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A166079
Given a row of n payphones, all initially unused, how many people can use the payphones, assuming (1) each always chooses one of the most distant payphones from those in use already, (2) the first person takes a phone at the end, and (3) no people use adjacent phones?
3
1, 1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 5, 6, 7, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 11, 12, 13, 14, 15, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33
OFFSET
1,3
LINKS
Julian Zbigniew Kuryllowicz-Kazmierczak, Table of n, a(n) for n = 1..20000
H.-K. Hwang, S. Janson, and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications. ACM Transactions on Algorithms, 13:4 (2017), #47. DOI:10.1145/3127585
Simon Wundling, About a combinatorial problem with n seats and n people, arXiv:2303.18175 [math.CO], 2023. (German)
FORMULA
a(n) = 1 + 2^floor(log_2(n-2) - 1) + max(0, n - (3/2) * 2^floor(log_2(n-2)) - 1).
A recurrence is: a(n) = a(m) + a(n-m+1) - 1, with a(1) = a(2) = 1 and a(3)=2, where m = ceiling(n/2). - John W. Layman, Feb 05 2011
a(n) = n - b(n,1) (see A095236 for definition and calculation of b(n,1)). - Simon Wundling, May 21 2023
EXAMPLE
From Julian Zbigniew Kuryllowicz-Kazmierczak, Feb 20 2024: (Start)
a(8)=4:
1st person takes payphone at the end: .......1
2nd person takes most distant, the 1st: 2......1
3rd person takes 4th or 5th payphone: 2..3...1 or 2...3..1
4th person must take 6th or 3rd, respectively: 2..3.4.1 or 2.4.3..1
Now each payphone is in use or adjacent to one in use, so a(8)=4.
(End)
MATHEMATICA
a[n_]:= If[n<3, 1, 1 + 2^Floor[Log2[n-2] - 1] + Max[0, n - (3/2) * 2^Floor[Log2[n-2]]- 1]]; Array[a, 78] (* Stefano Spezia, Oct 05 2024 *)
PROG
(PARI) A000523(n)=my(t=floor(sizedigit(n)*3.32192809)-5); n>>=t; while(n>3, n>>=2; t+=2); if(n==1, t, t+1);
a(n)=my(t=1<<(A000523(n-2)-1)); max(t+1, n-t-t)
(PARI) a(n) = if(n<3, return(1)); my(L=logint(n-2, 2)-1); 1 + 2^L + max(0, n - 3*2^L - 1) \\ Charles R Greathouse IV, Jan 27 2016
CROSSREFS
KEYWORD
easy,look,nonn
AUTHOR
STATUS
approved