%I #28 Jan 11 2021 22:37:25
%S 1,4,5,10,13,20,25,34,41,52,61,74,85,100,113,130,145,164,181,202,221,
%T 244,265,290,313,340,365,394,421,452,481,514,545,580,613,650,685,724,
%U 761,802,841,884,925,970,1013,1060,1105,1154,1201,1252
%N The number of orbits of triples of {1,2,...,n} under the action of the dihedral group of order 2n.
%C In other words, a(n) is the number of equivalence classes of length 3 words with an alphabet of size n where equivalence is up to rotation or reflection of the alphabet. For example when n is 3, the word 112 is equivalent to 223 and 331 by rotation of the alphabet, and these are equivalent to 332, 221 and 113 by reflection of the alphabet. - _Andrew Howroyd_, Jan 17 2020
%H Andrew Howroyd, <a href="/A236283/b236283.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).
%F Conjectures from _Colin Barker_, Jan 21 2014: (Start)
%F a(n) = (5 + 3*(-1)^n + 2*n^2)/4.
%F a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
%F G.f.: -x*(2*x^3-3*x^2+2*x+1) / ((x-1)^3*(x+1)).
%F (End)
%F From _Andrew Howroyd_, Jan 17 2020: (Start)
%F The above conjectures are true and can be derived from the following formulas for even and odd n.
%F a(n) = (n-2)*(n + 2)/2 + 4 for even n.
%F a(n) = (n-1)*(n + 1)/2 + 1 for odd n.
%F (End)
%F a(n) = A081352(n - 1) - A116940(n - 1). - _Miko Labalan_, Nov 12 2016
%e For n = 3 there are 5 orbits of triples:
%e [[1,1,1], [2,2,2], [3,3,3]],
%e [[1,1,2], [2,2,3], [1,1,3], [3,3,1], [3,3,2], [2,2,1]],
%e [[1,2,1], [2,3,2], [1,3,1], [3,1,3], [3,2,3], [2,1,2]],
%e [[1,2,2], [2,3,3], [1,3,3], [3,1,1], [3,2,2], [2,1,1]],
%e [[1,2,3], [2,3,1], [1,3,2], [3,1,2], [3,2,1], [2,1,3]].
%o (GAP)
%o a:=function(n)
%o local g,orbs;
%o g:=DihedralGroup(IsPermGroup,2*n);
%o orbs := OrbitsDomain(g, Tuples( [ 1 .. n ], 3), OnTuples );
%o return Size(orbs);
%o end;;
%o (PARI) a(n) = {(5 + 3*(-1)^n + 2*n^2)/4} \\ _Andrew Howroyd_, Jan 17 2020
%Y Cf. A236332 (4-tuples).
%K nonn,easy
%O 1,2
%A _W. Edwin Clark_, Jan 21 2014