OFFSET
7,1
COMMENTS
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
S_1 has trees of orders [5,1,1,...,1],
S_2 has trees of orders [4,2,1,...,1],
S_3 has trees of orders [3,3,1,...,1],
S_4 has trees of orders [3,2,2,1,...,1] and
S_5 has trees of orders [2,2,2,2,1,...,1].
|S_1| = 9 since there are 9 rooted trees with 5 points.
|S_2| = 4floor(m/2).
|S_3| = 3floor(m/2). We consider the 3 2-combinations (with repetition) of the 2 distinct rooted trees of order 3.
|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.
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.
This sequence is the fifth column of table T of A058879.
LINKS
Washington Bomfim, Table of n, a(n) for n = 7..100
Washington Bomfim, The 32 unicyclic graphs of order 9 with a pentagon..
FORMULA
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.
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]
EXAMPLE
E.g. a(9)=32. Click the link to see an illustration of the 32 unicyclic graphs of order 9 with a pentagon.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Washington Bomfim, Jul 31 2008
STATUS
approved