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

1,3

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

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.

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

Example

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).

Mathematica

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

Crossrefs

Sequence in context: A010563 A114310 A221954 * A011555 A122722 A127946

Adjacent sequences: A167481 A167482 A167483 * A167485 A167486 A167487

Keyword

easy,nonn

Author

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

Extensions

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

More terms from Ilya Gutkovskiy, Dec 18 2016

Status

approved