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%I #10 Feb 18 2013 11:30:41
%S 18,28,32,45,52,69,79,100,114,140,158,189,212,249,277,320,354,404,444,
%T 501,548,613,667,740,802,884,954,1045,1124,1225,1313,1424,1522,1644,
%U 1752,1885,2004,2149,2279,2436,2578,2748,2902,3085,3252
%N Number of connected graphs with one cycle of length m = n-4 and n nodes.
%C We have unicyclic graphs of order n = m+4 with a cycle of length m. Only 4 nodes of those graphs belong to the rooted trees attached to the cycle, so the orders of those trees can be only 1,2,3,4, or 5. The set of graphs can be divided in five subsets S_1, S_2, S_3, S_4 and S_5, such that
%C S_1 has trees of orders [5,1,1,...,1],
%C S_2 has trees of orders [4,2,1,...,1],
%C S_3 has trees of orders [3,3,1,...,1],
%C S_4 has trees of orders [3,2,2,1,...,1] and
%C S_5 has trees of orders [2,2,2,2,1,...,1].
%C |S_1| = 9 since there are 9 rooted trees with 5 points.
%C |S_2| = 4floor(m/2).
%C |S_3| = 3floor(m/2). We consider the 3 2-combinations (with repetition) of the 2 distinct rooted trees of order 3.
%C |S_4| = 2floor((m-1)^2/4) since floor((m-1)^2/4) is the number of bracelets with m beads, 2 of which are red, 1 of which is blue.
%C With x=m-4, |S_5| = <(x^3 +9x^2 +(32-9(x mod 2))x)/48 +0.6>. The value of |S_5| is equal to the number of m-bead bracelets with 4 red beads.
%C This sequence is the fifth column of table T of A058879.
%H Washington Bomfim, <a href="/A141782/b141782.txt">Table of n, a(n) for n = 7..100</a>
%H Washington Bomfim, <a href="http://commons.wikimedia.org/wiki/Image:FiguraA141782.PNG">The 32 unicyclic graphs of order 9 with a pentagon.</a>.
%F With m = n-4 and x = m-4, a(n) = <(x^3 +9x^2 +(32-9(x mod 2))x)/48 +0.6> + 2floor((m-1)^2/4) + 7floor(m/2) + 9. Empirically for n odd a(n) = (n^3 +9n^2 -n +87)/48 Empirically for n even a(n) = (n^3 +9n^2 +8n +192-n%4*6)/48.
%F Empirical g.f.: -x^7*(16*x^7-23*x^6-9*x^5+18*x^4-17*x^3+24*x^2+8*x-18) / ((x-1)^4*(x+1)^2*(x^2+1)). [_Colin Barker_, Feb 18 2013]
%e E.g. a(9)=32. Click the link to see an illustration of the 32 unicyclic graphs of order 9 with a pentagon.
%o (PARI) m=n-4 x=m-4 a(n) = round((x^3+9*x^2+(32-9*(x%2))*x)/48+0.6)+2*floor((m-1)^2/4)+7*floor(m/2)+9
%Y Cf. A058879, A005232, A000081, A002620.
%K nonn
%O 7,1
%A _Washington Bomfim_, Jul 31 2008
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