A167484 For n people on one side of a river, the number of ways they can all travel to the opposite side following the pattern of 2 sent, 1 returns, 2 sent, 1 returns, ..., 2 sent.

1, 1, 6, 108, 4320, 324000, 40824000, 8001504000, 2304433152000, 933295426560000, 513312484608000000, 372664863825408000000, 348814312540581888000000, 412647331735508373504000000, 606591577651197309050880000000, 1091864839772155156291584000000000, 2375897891344209620090486784000000000

Offset: 1

Author

Ron Smith (ron.smith(AT)henryschein.com), Nov 04 2009

Keywords

• easy
• nonn

Comments

This problem might arise if there was only a two-person boat available.

Also the number of ranked tree-child networks. - Michael Fuchs, May 29 2021

Examples

			For n=3 there are 6 ways. Let a,b,c start on one side. We have:
1) Send (a,b), return(a), send(a,c);
2) Send (a,b), return(b), send(b,c);
3) Send (b,c), return(b), send(a,b);
4) Send (b,c), return(c), send(a,c);
5) Send (a,c), return(a), send(a,b);
6) Send (a,c), return(c), send(b,c).
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..190
• Francois Bienvenu, Amaury Lambert, and Mike Steel, Combinatorial and stochastic properties of ranked tree-child networks, arXiv:2007.09701 [math.PR], 2021.
• Alessandra Caraceni, Michael Fuchs, and Guan-Ru Yu, Bijections for ranked tree-child networks, arXiv:2105.10137 [math.CO], 2021.

Programs

Mathematica

f[n_] := n! (n - 1)!^2/2^(n - 1); Array[f, 15] (* Robert G. Wilson v, Dec 17 2016 *)

Formula

a(n) = n!*((n-1)!)^2/((2!)^(n-1)).

a(n) ~ 4*sqrt(2)*Pi^(3/2)*n^(3*n-1/2)/(2^n*exp(3*n)). - Ilya Gutkovskiy, Dec 17 2016

Extensions

a(13) and a(14) corrected by Ilya Gutkovskiy, Dec 17 2016

More terms from Ilya Gutkovskiy, Dec 18 2016