OFFSET
1,2
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.
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)
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)
EXAMPLE
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.
CROSSREFS
KEYWORD
nonn
AUTHOR
Leroy Quet, Jul 03 2004
EXTENSIONS
Edited by Don Reble, Jul 04 2004
STATUS
approved