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A364746
Irregular triangle read by rows T(n,k), n >= 1, 1 <= k <= A002378(n), which is mentioned in the conjecture of A364639 (see Comments lines for definition).
1
1, 0, 0, 1, -1, 1, 0, 0, 0, 0, 1, -1, 0, 1, 0, -1, 1, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 1, 0, -1, 0, 1, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 1, 0, -1, 0, 0, 1, 0, 0, -1, 0, 1, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 1, 0, -1, 0, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0, 0, -1, 0, 1
OFFSET
1
COMMENTS
It is conjectured that 1 together with infinitely many copies of the n-th row gives the n-th column of the triangle A364639.
Row n has n blocks and every block has n + 1 terms.
In the n-th row its m-th block lists n - m zeros, 1, m - 1 zeros, and the last term of the block is -1 except in the n-th block, in which the last term is zero, with 1 <= m <= n.
Thus in the n-th row the number of -1's is equal to n - 1, the number of zeros is equal to A002061(n) and the number of 1's is equal to n.
EXAMPLE
Triangle begins:
1,0;
0,1,-1,1,0,0;
0,0,1,-1,0,1,0,-1,1,0,0,0;
0,0,0,1,-1,0,0,1,0,-1,0,1,0,0,-1,1,0,0,0,0;
0,0,0,0,1,-1,0,0,0,1,0,-1,0,0,1,0,0,-1,0,1,0,0,0,-1,1,0,0,0,0,0;
...
Also the triangle can be written showing the n blocks in the n-th row as shown below:
1, 0;
0,1,-1, 1,0, 0;
0,0,1,-1, 0,1,0,-1, 1,0,0, 0;
0,0,0,1,-1, 0,0,1,0,-1, 0,1,0,0,-1, 1,0,0,0, 0;
0,0,0,0,1,-1, 0,0,0,1,0,-1, 0,0,1,0,0,-1, 0,1,0,0,0,-1, 1,0,0,0,0, 0;
...
CROSSREFS
Row sums give A000012.
Column 1 gives A000007.
Right border gives A000004.
The length of row n is A002378(n).
Sequence in context: A288929 A285083 A266982 * A051341 A057211 A120531
KEYWORD
sign,tabf
AUTHOR
Omar E. Pol, Aug 05 2023
STATUS
approved