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Square array read by antidiagonals: T(n,k) is the number of k-edge colored trees on vertex set [n] (n>=2, k>=2).
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%I #15 Aug 26 2024 09:55:33

%S 2,3,6,4,18,24,5,36,168,120,6,60,528,2160,720,7,90,1200,10920,35640,

%T 5040,8,126,2280,34200,293760,720720,40320,9,168,3864,82800,1275120,

%U 9767520,17297280,362880,10,216,6048,170520,3946320,58968000,387636480,481178880,3628800

%N Square array read by antidiagonals: T(n,k) is the number of k-edge colored trees on vertex set [n] (n>=2, k>=2).

%C T(n,2) = n! = A000142(n).

%D R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999, Exercise 5.28, pp. 81, 124.

%H M. Cho, D. Kim, S. Seo, and H. Shin, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v11i1n10">Colored Prufer codes for k-edge colored trees</a>, The Electronic Journal of Combinatorics, 11 (2004), #N10.

%F T(n,k) = k(n-2)!*binomial((k-1)n, n-2).

%e T(3,3) = 18; indeed, a 3-vertex tree ABC can be labeled in 6 ways and for each labeled tree the 2 edges can be colored in 3 ways (a and b, a and c, b and c).

%e The antidiagonals start:

%e 2;

%e 3, 6;

%e 4, 18, 24;

%e 5, 36, 168, 120;

%e 6, 60, 528, 2160, 720;

%p T := proc(n,k) options operator, arrow: k*factorial(n-2)*binomial((k-1)*n, n-2) end proc: seq(seq(T(i,j-i), i = 2 .. j-2), j = 4 .. 12); # the command T(n,k) yields T(n,k).

%t T[n_, k_] := k (n - 2)! * Binomial[(k - 1) n, n - 2];

%t Table[T[n - k + 2, k], {n, 2, 10}, {k, n, 2, -1}] // Flatten (* _Jean-François Alcover_, Aug 26 2024 *)

%Y Cf. A000142.

%K nonn,tabl

%O 2,1

%A _Emeric Deutsch_, Oct 27 2014