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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
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
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
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