%I #36 Sep 27 2022 10:36:04
%S 1,1,1,3,2,1,7,6,3,1,15,16,10,4,1,31,40,30,15,5,1,63,96,84,50,21,6,1,
%T 127,224,224,154,77,28,7,1,255,512,576,448,258,112,36,8,1,511,1152,
%U 1440,1248,810,405,156,45,9,1,1023,2560,3520,3360,2420,1362,605,210,55,10,1
%N Triangle read by rows: CL(n,k) is the number of labeled subgraphs with k edges of the n-cycle C_n.
%H Andrew Howroyd, <a href="/A277919/b277919.txt">Table of n, a(n) for n = 0..1274</a>
%H Thomas Selig, <a href="https://arxiv.org/abs/2202.06487">Combinatorial aspects of sandpile models on wheel and fan graphs</a>, arXiv:2202.06487 [math.CO], 2022.
%F The identity CL(n,k) = 2^(n-2*k) * CL(n,n-k) can be proved combinatorially.
%F G.f.: (1 - 2*x + 2*x^2)/((1-x)*(1 - y*x - 2*x + y*x^2)). - _Andrew Howroyd_, Sep 27 2019
%e For row 3 of the triangle below: there are 7 labeled subgraphs of the triangle C_3 with 0 edges, 6 with 1 edge, 3 with 2 edges, and 1 with 3 edges (C_3 itself).
%e Triangle begins:
%e 1;
%e 1, 1;
%e 3, 2, 1;
%e 7, 6, 3, 1;
%e 15, 16, 10, 4, 1;
%e 31, 40, 30, 15, 5, 1;
%e 63, 96, 84, 50, 21, 6, 1;
%e 127, 224, 224, 154, 77, 28, 7, 1;
%e 255, 512, 576, 448, 258, 112, 36, 8, 1;
%e 511, 1152, 1440, 1248, 810, 405, 156, 45, 9, 1;
%e 1023, 2560, 3520, 3360, 2420, 1362, 605, 210, 55, 10, 1;
%e ...
%o (PARI) T(n)={[Vecrev(p) | p<-Vec((1 - 2*x + 2*x^2)/((1-x)*(1 - y*x - 2*x + y*x^2)) + O(x*x^n))]}
%o { my(A=T(12)); for(n=1, #A, print(A[n])) } \\ _Andrew Howroyd_, Sep 27 2019
%Y Row sums give A005592.
%Y Middle diagonal gives A110170.
%K nonn,tabl
%O 0,4
%A _John P. McSorley_, Nov 03 2016
%E More terms from _John P. McSorley_, Nov 17 2016