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A342268
Irregular triangle read by rows: Take a regular (2*n)-sided polygon (n>=2) with all diagonals drawn, as in A007678. Then T(n,k) = (1/(2*n))*(number of k-sided polygons in that figure) for k = 3, 4, ..., max_k.
2
1, 3, 1, 7, 3, 12, 9, 1, 23, 14, 34, 27, 7, 53, 42, 8, 3, 78, 53, 4, 1, 1, 110, 79, 29, 6, 0, 1, 136, 130, 37, 3, 2, 3, 184, 154, 35, 3, 184, 154, 35, 3, 297, 273, 76, 34, 4, 1, 389, 264, 48, 15, 449, 403, 153, 46, 7, 547, 497, 163, 69, 9, 3, 679, 519, 207, 59, 5, 2, 759, 717, 268, 71, 22, 5
OFFSET
2,2
COMMENTS
This is a version of A331450: take the even-indexed rows and divide by the number of vertices. That is, we only consider one sector (or pizza slice).
EXAMPLE
Triangle begins:
1;
3, 1;
7, 3;
12, 9, 1;
23, 14;
34, 27, 7;
53, 42, 8, 3;
78, 53, 4, 1, 1;
110, 79, 29, 6, 0, 1;
136, 130, 37, 3, 2, 2;
184, 154, 35, 3;
242, 195, 81, 15, 4, 1;
297, 273, 76, 34, 4, 1;
389, 264, 48, 15;
449, 403, 153, 46, 7;
547, 497, 163, 69, 9, 3;
679, 519, 207, 59, 5, 2;
759, 717, 268, 71, 22, 5;
900, 819, 329, 100, 16, 5, 0, 0, 0, 1;
1079, 885, 271, 82, 9, 1;
...
CROSSREFS
Cf. A007678.
Row sums give A341734.
Sequence in context: A329369 A083239 A333847 * A316742 A189050 A095868
KEYWORD
nonn,tabf
AUTHOR
STATUS
approved