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%I #31 Dec 07 2019 12:18:26
%S 1,3,15,123,1635,35043,1206915,66622083,5884188675,830476531203,
%T 187106645932035,67241729173555203,38521811621470420995,
%U 35161184767296890265603,51112793797110111859802115,118291368253025368001553530883,435713124846749574718274002747395,2553666761436949125065383836043837443
%N Number of simple labeled graphs with (at most) 3-colored nodes such that no edge connects two nodes of the same color.
%C Cf. A213442, which counts colorings of labeled graphs on n vertices using exactly 3 colors. For 3-colorable labeled graphs on n vertices see A076315. - _Peter Bala_, Apr 12 2013
%D F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, 1973, 16-18.
%H Andrew Howroyd, <a href="/A191371/b191371.txt">Table of n, a(n) for n = 0..50</a>
%H S. R. Finch, <a href="/A191371/a191371.pdf">Bipartite, k-colorable and k-colored graphs</a>, June 5, 2003. [Cached copy, with permission of the author]
%H R. C. Read, <a href="http://cms.math.ca/10.4153/CJM-1960-035-0">The number of k-colored graphs on labelled nodes</a>, Canad. J. Math., 12 (1960), 410—414.
%H R. P. Stanley, <a href="http://www-math.mit.edu/~rstan/pubs/pubfiles/18.pdf">Acyclic orientation of graphs</a>, Discrete Math. 5 (1973), 171-178. North Holland Publishing Company.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/k-ColorableGraph.html">k-Colorable Graph</a>
%F a(n) = Sum(C(n,k)*C(n-k,r)*2^(k*r+k*m+r*m)) where the sum is taken over all nonnegative solutions to k + r + m = n.
%F From _Peter Bala_, Apr 12 2013: (Start)
%F a(n) = Sum_{k = 0..n} binomial(n,k)*2^(k*(n-k))*A047863(k).
%F a(n)= 3*A000685(n) for n >= 1.
%F Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)). Then a generating function for this sequence is E(x)^3 = sum {n >= 0} a(n)*x^n/(n!*2^C(n,2)) = 1 + 3*x + 15*x^2/(2!*2) + 123*x^3/(3!*2^3) + 1635*x^4/(4!*2^6) + ....
%F In general, for k = 1, 2, ..., E(x)^k is a generating function for labeled k-colored graphs (see Read). For other examples see A047863 (k = 2) and A223887 (k = 4). (End)
%e a(2) = 15: There are two labeled 3-colorable graphs on 2 nodes, namely
%e A) 1 2 B) 1 2
%e o o o----o
%e Using 3 colors there are 9 ways to color the graph of type A and 3*2 = 6 ways to color the graph of type B so that adjacent vertices do not share the same color. Thus there are in total 15 labeled 3-colored graphs on 2 vertices.
%t f[{k_,r_,m_}]:= Binomial[m+r+k,k] Binomial[m+r,r] 2^ (k r + k m + r m); Table[Total[Map[f,Compositions[n,3]]],{n,0,20}]
%o (Python)
%o from sympy import binomial
%o def a047863(n): return sum([binomial(n, k)*2**(k*(n - k)) for k in range(n + 1)])
%o def a(n): return sum([binomial(n, k)*2**(k*(n - k))*a047863(k) for k in range(n + 1)]) # _Indranil Ghosh_, Jun 03 2017
%o (PARI) seq(n)={my(p=(sum(j=0, n, x^j/(j!*2^binomial(j, 2))) + O(x*x^n))^3); Vecrev(sum(j=0, n, j!*2^binomial(j,2)*polcoef(p,j)*x^j))} \\ _Andrew Howroyd_, Dec 03 2018
%Y Column k=3 of A322280.
%Y Cf. A047863. A000685, A076315, A223887.
%K nonn,easy
%O 0,2
%A _Geoffrey Critzer_, Jun 01 2011
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