OFFSET
0,3
COMMENTS
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.
LINKS
Alois P. Heinz, Rows n = 0..200, flattened
FORMULA
T(n,k) = A288531(n+1, k+1) - 1.
T(n,n) = n.
EXAMPLE
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;
...
CROSSREFS
AUTHOR
Omar E. Pol, Jun 10 2017
STATUS
approved