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A095236 Given a row of n pay-phones (or phone booths), all initially unused, how many ways are there for n people to choose the pay-phones, assuming each always chooses one of the most distant pay-phones from those in use already?. 13
1, 2, 4, 8, 16, 36, 136, 216, 672, 2592, 10656, 35904, 167808, 426240, 1866240, 15287040, 35573760, 147640320, 1323970560, 3104317440, 64865525760, 352235520000, 1891946004480, 11505792614400 (list; graph; refs; listen; history; text; internal format)



More precisely: The first person chooses any pay-phone. Thereafter, each person chooses the middle of a largest span of unused phones; but a span of length L at the end of the row is taken to have length 2L-1 and its "middle" is the outermost phone. If a span has even length, either middle may be chosen.

Each person continues to use his pay-phone until all are in use.

The problem was originally stated in terms of urinals in a mens-room.


Table of n, a(n) for n=1..24.


From 6 pay-phones: A may pick any of the 6; he picks #4. B must pick #1. C must pick #6, since the others all are adjacent to A or B. D may pick #2 or #3; he picks #2. E may pick #3 or #5; he picks #5. F must pick #3. That gives the permutation (4,1,6,2,5,3), one of 36 possible permutations.


Cf. A095239, A095240, A095923, A037256.

Sequence in context: A180414 A034343 A002876 * A018536 A162428 A028497

Adjacent sequences:  A095233 A095234 A095235 * A095237 A095238 A095239




Leroy Quet, Jul 03 2004


Edited by Don Reble, Jul 04 2004



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Last modified November 20 12:23 EST 2017. Contains 294971 sequences.