%I #30 Feb 16 2025 08:33:18
%S 0,1,10,450,46440,8580600,2485501200,1038647610000,591422976144000,
%T 440175696904944000,414834426527320800000,482828797838174467680000,
%U 680160665982184667280000000,1140497273795065245115056000000,2244756232031112064775686176000000
%N Number of assembly trees for complete bipartite graph K_{n,n}.
%H Andrew Vince and Miklos Bona, <a href="https://arxiv.org/abs/1204.3842">The Number of Ways to Assemble a Graph</a>, arXiv preprint, arXiv:1204.3842 [math.CO], 2012. See Theorem 23.
%H Andrew Vince and Miklos Bona, <a href="https://doi.org/10.37236/2644">The Number of Ways to Assemble a Graph</a>, The Electronic Journal of Combinatorics, Volume 19, Issue 4 (2012), Article P54.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/AssemblyNumber.html">Assembly Number</a>
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CompleteBipartiteGraph.html">Complete Bipartite Graph</a>
%F Define b(n) = (2*(6*n^2 - 12*n + 5)/n^2)*b(n-1) - ((4*n^3 - 20*n^2 + 29*n - 10)/(n^3 - n^2))*b(n-2), with b(0) = 0, b(1) = 1 and b(2) = 5/2. Then a(n) = n!*n!*b(n). - _Franck Maminirina Ramaharo_, Jan 28 2019
%t Table[n!^2 SeriesCoefficient[1 - Sqrt[(1 - x)^2 + (1 - y)^2 - 1], {x, 0, n}, {y, 0, n}], {n, 10}] (* _Eric W. Weisstein_, Mar 01 2023 *)
%t With[{nterms = 10}, RecurrenceTable[{b[1] == 1, b[2] == 5/2, b[n] == 2 (6 n^2 - 12 n + 5)/n^2 b[n - 1] - (4 n^3 - 20 n^2 + 29 n - 10)/(n^3 - n^2) b[n - 2]}, b, {n, nterms}] Range[nterms]!^2] (* _Eric W. Weisstein_, Mar 01 2023 *)
%o (Maxima)
%o (b[0] : 0, b[1] : 1, b[2] : 5/2, b[n] := (2*(6*n^2 - 12*n + 5)/n^2)*b[n - 1] - ((4*n^3 - 20*n^2 + 29*n - 10)/(n^3 - n^2))*b[n - 2])$
%o a(n) := n!*n!*b[n]$ /* _Franck Maminirina Ramaharo_, Jan 28 2019 */
%Y Cf. A361072 (number of assembly trees for K_{n,n,n}).
%K nonn,changed
%O 0,3
%A _N. J. A. Sloane_, Oct 08 2012
%E More terms from _Franck Maminirina Ramaharo_, Jan 28 2019