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%I #32 May 25 2023 08:55:34
%S 1,1,1,1,2,1,1,3,3,1,1,4,6,4,3,1,5,10,10,15,12,1,6,15,20,45,72,60,1,7,
%T 21,35,105,252,420,360,1,8,28,56,210,672,1680,2880,2520,1,9,36,84,378,
%U 1512,5040,12960,22680,20160,1,10,45,120,630,3024,12600,43200,113400,201600,181440
%N Triangle T(n,k), n>=0, 0<=k<=n, read by rows: T(n,k) = number of ways an undirected cycle of length k can be built from n labeled nodes.
%H Alois P. Heinz, <a href="/A144151/b144151.txt">Rows n = 0..140, flattened</a>
%F T(n,k) = C(n,k) if k<=2, else T(n,k) = C(n,k)*(k-1)!/2.
%F E.g.f.: exp(x)*(log(1/(1 - y*x))/2 + 1 + y*x/2 + (y*x)^2/4). - _Geoffrey Critzer_, Jul 22 2016
%e T(4,3) = 4, because 4 undirected cycles of length 3 can be built from 4 labeled nodes:
%e .1.2. .1.2. .1-2. .1-2.
%e ../|. .|\.. ..\|. .|/..
%e .3-4. .3-4. .3.4. .3.4.
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 3, 3, 1;
%e 1, 4, 6, 4, 3;
%e 1, 5, 10, 10, 15, 12;
%e ...
%p T:= (n,k)-> if k<=2 then binomial(n,k) else mul(n-j, j=0..k-1)/k/2 fi:
%p seq(seq(T(n,k), k=0..n), n=0..12);
%t t[n_, k_ /; k <= 2] := Binomial[n, k]; t[n_, k_] := Binomial[n, k]*(k-1)!/2; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 18 2013 *)
%t CoefficientList[Table[1 + n x (2 + (n - 1) x + 2 HypergeometricPFQ[{1, 1, 1 - n}, {2}, -x])/4, {n, 0, 10}], x] (* _Eric W. Weisstein_, Apr 06 2017 *)
%Y Columns 0-4 give: A000012, A000027, A000217, A000292, A050534.
%Y Diagonal gives: A001710.
%Y Cf. A000142, A007318.
%Y Row sums are in A116723. - _Alois P. Heinz_, Jun 01 2009
%Y Excluding columns k=0,1,and 2 the row sums are A002807. - _Geoffrey Critzer_, Jul 22 2016
%Y Cf. A284947 (k-cycle counts for k >= 3 in the complete graph K_n). - _Eric W. Weisstein_, Apr 06 2017
%Y T(2n,n) gives A006963(n+1) for n>=3.
%K nonn,tabl
%O 0,5
%A _Alois P. Heinz_, Sep 12 2008
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