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
Keywords
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)
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
Comments