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%I #46 Jun 14 2017 20:12:44
%S 1,2,3,3,1,4,4,2,5,6,5,6,1,3,2,6,4,2,5,7,8,7,5,3,1,4,9,10,8,9,10,2,5,
%T 6,11,7,9,7,6,4,1,3,8,10,12,10,8,11,12,3,7,9,13,4,5,11,12,7,9,8,1,13,
%U 14,15,6,16,12,10,8,11,6,2,7,9,16,17,13,14,13,11,9,7,10,4,1,12,6,8,14,15,17
%N Triangle read by rows in reverse order: T(n,k), (1<=k<=n), in which each term is the least positive integer such that no row, column, diagonal, or antidiagonal contains a repeated term.
%C Note that the n-th row of this triangle is constructed from right to left, starting at the column n and ending at the column 1.
%C Theorem 1: the middle diagonal gives A000012, the all 1's sequence.
%C Theorem 2: all 1's are in the middle diagonal.
%C For the proofs of the theorems 1 and 2 see the proofs of the theorems 1 and 2 of A274650, because this is essentially the same problem.
%C Conjecture 3: every column is a permutation of the positive integers.
%C Conjecture 4: every diagonal is a permutation of the right border which gives the positive integers.
%H Alois P. Heinz, <a href="/A288531/b288531.txt">Rows n = 1..201, flattened</a>
%F T(n,k) = A288530(n-1,k-1) + 1.
%F T(n,n) = n.
%e Note that every row of the triangle is constructed from right to left, so the sequence is 1, 2, 3, 3, 1, 4,... (see below):
%e 1,
%e 3, 2,
%e 4, 1, 3,
%e 6, 5, 2, 4,
%e 2, 3, 1, 6, 5, Every row is constructed
%e 8, 7, 5, 2, 4, 6, <--- from right to left.
%e 10, 9, 4, 1, 3, 5, 7,
%e 7, 11, 6, 5, 2, 10, 9, 8,
%e 12, 10, 8, 3, 1, 4, 6, 7, 9,
%e 5, 4, 13, 9, 7, 3, 12, 11, 8, 10,
%e 16, 6, 15, 14, 13, 1, 8, 9, 7, 12, 11,
%e 14, 13, 17, 16, 9, 7, 2, 6, 11, 8, 10, 12,
%e 17, 15, 14, 8, 6, 12, 1, 4, 10, 7, 9, 11, 13,
%e ...
%e The triangle may be reformatted as an isosceles triangle so that the all 1's sequence (A000012) appears in the central column (but note that this is NOT the way the triangle is constructed!):
%e .
%e . 1,
%e . 3, 2,
%e . 4, 1, 3,
%e . 6, 5, 2, 4,
%e . 2, 3, 1, 6, 5,
%e . 8, 7, 5, 2, 4, 6,
%e . 10, 9, 4, 1, 3, 5, 7,
%e ...
%e Also the triangle may be reformatted for reading from left to right:
%e .
%e . 1;
%e . 2, 3;
%e . 3, 1, 4;
%e . 4, 2, 5, 6;
%e . 5, 6, 1 , 3, 2;
%e . 6, 4, 2, 5, 7, 8;
%e . 7, 5, 3, 1, 4, 9, 10;
%e ...
%Y Middle diagonal gives A000012.
%Y Right border gives A000027.
%Y Indices of the 1's are in A001844.
%Y Cf. A288530 is the same triangle but with every entry minus 1.
%Y Other sequences of the same family are A269526, A274528, A274650, A274651, A274820, A274821, A286297.
%K nonn,tabl
%O 1,2
%A _Omar E. Pol_, Jun 10 2017
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