A095236 Given a row of n payphones (or phone booths), all initially unused, sequence gives number of ways for n people to choose the payphones assuming each always chooses one of the most distant payphones from those in use already.

1, 2, 4, 8, 16, 36, 136, 216, 672, 2592, 10656, 35904, 167808, 426240, 1866240, 15287040, 35573760, 147640320, 1323970560, 3104317440, 64865525760, 352235520000, 1891946004480, 11505792614400

Offset: 1

Author

Leroy Quet, Jul 03 2004

Keywords

• nonn

Comments

More precisely: The first person chooses any payphone. 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 payphone until all are in use.

The problem was originally stated in terms of urinals in a men's room.

Examples

			From 6 payphones: 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.
		

Links

• Max Alekseyev, Table of n, a(n) for n = 1..100
• Max A. Alekseyev, Enumeration of Payphone Permutations, arXiv:2304.04324 [math.CO], 2023.
• Simon Wundling, About a combinatorial problem with n seats and n people, arXiv:2303.18175 [math.CO], 2023. (German)

Crossrefs

Cf. A095239, A095240, A095923, A037256.

Formula

From Simon Wundling, Apr 12 2023: (Start)

Let adjacent payphones have the distance 1. We now look at the situation with p payphones and the first person choosing the payphone at the left end. Then let b(p,k) be the number of people who choose a payphone with distance k and let d(p,k) be the number of different sets of two adjacent payphones which both have at one time the distance k.

1) Calculation of b(p,k) for k >= 2 and all p (m = floor(log_2((p-1)/2k)) for p >= 5):

For p < k + 1: 0.

For p = k + 1: 1.

For k + 1 < p < 1 + 2k: 0.

For 1 + 2^m*2k <= p <= 1 + 2^m*(2k+1): 2^m.

For 1 + 2^m*(2k+1) < p <= 1 + 2^m*(2k+2): 1 + 2^m*(2k+2) - p.

For 1 + 2^m*(2k+2) < p <= 1 + 2^m*(4k-2): 0.

For 1 + 2^m*(4k-2) < p < 1 + 2^(m+1)*2k: p - 1 - 2^m*(4k-2).

2) Calculation of b(p,k) for k = 1 and all p (m = floor(log_2((p-1)/3)) for p >= 4):

For p = 1: 0.

For p = 2 or p = 3: 1.

For 1 + 2^m*3 <= p <= 1 + 2^m*4: 2^(m+1).

For 1 + 2^m*4 < p < 1 + 2^(m+1)*3: p - 1 - 2^(m+1).

3) Calculation of d(p,k) for k >= 2 and all p (m = floor(log_2((p-1)/2k)) for p >= 5):

For p < 1 + 2k: 0.

For 1 + 2^m*2k <= p <= 1 + 2^m*(2k+1): p - 1 - 2^m*2k.

For 1 + 2^m*(2k+1) < p <= 1 + 2^m*(2k+2): 1 + 2^m*(2k+2) - p.

For 1 + 2^m*(2k+2) < p < 1 + 2^(m+1)*2k: 0.

4) Calculation of d(p,k) for k = 1 and all p (m = floor(log_2((p-1)/3)) for p >= 4):

For p < 4: 0.

For 1 + 2^m*3 <= p <= 1 + 2^m*4: 1 + 2^m*4 - p.

For 1 + 2^m*4 < p < 1 + 2^(m+1)*3: p - 1 - 2^m*4.

Now you can give a formula for a(n):

a(n) = Sum_{i=1..n} Product_{j=1..n-1} 2^(d(i,j) + d(n+1-i,j)) * (d(i,j) + d(n+1-i,j))! * (b(i,j) + b(n+1-i,j) - d(i,j) - d(n+1-i,j))!. (End)

Extensions

Edited by Don Reble, Jul 04 2004