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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.

%I A342268 #17 Mar 30 2024 23:08:29
%S A342268 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,
%T A342268 136,130,37,3,2,3,184,154,35,3,184,154,35,3,297,273,76,34,4,1,389,264,
%U A342268 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
%N 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.
%C A342268 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).
%H A342268 Scott R. Shannon, <a href="/A342268/a342268.txt">Rows 2 through 70</a>.
%e A342268 Triangle begins:
%e A342268 1;
%e A342268 3, 1;
%e A342268 7, 3;
%e A342268 12, 9, 1;
%e A342268 23, 14;
%e A342268 34, 27, 7;
%e A342268 53, 42, 8, 3;
%e A342268 78, 53, 4, 1, 1;
%e A342268 110, 79, 29, 6, 0, 1;
%e A342268 136, 130, 37, 3, 2, 2;
%e A342268 184, 154, 35, 3;
%e A342268 242, 195, 81, 15, 4, 1;
%e A342268 297, 273, 76, 34, 4, 1;
%e A342268 389, 264, 48, 15;
%e A342268 449, 403, 153, 46, 7;
%e A342268 547, 497, 163, 69, 9, 3;
%e A342268 679, 519, 207, 59, 5, 2;
%e A342268 759, 717, 268, 71, 22, 5;
%e A342268 900, 819, 329, 100, 16, 5, 0, 0, 0, 1;
%e A342268 1079, 885, 271, 82, 9, 1;
%e A342268 ...
%Y A342268 Cf. A007678.
%Y A342268 Row sums give A341734.
%K A342268 nonn,tabf
%O A342268 2,2
%A A342268 _Scott R. Shannon_ and _N. J. A. Sloane_, Mar 07 2021