A274651 -id:A274651 - cp's OEIS frontend

A269526 Square array T(n,k) (n>=1, k>=1) read by antidiagonals upwards in which each term is the least positive integer satisfying the condition that no row, column, diagonal, or antidiagonal contains a repeated term.

1, 2, 3, 3, 4, 2, 4, 1, 5, 6, 5, 2, 6, 1, 4, 6, 7, 3, 2, 8, 5, 7, 8, 1, 5, 9, 3, 10, 8, 5, 9, 4, 1, 7, 6, 11, 9, 6, 4, 7, 2, 8, 5, 12, 13, 10, 11, 7, 3, 5, 6, 9, 4, 14, 8, 11, 12, 8, 9, 6, 10, 3, 7, 15, 16, 14, 12, 9, 13, 10, 11, 14, 4, 15, 16, 17, 7, 18, 13, 10, 14, 11, 3, 4, 8, 16, 9, 6, 12, 15, 7

Offset: 1

Alec Jones, Apr 07 2016

An infinite Sudoku-type array.
In the definition, "diagonal" means a diagonal line of slope -1, and "antidiagonal" means a diagonal line of slope +1.
Theorem C (Bob Selcoe, Jul 01 2016): Every column is a permutation of the natural numbers.
Proof: Fix k, and suppose j is the smallest number missing from that column. For this to happen, every entry T(n,k) for sufficiently large n in that column must see a j in the NW diagonal through that cell or in the row to the W of that cell. But there are at most k-1 copies of j in the columns to the left of the k-th column, and if n is very large the entry T(n,k) will be unaffected by those j's, and so T(n,k) would then be set to j, a contradiction. QED
Theorem R (Rob Pratt, Bob Selcoe, N. J. A. Sloane, Jul 02 2016): Every row is a permutation of the natural numbers.
Proof: Fix n, and suppose j is the smallest number missing from that row. For this to happen, every entry T(n,k) for sufficiently large k in that row must see a j in the column to the N, or in the NW diagonal through that cell or in the SW diagonal through that cell.
Rows 1 through n-1 contain at most n-1 copies of j, and their influence on the entries in the n-th row only extend out to the entry T(n,k_0), say. We take k to be much larger than k_0 and consider the entry T(n,k). We will show that for large enough k it can (and therefore must) be equal to j, which is a contradiction.
Consider the triangle bounded by row n, column 1, and the SW antidiagonal through cell (n,k). Replace every copy of j in this triangle by a queen and think of these cells as a triangular chessboard. These are non-attacking queens, by definition of the sequence, and by the result in A274616 there can be at most 2*k/3 + 1 such queens. However, there are k-k_0 cells in row n that have to be attacked, and for large k this is impossible since k-k_0 > 2*k/3+1. If a cell (n,k) is not attacked by a queen, then T(n,k) can take the value j. QED
Presumably every diagonal is also a permutation of the natural numbers, but the proof does not seem so straightforward. Of course the antidiagonals are not permutations of the natural numbers, since they are finite in length. - N. J. A. Sloane, Jul 02 2016
For an interpretation of this array in terms of Sprague-Grundy values, see A274528.
From Don Reble, Jun 30 2016: (Start)
Let b(n) be the position in column n where 1 appears, i.e., such that T(b(n),n) = 1. Then b(n) is A065188, which is Antti Karttunen's "Greedy Queens" permutation.
Let b'(n) be the position in row n where 1 appears, i.e., such that T(n,b'(n)) = 1. Then b'(n) is A065189, the inverse "Greedy Queens" permutation. (End)
The same sequence arises if we construct a triangle, by reading from left to right in each row, always choosing the smallest positive number which does not produce a duplicate number in any row or diagonal. - N. J. A. Sloane, Jul 02 2016
It appears that the numbers generally appear for the first time in or near the first few rows. - Omar E. Pol, Jul 03 2016
The last comment in the FORMULA section seems wrong: It seems that columns 4, 5, 6, 7, 8, 9, ...(?) all have first differences which become 16-periodic from, respectively, term 8, 17, 52, 91, 92, 131, ... on, rather than having period 4^(k-1) from term k on. - M. F. Hasler, Sep 26 2022

			The array is constructed along its antidiagonals, in the following way:
  a(1)  a(3)  a(6)  a(10)
  a(2)  a(5)  a(9)
  a(4)  a(8)
  a(7)
See the link from Peter Kagey for an animated example.
The beginning of the square array is:
   1,  3,  2,  6,  4,  5, 10, 11, 13,  8, 14, 18,  7, 20, 19,  9, 12, ...
   2,  4,  5,  1,  8,  3,  6, 12, 14, 16,  7, 15, 17,  9, 22, 21, 11, ...
   3,  1,  6,  2,  9,  7,  5,  4, 15, 17, 12, 19, 18, 21,  8, 10, 23, ...
   4,  2,  3,  5,  1,  8,  9,  7, 16,  6, 18, 17, 11, 10, 23, 22, 14, ...
   5,  7,  1,  4,  2,  6,  3, 15,  9, 10, 13,  8, 20, 14, 12, 11, 17, ...
   6,  8,  9,  7,  5, 10,  4, 16,  2,  1,  3, 11, 22, 15, 24, 13, 27, ...
   7,  5,  4,  3,  6, 14,  8,  9, 11, 18,  2, 21,  1, 16, 10, 12, 20, ...
   8,  6,  7,  9, 11,  4, 13,  3, 12, 15,  1, 10,  2,  5, 26, 14, 18, ...
   9, 11,  8, 10,  3,  1, 14,  6,  7, 13,  4, 12, 24, 18,  2,  5, 19, ...
  10, 12, 13, 11, 16,  2, 17,  5, 20,  9,  8, 14,  4,  6,  1,  7,  3, ...
  11,  9, 14, 12, 10, 15,  1,  8, 21,  7, 16, 20,  5,  3, 18, 17, 32, ...
  12, 10, 11,  8,  7,  9,  2, 13,  5, 23, 25, 26, 14, 17, 16, 15, 33, ...
...
  - _N. J. A. Sloane_, Jun 29 2016

Peter Kagey, Table of n, a(n) for n = 1..10000
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
Peter Kagey, Animated example illustrating the first fifteen terms.

First 4 rows are A274315, A274316, A274317, A274791.
Main diagonal is A274318.
Column 1 is A000027, column 2 is A256008(n) = A004443(n-1)+1 = 1 + (nimsum of n-1 and 2), column 3 is A274614 (or equally, A274615 + 1), and column 4 is A274617 (or equally, A274619 + 1).
Antidiagonal sums give A274530. Other properties of antidiagonals: A274529, A275883.
Cf. A274080 (used in Haskell program), A274616.
A065188 and A065189 say where the 1's appear in successive columns and rows.
If all terms are reduced by 1 and the offset is changed to 0 we get A274528.
A274650 and A274651 are triangles in the shape of a right triangle and with a similar definition.
See A274630 for the case where both queens' and knights' moves must avoid duplicates.

import Data.List ((\\))
a269526 n = head $ [1..] \\ map a269526 (a274080_row n)
-- Peter Kagey, Jun 10 2016

# The following Maple program was provided at my request by Alois P. Heinz, who said that he had not posted it himself because it stores the data in an inefficient way. - N. J. A. Sloane, Jul 01 2016
A:= proc(n, k) option remember; local m, s;
         if n=1 and k=1 then 1
       else s:= {seq(A(i,k), i=1..n-1),
                 seq(A(n,j), j=1..k-1),
                 seq(A(n-t,k-t), t=1..min(n,k)-1),
                 seq(A(n+j,k-j), j=1..k-1)};
            for m while m in s do od; m
         fi
     end:
[seq(seq(A(1+d-k, k), k=1..d), d=1..15)];

A[n_, k_] := A[n, k] = If[n == 1 && k == 1, 1, s = {Table[A[i, k], {i, 1, n-1}], Table[A[n, j], {j, 1, k-1}], Table[A[n-t, k-t], {t, 1, Min[n, k] - 1}], Table[A[n+j, k-j], {j, 1, k-1}]} // Flatten; For[m = 1, True, m++, If[FreeQ[s, m], Return[m]]]];
Table[Table[A[1+d-k, k], {k, 1, d}], {d, 1, 15}] // Flatten (* Jean-François Alcover, Jul 21 2016, translated from Maple *)

{M269526=Map(); A269526=T(r,c)=c>1 && !mapisdefined(M269526, [r,c], &r) && mapput(M269526, [r,c], r=sum(k=1, #c=Set(concat([[T(r+k,c+k)|k<-[1-min(r, c)..-1]], [T(r,k)|k<-[1..c-1]], [T(k,c)|k<-[1..r-1]], [T(r+c-k,k)|k<-[1..c-1]]])), c[k]==k)+1); r} \\ M. F. Hasler, Sep 26 2022

Theorem 1: T(n,1) = n.
Proof by induction. T(1,1)=1 by definition. When calculating T(n,1), the only constraint is that it be different from all earlier entries in the first column, which are 1,2,3,...,n-1. So T(n,1)=n. QED
Theorem 2 (Based on a message from Bob Selcoe, Jun 29 2016): Write n = 4t+i with t >= 0, i=1,2,3, or 4. Then T(n,2) = 4t+3 if i=1, 4t+4 if i=2, 4t+1 if i=3, 4t+2 if i=4. This implies that the second column is the permutation A256008.
Proof: We check that the first 4 entries in column 2 are 2,5,6,3. From then on, to calculate the entry T(n,2), we need only look to the N, NW, W, and SW (we need never look to the East). After we have found the first 4t entries in the column, the column contains all the numbers from 1 to 4t. The four smallest free numbers are 4t+1, 4t+2, 4t+3, 4t+4. Entry T(4t+1,2) cannot be 4t+1 or 4t+2, but it can (and therefore must) be 4t+3. Similarly T(4t+2,2)=4t+4, T(4t+3,2)=4t+1, and T(4t+4,2)=4t+2. The column now contains all the numbers from 1 to 4t+4. Repeating this argument established the theorem. QED
Comments from Bob Selcoe, Jun 29 2016: (Start)
From Theorem 2, column 2 (i.e., terms a((j^2+j+4)/2), j>=1) is a permutation. After a(3)=3, the differences of successive terms follow the pattern a(n) = 3 [+1, -3, +1, +5], so a(5)=4, a(8)=1, a(12)=2, a(17)=7, a(23)=8, a(30)=5...
Similarly, column 3 (i.e., terms a((j^2+j+6)/2), j>=2) appears to be a permutation, but with the pattern after a(6)=2 and a(9)=5 being 5 [+1, -3, -2, +8, -5, +3, +1, +5, +1, -3, +1, -2, +8, -3, +1, +5]. (See A274614 and A274615.)
I conjecture that other similar cyclical difference patterns should hold for any column k (i.e., terms a((j^2+j+2*k)/2), j>=k-1), so that each column is a permutation.
Also, the differences in column 1 are a 1-cycle ([+1]), in column 2 a 4-cycle after the first term, and in column 3 a 16-cycle after the second term. Perhaps the cycle lengths are 4^(k-1) starting after j=k-1. (End)  WARNING: These comments may be wrong - see COMMENTS section. - N. J. A. Sloane, Sep 26 2022

Definition clarified by Omar E. Pol, Jun 29 2016

A274650 Triangle read by rows: T(n,k), (0 <= k <= n), in which each term is the least nonnegative integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

Original entry on oeis.org

0, 1, 2, 3, 0, 1, 2, 4, 3, 5, 5, 1, 0, 2, 3, 4, 3, 5, 1, 6, 7, 6, 7, 2, 0, 5, 4, 8, 8, 5, 9, 4, 7, 2, 10, 6, 7, 10, 8, 3, 0, 6, 9, 5, 4, 11, 6, 12, 7, 1, 8, 3, 10, 9, 13, 9, 8, 4, 11, 2, 0, 1, 12, 6, 7, 10, 10, 11, 7, 12, 4, 3, 2, 9, 8, 14, 13, 15, 12, 9, 10, 6, 8, 1, 0, 11, 7, 4, 16, 14, 17

Offset: 0

Omar E. Pol, Jul 02 2016

Similar to A274528, but the triangle here is a right triangle.
The same rule applied to an equilateral triangle gives A274528.
By analogy, the offset for both rows and columns is the same as the offset of A274528.
We construct the triangle by reading from left to right in each row, starting with T(0,0) = 0.
Presumably every diagonal and every column is also a permutation of the nonnegative integers, but the proof does not seem so straightforward. Of course neither the rows nor the antidiagonals are permutations of the nonnegative integers, since they are finite in length.
Omar E. Pol's conjecture that every column and every diagonal of the triangle is a permutation of the nonnegative integers is true: see the link. - N. J. A. Sloane, Jun 07 2017
It appears that the numbers generally appear for the first time in or near the right border of the triangle.
Theorem 1: the middle diagonal gives A000004 (the zero sequence).
Proof: T(0,0) = 0 by definition. For the next rows we have that in row 1 there are no zeros because the first term belongs to a column that contains a zero and the second term belongs to a diagonal that contains a zero. In row 2 the unique zero is T(2,1) = 0 because the preceding term belongs to a column that contains a zero and the following term belongs to a diagonal that contains a zero. Then we have two recurrences for all rows of the triangle:
a) If T(n,k) = 0 then row n+1 does not contain a zero because every term belongs to a column that contains a zero or it belongs to a diagonal that contains a zero.
b) If T(n,k) = 0 the next zero is T(n+2,k+1) because every preceding term in row n+2 is a positive integer because it belongs to a column that contains a zero and, on the other hand, the column, the diagonal and the antidiagonal of T(n+2,k+1) do not contain zeros.
Finally, since both T(n,k) = 0 and T(n+2,k+1) = 0 are located in the middle diagonal, the other terms of the middle diagonal are zeros, or in other words: the middle diagonal gives A000004 (the zero sequence). QED
Theorem 2: all zeros are in the middle diagonal.
Proof: consider the first n rows of the triangle. Every element located above or at the right-hand side of the middle diagonal must be positive because it belongs to a diagonal that contains one of the zeros of the middle diagonal. On the other hand every element located below the middle diagonal must be positive because it belongs to a column that contains one of the zeros of the middle diagonal, hence there are no zeros outside of the middle diagonal, or in other words: all zeros are in the middle diagonal. QED
From Hartmut F. W. Hoft, Jun 12 2017: (Start)
T(2k,k) = 0, for all k >= 0, and T(n,{(n-1)/2,(n+2)/2,(n-2)/2,(n+1)/2}) = 1, for all n >= 0 with n mod 8 = {1,2,4,5} respectively, and no 0's or 1's occur in other positions. The triangle of positions of 0's and 1's for this sequence is the triangle in the Comment section of A274651 with row and column indices and values shifted down by one.
The sequence of rows containing 1's is A047613 (n mod 8 = {1,2,4,5}), those containing only 1's is A016813 (n mod 8 = {1,5}), those containing both 0's and 1's is A047463 (n mod 8 = {2,4}), those containing only 0's is A047451 (n mod 8 = {0,6}), and those containing neither 0's nor 2's is A004767 (n mod 8 = {3,7}).
(End)

			Triangle begins:
   0;
   1,  2;
   3,  0,  1;
   2,  4,  3,  5;
   5,  1,  0,  2,  3;
   4,  3,  5,  1,  6,  7;
   6,  7,  2,  0,  5,  4,  8;
   8,  5,  9,  4,  7,  2, 10,  6;
   7, 10,  8,  3,  0,  6,  9,  5,  4;
  11,  6, 12,  7,  1,  8,  3, 10,  9, 13;
   9,  8,  4, 11,  2,  0,  1, 12,  6,  7, 10;
  10, 11,  7, 12,  4,  3,  2,  9,  8, 14, 13, 15;
  12,  9, 10,  6,  8,  1,  0, 11,  7,  4, 16, 14, 17;
  ...
From _Omar E. Pol_, Jun 07 2017: (Start)
The triangle may be reformatted as an isosceles triangle so that the zero sequence (A000004) appears in the central column (but note that this is NOT the way the triangle is constructed!):
.
.                    0;
.                  1,  2;
.                3,  0,  1;
.              2,  4,  3,  5;
.            5,  1,  0,  2,  3;
.          4,  3,  5,  1,  6,  7;
.        6,  7,  2,  0,  5,  4,  8;
.     8,   5,  9,  4,  7,  2, 10,  6;
.   7,  10,  8,  3,  0,  6,  9,  5,  4;
...
(End)

Alois P. Heinz, Rows n = 0..200, flattened
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
Rémy Sigrist, Colored representation of the rows n = 0..999
Rémy Sigrist, PARI program for A274650
N. J. A. Sloane, Notes on A274650 and Proof by Non-Attacking Queens

Cf. A000004 (middle diagonal).
Cf. A046092 (indices of the zeros).
Every diagonal and every column of the right triangle is a permutation of A001477.
The left and right edges of the right triangle give A286294 and A286295.
Cf. A274651 is the same triangle but with 1 added to every entry.
Other sequences of the same family are A269526, A274528, A274820, A274821, A286297, A288530, A288531.
Sequences mentioned in N. J. A. Sloane's proof are A000170, A274616 and A287864.
Cf. A288384.
Cf. A004767, A016813, A047451, A047463, A047613. - Hartmut F. W. Hoft, Jun 12 2017
See A308179, A308180 for a very similar triangle.

(* function a274651[] is defined in A274651 *)
(* computation of rows 0 ... n-1 *)
a274650[n_] := a274651[n]-1
Flatten[a274650[13]] (* data *)
TableForm[a274650[13]] (* triangle *)
(* Hartmut F. W. Hoft, Jun 12 2017 *)

PARI
```
See Links section.
```

T(n,k) = A274651(n+1,k+1) - 1.

A274821 Hexagonal spiral constructed on the nodes of the infinite triangular net in which each term is the least positive integer such that no diagonal contains a repeated term.

Original entry on oeis.org

1, 2, 3, 2, 3, 2, 3, 1, 4, 1, 5, 4, 6, 4, 5, 6, 4, 5, 7, 6, 7, 8, 5, 7, 6, 8, 7, 4, 1, 7, 6, 8, 1, 7, 8, 6, 5, 9, 2, 4, 7, 9, 2, 10, 8, 9, 3, 5, 10, 9, 3, 11, 12, 9, 10, 11, 13, 4, 9, 10, 8, 11, 10, 3, 5, 9, 6, 11, 3, 12, 10, 12, 1, 11, 8, 9, 7, 1, 10, 8, 11, 13, 8, 2, 5, 9, 6, 12, 2, 11, 13, 10, 6, 12, 11, 14, 13, 12, 14, 15

Offset: 0

Omar E. Pol, Jul 09 2016

nonn

Also spiral constructed on the infinite hexagonal grid in which each term is the least positive integer such that no diagonal of successive adjacent cells contains a repeated term. Every number is located in the center of a hexagonal cell. Every cell is also the center of three diagonals of successive adjacent cells.

			Illustration of initial terms as a spiral:
.
.                  10 - 5 - 3 - 9 - 8
.                  /                 \
.                 9   4 - 7 - 8 - 6   10
.                /   /             \   \
.               3   1   6 - 4 - 5   7   2
.              /   /   /         \   \   \
.            11   7   4   2 - 3   1   5   9
.            /   /   /   /     \   \   \   \
.          12   6   5   3   1 - 2   4   8   7
.            \   \   \   \         /   /   /
.             9   8   6   2 - 3 - 1   7   4
.              \   \   \             /   /
.              10   1   4 - 5 - 7 - 6   2
.                \   \                 /
.                11   7 - 8 - 6 - 5 - 9
.                  \
.                  13 - 4 - 9 - 10 - 8
.

Cf. A269526 (square array), A274640 (square spiral), A274651 (right triangle), A274820, A274920, A274921, A275606, A275610.

a(n) = A274820(n) + 1.

A286294 Leading column of right triangle in A274650.

Original entry on oeis.org

0, 1, 3, 2, 5, 4, 6, 8, 7, 11, 9, 10, 12, 13, 15, 14, 16, 18, 17, 19, 20, 22, 21, 23, 25, 24, 27, 26, 28, 29, 31, 30, 32, 33, 35, 34, 36, 38, 37, 40, 39, 41, 44, 42, 43, 45, 46, 47, 49, 48, 50, 51, 53, 52, 54, 55, 57, 56, 58, 60, 59, 62, 61, 64, 65, 63, 66, 68, 67, 70, 71, 69, 72, 73, 75, 74, 76, 77, 78

Offset: 0

N. J. A. Sloane, May 30 2017

nonn

Theorem: This is a permutation of the positive integers. For proof, see link in A274650. - N. J. A. Sloane, Jun 07 2017

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A274650, A274651, A286295, A288384, A288424.

(* function a274651[] is defined in A274651 for n>=1 *)
a286294[n_] := Map[First, a274651[n+1]-1]
a286294[78] (* data *) (* Hartmut F. W. Hoft, Jun 13 2017 *)

a(n) = A274650(n,0) = A274651(n+1,1) - 1. - Hartmut F. W. Hoft, Jun 13 2017

A288530 Triangle read by rows in reverse order: T(n,k), (0 <= k <= n), in which each term is the least nonnegative integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

Original entry on oeis.org

0, 1, 2, 2, 0, 3, 3, 1, 4, 5, 4, 5, 0, 2, 1, 5, 3, 1, 4, 6, 7, 6, 4, 2, 0, 3, 8, 9, 7, 8, 9, 1, 4, 5, 10, 6, 8, 6, 5, 3, 0, 2, 7, 9, 11, 9, 7, 10, 11, 2, 6, 8, 12, 3, 4, 10, 11, 6, 8, 7, 0, 12, 13, 14, 5, 15, 11, 9, 7, 10, 5, 1, 6, 8, 15, 16, 12, 13, 12, 10, 8, 6, 9, 3, 0, 11, 5, 7, 13, 14, 16

Offset: 0

Omar E. Pol, Jun 10 2017

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 0.
Theorem 1: the middle diagonal gives A000004, the all-zeros sequence.
Theorem 2: all zeros are in the middle diagonal.
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.
Conjecture 3: every column is a permutation of the nonnegative integers.
Conjecture 4: every diagonal is a permutation of the right border which gives the nonnegative integers.

			Note that every row of the triangle is constructed from right to left, so the sequence is 0, 1, 2, 2, 0, 3, ... (see below):
0,
2,   1,
3,   0,  2,
5,   4,  1,  3,
1,   2,  0,  5,  4,                      Every row is constructed
7,   6,  4,  1,  3,  5,              <---   from right to left.
9,   8,  3,  0,  2,  4,  6,
6,  10,  5,  4,  1,  9,  8,  7,
11,  9,  7,  2,  0,  3,  5,  6,  8,
4,   3, 12,  8,  6,  2, 11, 10,  7,  9,
15,  5, 14, 13, 12,  0,  7,  8,  6, 11, 10,
13, 12, 16, 15,  8,  6,  1,  5, 10,  7,  9, 11,
16, 14, 13,  7,  5, 11,  0,  3,  9,  6,  8, 10, 12,
...
The triangle may be reformatted as an isosceles triangle so that the all-zeros sequence (A000004) appears in the central column (but note that this is NOT the way the triangle is constructed!):
.
.              0,
.            2,  1,
,          3,  0,  2,
.        5,  4,  1,  3,
.      1,  2,  0,  5,  4,
.    7,  6,  4,  1,  3,  5,
.  9,  8,  3,  0,  2,  4,  6,
...
Also the triangle may be reformatted for reading from left to right:
.
.                           0;
.                       1,  2;
.                   2,  0,  3;
.               3,  1,  4,  5;
.           4,  5,  0 , 2,  1;
.       5,  3,  1,  4,  6,  7;
.   6,  4,  2,  0,  3,  8,  9;
...

Alois P. Heinz, Rows n = 0..200, flattened

Middle diagonal gives A000004.
Right border gives A001477.
Indices of the zeros are in A046092.
Cf. A288531 is the same triangle but with 1 added to every entry.
Other sequences of the same family are A269526, A274528, A274650, A274651, A274820, A274821, A286297.

T(n,k) = A288531(n+1, k+1) - 1.

T(n,n) = n.

A288531 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.

Original entry on oeis.org

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, 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, 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

Offset: 1

Omar E. Pol, Jun 10 2017

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.
Theorem 1: the middle diagonal gives A000012, the all 1's sequence.
Theorem 2: all 1's are in the middle diagonal.
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.
Conjecture 3: every column is a permutation of the positive integers.
Conjecture 4: every diagonal is a permutation of the right border which gives the positive integers.

			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):
1,
3,   2,
4,   1,  3,
6,   5,  2,  4,
2,   3,  1,  6,  5,                      Every row is constructed
8,   7,  5,  2,  4,  6,              <---   from right to left.
10,  9,  4,  1,  3,  5,  7,
7,  11,  6,  5,  2, 10,  9,  8,
12, 10,  8,  3,  1,  4,  6,  7,  9,
5,   4, 13,  9,  7,  3, 12, 11,  8, 10,
16,  6, 15, 14, 13,  1,  8,  9,  7, 12, 11,
14, 13, 17, 16,  9,  7,  2,  6, 11,  8, 10, 12,
17, 15, 14,  8,  6, 12,  1,  4, 10,  7,  9, 11, 13,
...
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!):
.
.                1,
.              3,  2,
.            4,  1,  3,
.          6,  5,  2,  4,
.        2,  3,  1,  6,  5,
.      8,  7,  5,  2,  4,  6,
.   10,  9,  4,  1,  3,  5,  7,
...
Also the triangle may be reformatted for reading from left to right:
.
.                           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;
...

Alois P. Heinz, Rows n = 1..201, flattened

Middle diagonal gives A000012.
Right border gives A000027.
Indices of the 1's are in A001844.
Cf. A288530 is the same triangle but with every entry minus 1.
Other sequences of the same family are A269526, A274528, A274650, A274651, A274820, A274821, A286297.

T(n,k) = A288530(n-1,k-1) + 1.

T(n,n) = n.

A288384 a(n) is the n-th term of the leading column of the right triangle in A274650 minus n.

Original entry on oeis.org

0, 0, 1, -1, 1, -1, 0, 1, -1, 2, -1, -1, 0, 0, 1, -1, 0, 1, -1, 0, 0, 1, -1, 0, 1, -1, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 1, -1, 1, -1, 0, 2, -1, -1, 0, 0, 0, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 1, -1, 1, -1, 1, 1, -2, 0, 1, -1, 1, 1, -2, 0, 0, 1, -1, 0, 0, 0, 1, -1, 2, 0, -2, 0, 0, 1, -1, 1, -1, 0, 1, -1, 1, -1, 0, 0, 1, -1, 0, 0, 1, -1, 0, 1

Offset: 0

Omar E. Pol, Jun 08 2017

sign

It appears that this sequence has infinitely many negative terms, infinitely many zeros and infinitely many positive terms.

Inspired by A286294.
Partial sums give A288424.
Cf. A274650, A274651.

(* function a286294[] is defined in A286294 using A274651 *)
a288384[n_] := a286294[n] - Range[0,n]
a288384[104] (* data *) (* Hartmut F. W. Hoft, Jun 13 2017 *)

a(n) = A274650(n,0) - n = A286294(n) - n, n >= 0.

Signs reversed at the suggestion of Hartmut F. W. Hoft and edited by Omar E. Pol, Jun 13 2017

A284145 Triangle read by rows T(n,k) in which each term is the least positive integer not yet appearing in the triangle that is coprime to all the terms in its associated row, column, diagonal and antidiagonal.

Original entry on oeis.org

1, 2, 3, 5, 7, 4, 9, 11, 13, 17, 19, 8, 21, 23, 25, 29, 31, 37, 16, 27, 41, 43, 47, 53, 35, 59, 61, 67, 49, 71, 73, 33, 79, 83, 85, 89, 97, 101, 95, 103, 14, 107, 109, 113, 121, 127, 131, 137, 139, 143, 115, 149, 151, 157, 133, 163, 65, 167, 173, 179, 6, 181, 187, 191, 193, 197

Offset: 1

Bob Selcoe, Mar 20 2017

Conjecture 1: The triangle is a permutation of the natural numbers.
Let F(k) and G(n) be the set of prime factors of all terms in a given column k or diagonal n (diagonal n originates at T(n,1)).
Conjecture 2: Each F(k) and G(n) is a permutation of the prime numbers (except F(1) and G(1), which obviously also contain 1).
Let S be a set of terms whose members have certain specified characteristics (e.g., even numbers or prime numbers). Sets S whose members appear in due course in ascending order include:
(a) Prime numbers (so 2 appears first, followed by 3, 5, 7, 11, ...);
(b) Numbers which have exactly the same prime factors (so for example: {6, 12, 18, 24, 36, 48, 54, 72, ...} appear ascending order because their prime factors are {2,3});
(c) Powers of prime(j), because they are a subcategory of (b) (so for example: 5 appears first, followed by 25, 125, 625, 3125, ...).

			Triangle begins:
    1
    2   3
    5   7   4
    9  11  13  17
   19   8  21  23  25
   29  31  37  16  27  41
   43  47  53  35  59  61  67
   49  71  73  33  79  83  85  89
   97 101  95 103  14 107 109 113 121
  127 131 137 139 143 115 149 151 157 133
  163  65 167 173 179   6 181 187 191 193 197
T(7,4) = 35 because terms with prime factor 2 already appear in the diagonal (and column), and terms with prime factor 3 already appear in the diagonal (and antidiagonal) to T(7,4); no terms with prime factors 5 or 7 appear in any row, column, diagonal or antidiagonal to T(7,4); and terms 5, 7, and 25 already appear in the triangle.

Rémy Sigrist, Table of n, a(n) for n = 1..20100; rows 1..50 in flattened form
Rémy Sigrist, PARI program for A284145
Rémy Sigrist, Representation of prime numbers among the first 100 rows

Cf. A274651.

cp's OEIS Frontend

A269526 Square array T(n,k) (n>=1, k>=1) read by antidiagonals upwards in which each term is the least positive integer satisfying the condition that no row, column, diagonal, or antidiagonal contains a repeated term.

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A274650 Triangle read by rows: T(n,k), (0 <= k <= n), in which each term is the least nonnegative integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

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A274821 Hexagonal spiral constructed on the nodes of the infinite triangular net in which each term is the least positive integer such that no diagonal contains a repeated term.

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A286294 Leading column of right triangle in A274650.

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A288530 Triangle read by rows in reverse order: T(n,k), (0 <= k <= n), in which each term is the least nonnegative integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

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A288531 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.

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A288384 a(n) is the n-th term of the leading column of the right triangle in A274650 minus n.

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A284145 Triangle read by rows T(n,k) in which each term is the least positive integer not yet appearing in the triangle that is coprime to all the terms in its associated row, column, diagonal and antidiagonal.

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