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%I #12 Mar 08 2021 02:16:37
%S 0,1,4,10,22,37,68,106,137,225,310,376,538,685,716,1058,1288,1471,
%T 1842,2170,2327,2941,3388,3734,4412,4993,5444,6306,7042,7391,8680,
%U 9586,10289,11585,12682,13628,15078,16381,17440,19210,20740,21899,24038,25810,27245,29613,31648,33418,35992,38305
%N a(n) = A007678(2*n)/(2*n).
%C This is the number of cells in a 1/(2*n)-th sector of a regular (2*n)-gon with all diagonals drawn. See Rubinstein's illustrations in A007678.
%H Scott R. Shannon, <a href="/A341734/b341734.txt">Table of n, a(n) for n = 1..71</a>
%e If we divide a regular hexagon with all diagonals drawn into 6 sectors (or pizza slices), each sector contains three triangles and one quadrilateral (cf. A331450), so a(3) = A007678(6)/6 = 24/6 = 4.
%Y Cf. A007678, A331450, A341735.
%Y Row sums of triangle in A342268.
%K nonn
%O 1,3
%A _Scott R. Shannon_ and _N. J. A. Sloane_, Mar 07 2021