This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A274651 #136 Mar 07 2020 13:50:20
%S A274651 1,2,3,4,1,2,3,5,4,6,6,2,1,3,4,5,4,6,2,7,8,7,8,3,1,6,5,9,9,6,10,5,8,3,
%T A274651 11,7,8,11,9,4,1,7,10,6,5,12,7,13,8,2,9,4,11,10,14,10,9,5,12,3,1,2,13,
%U A274651 7,8,11,11,12,8,13,5,4,3,10,9,15,14,16,13,10,11,7,9,2,1,12,8,5,17,15,18
%N A274651 Triangle read by rows: 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 A274651 Analog of A269526, but note that this is a right triangle.
%C A274651 The same rule applied to an equilateral triangle gives A269526.
%C A274651 We construct the triangle by reading from left to right in each row, starting with T(1,1) = 1.
%C A274651 Presumably every diagonal and every column is also a permutation of the positive integers, but the proof does not seem so straightforward. Of course, neither the rows nor the antidiagonals are permutations of the positive integers, since they are finite in length.
%C A274651 _Omar E. Pol_'s conjecture that every column and every diagonal of the triangle is a permutation of the positive integers is true: see the link in A274650 (duplicated below). - _N. J. A. Sloane_, Jun 07 2017
%C A274651 It appears that the numbers generally appear for the first time in or near the right border of the triangle.
%C A274651 Theorem 1: the middle diagonal gives A000012 (the all 1's sequence).
%C A274651 Theorem 2: all 1's are in the middle diagonal.
%C A274651 For the proofs of the theorems 1 and 2 see the proofs of the theorems 1 and 2 of A274650 since both sequences are essentially the same.
%C A274651 From _Bob Selcoe_, Feb 15 2017: (Start)
%C A274651 The columns and diagonal are permutations of the natural numbers. The proofs are essentially the same as the proofs given for the columns and rows (respectively) in A269526.
%C A274651 All coefficients j <= 4 eventually populate in a repeating pattern toward the "middle diagonal" (i.e., relatively near the 1's); this is because we can build the triangle by j in ascending order; that is, we can start by placing all the 1's in the proper cells, then add the 2's, 3's, 4's, 5's, etc. So for i >= 0: since the 1's appear at T(1+2i, 1+i), the 2's appear at T(2+8i, 1+4i), T(3+8i, 3+4i), T(5+8i, 2+4i) and T(6+8i, 4+4i). Accordingly, after the first five 3's appear (at T(2,2), T(4,1), T(5,4), T(7,3) and T(8,6)), the remaining 3's appear at T(11+8i, 5+4i), T(12+8i, 7+4i), T(16+8i, 8+4i) and T(17+8i, 10+4i). Similarly for 4's, after the first 21 appearances, 4's appear at T(44+8i, 21+4i), T(45+8i, 24+4i), T(47+8i, 23+4i) and T(48+8i, 26+4i). So starting at T(41,21), this 16-coefficient pattern repeats at T(41+8i, 21+4i):
%C A274651 n/k 21 22 23 24 25 26
%C A274651 41 1 3
%C A274651 42 2
%C A274651 43 3 1 2
%C A274651 44 4 3
%C A274651 45 2 1 4
%C A274651 46 2
%C A274651 47 4 1
%C A274651 48 3 4
%C A274651 where the next 1 appears at T(49,25), and the pattern repeats at that point from the top left (so T(49,26) = 3, T(50,25) = 2, etc.).
%C A274651 Conjecture: as n gets sufficiently large, all coefficients j>4 will appear in a repeating pattern, populating all rows and diagonals around smaller j's near the "middle diagonal" (while I can offer no formal proof, it appears very likely that this is the case). (End)
%C A274651 From _Hartmut F. W. Hoft_, Jun 12 2017: (Start)
%C A274651 T(2k+1,k+1) = 1, for all k>=0, and T(n,{n/2,(n+3)/2,(n-1)/2,(n+2)/2}) = 2, for all n>=1 with mod(n,8) = {2,3,5,6} respectively, and no 1's or 2's occur in other positions.
%C A274651 Proof by (recursive) picture:
%C A274651 Positions in the triangle that are empty and those containing the dots of the guiding diagonals contain numbers larger than two.
%C A274651 n\k 1 2 3 4 5 6 7 8 10 12 14 16 18 20 22 24
%C A274651 1 |1
%C A274651 2 |2 .
%C A274651 3 | 1 2
%C A274651 4 | .
%C A274651 5 | 2 1 .
%C A274651 6 | 2 .
%C A274651 7 | 1 .
%C A274651 8 |______________.
%C A274651 9 | |1 .
%C A274651 10 | |2 . .
%C A274651 11 | | 1 2 .
%C A274651 12 | | . .
%C A274651 13 | | 2 1 . .
%C A274651 14 | | 2 . .
%C A274651 15 | | 1 . .
%C A274651 16 |_______|______________._______.
%C A274651 17 | | |1 . .
%C A274651 18 | | |2 . . .
%C A274651 19 | | | 1 2 . .
%C A274651 20 | | | . . .
%C A274651 21 | | | 2 1 . . .
%C A274651 22 | | | 2 . . .
%C A274651 23 | | | 1 . . .
%C A274651 24 |_______|_______|______._______._______._______.
%C A274651 1 2 3 4 5 6 7 8 12 16 20 24
%C A274651 Consider the center of the triangle. In each octave of rows the columns in the first central quatrain contain a 1 and a 2, and the diagonals in the second central quatrain contain a 1 and a 2. Therefore, no 1's or 2's can occur in the respective downward quatrains of leading columns and trailing diagonals.
%C A274651 The sequence of rows containing 2's is A047447 (n mod 8 = {2,3,5,6}), those containing only 2's is A016825 (n mod 8 = {2,6}), those containing both 1's and 2's is A047621 (n mod 8 = {3,5}), those containing only 1's is A047522 (n mod 8 = {1,7}), and those containing neither 1's nor 2's is A008586 (n mod 8 = {0,4}).
%C A274651 (End)
%H A274651 Michel Marcus, <a href="/A274651/b274651.txt">Table of n, a(n) for n = 1..20100</a>
%H A274651 F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, <a href="https://www.combinatorics.org/ojs/index.php/eljc/article/view/v27i1p52/8039">Queens in exile: non-attacking queens on infinite chess boards</a>, Electronic J. Combin., 27:1 (2020), #P1.52.
%H A274651 N. J. A. Sloane, <a href="/A274650/a274650.txt">Notes on A274650 and Proof by Non-Attacking Queens</a>
%F A274651 T(n,k) = A274650(n-1,k-1) + 1.
%e A274651 Triangle begins:
%e A274651 1;
%e A274651 2, 3;
%e A274651 4, 1, 2;
%e A274651 3, 5, 4, 6;
%e A274651 6, 2, 1, 3, 4;
%e A274651 5, 4, 6, 2, 7, 8;
%e A274651 7, 8, 3, 1, 6, 5, 9;
%e A274651 9, 6, 10, 5, 8, 3, 11, 7;
%e A274651 8, 11, 9, 4, 1, 7, 10, 6, 5;
%e A274651 12, 7, 13, 8, 2, 9, 4, 11, 10, 14;
%e A274651 10, 9, 5, 12, 3, 1, 2, 13, 7, 8, 11;
%e A274651 11, 12, 8, 13, 5, 4, 3, 10, 9, 15, 14, 16;
%e A274651 13, 10, 11, 7, 9, 2, 1, 12, 8, 5, 17, 15, 18;
%e A274651 ...
%e A274651 From _Omar E. Pol_, Jun 07 2017: (Start)
%e A274651 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 A274651 .
%e A274651 . 1;
%e A274651 . 2, 3;
%e A274651 . 4, 1, 2;
%e A274651 . 3, 5, 4, 6;
%e A274651 . 6, 2, 1, 3, 4;
%e A274651 . 5, 4, 6, 2, 7, 8;
%e A274651 . 7, 8, 3, 1, 6, 5, 9;
%e A274651 . 9, 6, 10, 5, 8, 3, 11, 7;
%e A274651 . 8, 11, 9, 4, 1, 7, 10, 6, 5;
%e A274651 ...
%e A274651 (End)
%t A274651 f[1,1] = 1; (* for 1 < n and 1 <= k <= n *)
%t A274651 f[n_,k_] := f[n,k] = Module[{vals=Sort[Join[Map[f[n, #]&, Range[1, k-1]], Map[f[#, k]&, Range[k, n-1]], Map[f[n-k+#, #]&, Range[1, k-1]], Map[f[n-#, k+#]&, Range[1, Floor[(n-k)/2]]]]], c}, c=Complement[Range[1, Last[vals]], vals]; If[c=={}, Last[vals]+1, First[c]]]
%t A274651 (* computation of rows 1 ... n of triangle *)
%t A274651 a274651[n_] := Prepend[Table[f[i, j], {i, 2, n}, {j, 1, i}], {1}]
%t A274651 Flatten[a274651[13]] (* data *)
%t A274651 TableForm[a274651[13]] (* triangle *)
%t A274651 (* _Hartmut F. W. Hoft_, Jun 12 2017 *)
%Y A274651 Cf. A001844 (indices of the 1's).
%Y A274651 Cf. A000012 (middle diagonal).
%Y A274651 Every diagonal and every column of the right triangle is a permutation of A000027.
%Y A274651 Cf. A274650 is the same triangle but with every entry minus 1.
%Y A274651 Other sequences of the same family are A269526, A274528, A274820, A274821, A286297, A288530, A288531.
%Y A274651 Sequences mentioned in _N. J. A. Sloane_'s proof are A000170, A274616 and A287864.
%Y A274651 Cf. A008586, A016825, A047447, A047522, A047621. - _Hartmut F. W. Hoft_, Jun 12 2017
%K A274651 nonn,tabl,look
%O A274651 1,2
%A A274651 _Omar E. Pol_, Jul 02 2016