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%I #23 Dec 06 2019 21:43:32
%S 1,3,5,8,9,14,15,16,17,26,27,28,29,30,31,32,33,50,51,52,53,54,55,56,
%T 57,58,59,60,61,62,63,64,65,98,99,100,101,102,103,104,105,106,107,108,
%U 109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124
%N Payphone packing sequence.
%C Assume that the first person to use a bank of payphones selects one at the end, and all subsequent users select the phone which puts them farthest from the current phone users. U(n) is the smallest number of phones such that n may be used without any two adjacent phones being used.
%F a(n) is the index of the n-th record in A166079, which is given by the recurrence y(n) = y(m) + y(n-m+1) - 1, with y(1) = y(2) = 1 and y(3) = 2, where m = ceiling(n/2). - _John W. Layman_, Feb 05 2011
%F From _Nathaniel Johnston_, Apr 12 2011: (Start)
%F a(n) = a(n-1) + n - 1 if n = 2^k + 2 for some natural number k, a(n) = a(n-1) + 1 otherwise, for n >= 3.
%F a(n) = n + 2^(1+floor(log_2(n-2))) for n >= 3. (End)
%e For 4 phones, only the outer two will be used. For a fifth phone, however, a third person may come along and use the middle phone without any two being adjacent; thus U(3)=5. A seventh phone will not lead to a fourth being used without adjacent people, but an eighth will, hence U(4)=8.
%K easy,nonn
%O 1,2
%A _Craig B. Daniel_, Feb 04 2011
%E Terms 26,27,...,114 added by _John W. Layman_, Feb 05 2011
%E Edited by _N. J. A. Sloane_, Feb 07 2011
%E a(51) - a(60) from _Nathaniel Johnston_, Apr 12 2011
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