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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)

Write the sequence of odious odd numbers above the sequence of evil odd numbers connecting all that are 2 apart:

1 7 11-13 19-21 25 31 35-37 41 47-49 55 59-61 67-69 73 79-81 87 91-93 97

3-5 9 15-17 23 27-29 33 39 43-45 51-53 57 63-65 71 75-77 83-85 89 95 99-

Remove the connected numbers:

1 7 25 31 41 55 73 87 97

9 23 33 39 57 71 89 95

Define these as "isolated".

The sequence is the smaller of the remaining pairs that are 2 apart.

The 1 is not a member since there is no change in parity between 1 and 7.

All of the differences between adjacent numbers in both the evil and odious sequences are either 2, 4 or 6, with 4 being the indicator that a transition in parity occurs. The program provided utilizes that fact to produce the sequence.

The sequence that includes all numbers along this path is A047522 (numbers congruent to {1,7} mod 8). This is also the same as the odd terms of A199398 (XOR of the first n odd numbers).

This sequence is similar to A044449 (numbers n such that string 1,3 occurs in the base 4 representation of n but not of n+1), but it contains additional terms. An example is 119. Its base 4 representation is 1313 while the base 4 representation of 120 is 1320. It may be that another workable definition of the sequence is -- numbers n in base 4 representation such that string 1,3 occurs one less time in n+1 than n, but I have not been able to check this.

The difference between the numbers in the sequence is always either 8 or 16, however there appears to be no recurring repetitions in it. Writing the 8 as a 0 and the 16 as a 1 (or dividing the difference pattern by 2 and subtracting a 1) produces a difference pattern of: 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1... which is an infinite word.

A similar process writing Even Odious over Even Evils produces 6, 22, 30, 38, 54, 70... which is twice A131323 (Odd numbers n such that the binary expansion ends in an even number of 1's), with all numbers along the path given by A047451 (numbers congruent to {0,6} mod 8) and yields the same difference pattern which produces the same infinite word.

These are the values of n for which binomial(n,6) is odd. See Maple code. - Gary Detlefs, Nov 29 2011

This is the case k=7 of the form (n + sqrt(k))^2 + (n - sqrt(k))^2.

Equivalently, numbers m such that 2*m - 28 is a square.