

A276657


Number of ways to occupy n unlabeled phone booths in a circle one by one, each time picking a phone booth adjacent to the smallest number of previously occupied phone booths.


3



1, 1, 2, 2, 8, 28, 72, 432, 1728, 9792, 56448, 372096, 2419200, 19526400, 144806400, 1310515200, 11338444800, 112903372800, 1102226227200, 12163505356800, 131369759539200, 1589020051046400, 18899570737152000, 247773823008768000, 3220159580209152000, 45535430530695168000
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OFFSET

1,3


COMMENTS

Each phone booth has two adjacent ones, each of which may or may not be occupied. So, available phone booths may have from 0 to 2 adjacent ones that are occupied. Each time we pick a phone booth with the smallest number of those (0 is top priority, then 1, then 2).


LINKS

Max Alekseyev, Table of n, a(n) for n = 1..100


FORMULA

a(n) = A192009(n) / n.
For n > 1, a(n) = Sum (m+k1)!*binomial(m+k,m)*2^k*k!*(m+k)!, where the sum is taken over nonnegative m,k such that 2*m+3*k = n.


MATHEMATICA

r[n_] := {ToRules[Reduce[m >= 0 && k >= 0 && 2 m + 3 k == n, {m, k}, Integers]]}; f[{m_, k_}] := (m + k  1)!*Binomial[m + k, m]*2^k*k!*(m + k)!; a[n_] := Total[f /@ ({m, k} /. r[n])]; a[1] = 1; Array[a, 26] (* JeanFrançois Alcover, Sep 13 2016 *)


PROG

(PARI) { A276657(n) = my(r, k); if(n==1, return(1)); r=0; forstep(m=lift(Mod(n, 3)/2), n\2, 3, k=(n2*m)\3; r+=(m+k1)!*binomial(m+k, m)*2^k*k!*(m+k)!); r; }


CROSSREFS

Cf. A192009 (labeled case).
Cf. A095236, A192008.
Sequence in context: A287756 A287496 A003616 * A079895 A053047 A076143
Adjacent sequences: A276654 A276655 A276656 * A276658 A276659 A276660


KEYWORD

nonn


AUTHOR

Max Alekseyev, Sep 11 2016


STATUS

approved



