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Triangle read by rows: T(n,k) is the number of polygons with 2*n sides, of which k run through the center of a circle, on the circumference of which the 2*n vertices of the polygon are arranged at equal spacing.
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%I #47 Jun 13 2021 03:30:27

%S 0,0,1,1,0,2,16,24,12,8,744,960,576,192,48,56256,69120,39360,13440,

%T 2880,384,6385920,7580160,4204800,1420800,316800,46080,3840,

%U 1018114560,1178956800,642539520,216115200,49190400,7741440,806400,46080

%N Triangle read by rows: T(n,k) is the number of polygons with 2*n sides, of which k run through the center of a circle, on the circumference of which the 2*n vertices of the polygon are arranged at equal spacing.

%C Rotations and reflections are counted separately.

%C By "2*n-sided polygons" we mean the polygons that can be drawn by connecting 2*n equally spaced points on a circle.

%C T(0,0)=0 and T(0,1)=1 by convention.

%C The sequence is limited to even-sided polygons, since all odd-sided polygons have no side passing through the center.

%H Ludovic Schwob, <a href="/A330662/b330662.txt">Table of n, a(n) for n = 0..494</a>

%H Ludovic Schwob, <a href="/A330662/a330662.pdf">Illustration of T(3,k), 0≤k≤3</a>.

%F T(n,n) = 2^(n-1) * (n-1)! for all n >= 1.

%F T(n,0) = A307923(n) for all n>=1.

%F T(n,k) = binomial(n,k)* Sum_{i=k..n} (-1)^(i-k)*binomial(n-k,i-k)*(2n-1-i)!*2^(i-1), for n>=2 and 0<=k<=n.

%e Triangle begins:

%e 0;

%e 0, 1;

%e 1, 0, 2;

%e 16, 24, 12, 8;

%e 744, 960, 576, 192, 48;

%p T := (n, k) -> `if`(n<2, k, 2^(k-1)*binomial(n,k)*(2*n-k-1)!*hypergeom([k-n], [k-2*n+ 1], -2)):

%p seq(seq(simplify(T(n,k)), k=0..n),n=0..7); # _Peter Luschny_, Jan 07 2020

%Y Row sums give A001710(2*n-1) (number of polygons with 2*n sides).

%Y Cf. A000165 (diagonal).

%Y Star polygons: A014106, A055684, A102302.

%Y Cf. A309318.

%K nonn,tabl

%O 0,6

%A _Ludovic Schwob_, Dec 23 2019