

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.


4



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

1,3


COMMENTS

This problem might arise if there was only a twoperson boat available.
Also the number of ranked treechild 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 treechild networks, arXiv:2007.09701 [math.PR], 2021.
Alessandra Caraceni, Michael Fuchs, and GuanRu Yu, Bijections for ranked treechild networks, arXiv:2105.10137 [math.CO], 2021.


FORMULA

a(n) = n!*((n1)!)^2/((2!)^(n1)).
a(n) ~ 4*sqrt(2)*Pi^(3/2)*n^(3*n1/2)/(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



