O 1,2
%A _Eric Angelini_, May 21 2019