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A276657 Number of ways to occupy n unlabeled pay-phones on a circle one-by-one, each time picking a phone adjacent to the smallest number of previously occupied phones. 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

Each phone has two adjacent ones, each of which may be or be not occupied. So, available phones may have from 0 to 2 adjacent ones that are occupied. Each time we pick a phone 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+k-1)!*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] (* Jean-Fran├žois 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=(n-2*m)\3; r+=(m+k-1)!*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

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Last modified July 23 03:00 EDT 2019. Contains 325230 sequences. (Running on oeis4.)