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%I #16 Sep 08 2023 13:21:31
%S 1,1,1,4,11,49,204,984,4807,24739,130065,701584,3851316,21489836,
%T 121517768,695307888,4019338527,23446201495,137875318035,816646459860,
%U 4868576661795,29196022525905,176022384523440,1066433501134560,6490009520072676,39659537885087124
%N Number of nonequivalent noncrossing trees with n edges up to rotation.
%C The number of all noncrossing trees with n edges is given by A001764.
%C The number of nodes will be n + 1.
%C Rotational symmetry is only possible with an even number of nodes and with a rotation of 180 degrees (rotation by n/2 nodes). A tree with rotational symmetry will always include exactly one edge that connects diametrically opposite nodes.
%C The sequence satisfies a(2n) = A000139(2n)/2. - _F. Chapoton_, Sep 08 2023
%H Andrew Howroyd, <a href="/A296532/b296532.txt">Table of n, a(n) for n = 0..200</a>
%F a(2n) = A001764(2n)/(2n+1), a(2n-1) = (A001764(2n-1) + n*A006013(n-1))/(2n).
%e Case n=3:
%e o---o o---o o---o o---o
%e | | \ \ /
%e o---o o o o---o o---o
%e In total there are 4 distinct noncrossing trees up to rotation.
%t a[n_] := If[EvenQ[n], Binomial[3*n, n]/((n + 1)*(2*n + 1)), ((2*n + 1)*Binomial[(1/2)*(3*n - 1), (n - 1)/2] + Binomial[3*n, n]) / ((n + 1)*(2*n + 1))];
%t Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, Dec 27 2017, after _Andrew Howroyd_ *)
%o (PARI) a(n)={(binomial(3*n, n)/(2*n+1) + if(n%2, binomial((3*n-1)/2, (n-1)/2)))/(n+1)}
%Y Cf. A001764, A006013, A296533 (up to rotation and reflection), A000139.
%K nonn
%O 0,4
%A _Andrew Howroyd_, Dec 14 2017
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