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A239910 Number of forests with three connected components in the complete graph K_{n}. 5
0, 0, 1, 6, 45, 435, 5250, 76608, 1316574, 26100000, 587030895, 14780620800, 412069511139, 12604714327296, 419801484375000, 15123782440058880, 586049426860524300, 24307340986526810112, 1074495780444130114509, 50429952000000000000000 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
COMMENTS
Equation (47) of Liu-Chow (1984) also gives the analogous formulas for four and five components. (They should also be entered into the OEIS, in case someone wants to help.)
LINKS
C. J. Liu and Yutze Chow, On operator and formal sum methods for graph enumeration problems, SIAM J. Algebraic Discrete Methods, 5 (1984), no. 3, 384-406. MR0752043 (86d:05059).
FORMULA
From Harry Richman, Aug 17 2022: (Start)
a(n) = n^(n-6)*(n-1)*(n-2)*(n^2+13*n+60)/8.
E.g.f.: T(x)^{3}/3!, where T(x) is the e.g.f. for the number of spanning trees in K_{n} A000272, i.e., T(x) = Sum_{i>=1} i^(i-2)*x^i/i!. (End)
MAPLE
f := n-> (n-1)*(n-2)*n^(n-6)*(n^2+13*n+60)/8; [seq(f(n), n=3..20)];
MATHEMATICA
Table[(n-1)*(n-2) * n^(n - 6) * (n^2 + 13 n + 60)/8, {n, 1, 20}] (* Vincenzo Librandi, Apr 10 2014, simplified by Vaclav Kotesovec, Feb 20 2020 *)
PROG
(Magma) [(n-1)*(n-2)*n^(n-6)*(n^2+13*n+60)/8: n in [1..20]]; // Vincenzo Librandi, Apr 10 2014
CROSSREFS
Column m=3 of A105599. A diagonal of A138464. - Alois P. Heinz, Apr 10 2014
Sequence in context: A001879 A019577 A097814 * A228194 A331726 A084064
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Apr 09 2014
STATUS
approved

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Last modified March 18 22:56 EDT 2024. Contains 370952 sequences. (Running on oeis4.)