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It is conjectured that this and A296346 together are a permutation of the nonnegative numbers.

It is conjectured that this and A296345 together are a permutation of the nonnegative numbers.

It is conjectured that this and A296347 together are a permutation of the nonnegative numbers.

It is conjectured that this and A296348 together are a permutation of the nonnegative numbers.

It is conjectured that this and A296343 together are a permutation of the nonnegative numbers.

It is conjectured that this and A296344 together are a permutation of the nonnegative numbers.

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)

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.

Analog of A269526, but note that this is a right triangle.

The same rule applied to an equilateral triangle gives A269526.

We construct the triangle by reading from left to right in each row, starting with T(1,1) = 1.

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.

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

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 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 since both sequences are essentially the same.

From Bob Selcoe, Feb 15 2017: (Start)

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.

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):

n/k 21 22 23 24 25 26

41 1 3

42 2

43 3 1 2

44 4 3

45 2 1 4

46 2

47 4 1

48 3 4

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.).

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)

From Hartmut F. W. Hoft, Jun 12 2017: (Start)

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.

Proof by (recursive) picture:

Positions in the triangle that are empty and those containing the dots of the guiding diagonals contain numbers larger than two.

n\k 1 2 3 4 5 6 7 8 10 12 14 16 18 20 22 24

1 |1

2 |2 .

3 | 1 2

4 | .

5 | 2 1 .

6 | 2 .

7 | 1 .

8 |____________.

9 | |1 .

10 | |2 . .

11 | | 1 2 .

12 | | . .

13 | | 2 1 . .

14 | | 2 . .

15 | | 1 . .

16 |_____|______________._____.

17 | | |1 . .

18 | | |2 . . .

19 | | | 1 2 . .

20 | | | . . .

21 | | | 2 1 . . .

22 | | | 2 . . .

23 | | | 1 . . .

24 |_____|_______|____._______._____._______.

1 2 3 4 5 6 7 8 12 16 20 24

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.

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}).

(End)