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Additional info for Counting labelled trees (Canadian mathematical monographs)
Suppose we select one of the edges xy incident with the root x of a random tree Tn and proceed along it to y; then s(Tn) = 1 if and only if s(T*) = 1 - I, where T* is the subtree defined earlier. 4 that 2: n-1 pen, I) = pen, k)·P(k, 1 - I) k=1 n-1 = _1_ '" nn-2 L... (nk -- l)kk-2(n I _ k _ W-k-1P(k, 1 - I). k=1 I (nk -- II) kk-2(n pen, k) = -nn-2 - k _ l)n-k-1 . 2: n-1 The k nodes of T* can be chosen in (n k I) ways and, having chosen the nodes, T* can be formed in kk-2 ways. 2 there are /L(n) = IP(n, I) 1=1 n-1 =~ nn 2 '" (nk -- l)kk-2(n L...
N-2 m f(Cm, n) = m2( _n)l-m + '" ~ 1=2 m-I+1 '" ~ 1=1 (m - ~ + 1m-J-I+l 2)( -ny-m. (n - l)(n - 3) (n - 2)(n2 - 4n + 2) (n2 - 3n + 1)(n2 - 5n + 5) (n - 1)(n - 2)(n - 3)(n2 - 4n + 1) (n (n (n2 (n - 3)2 2)2(n - 4) 5n + 5)2 1)2(n - 3)2(n - 4) Bedrosian (1961, 1964b) derived formulas forf(Pm, n) andf(Cm, n) using the foldant algorithm of Nakagawa (1958) and he has expressed these polynomials in terms of a third family of polynomials the absolute values of whose coefficients add up to a Fibonacci number.
Kasai et ai. (1966a) used matrix methods to determine c(G) when G consists of isolated nodes and one of ten types of graphs A B where A and B are complete graphs, cycles, paths or complete 1 by s bipartite graphs (in a subsequent paper (I966b) they discussed the use of continuants in evaluating determinants that arise in these problems). The formulas that arise when A or B are paths or cycles are rather complicated so we shall derive polynomials of A B only when A and B are complete graphs (m) or complete bipartite graphs (I, s).