|
%I #37 Jun 19 2022 15:22:43
%S 1,3,7,13,21,31,39,57,73,91,95,133
%N Suppose there are equally spaced chairs around a round table. Then a(n) is the maximal number of chairs for which there exists a seating arrangement of n people around the table such that if a waiter puts two glasses (randomly) on the table in front of two (different) chairs, it is always possible to turn the table so that the two glasses end up in front of two seated persons.
%C a(n) <= n(n-1)+1. Moreover, a(n)=n(n-1)+1 iff A058241(n)>0, i.e., when a perfect difference set modulo n(n-1)+1 exists. In particular, a(12) = 133, a(14)=183, a(17)=273, etc.
%C This problem is a circular analog of an optimal ruler problem; see A004137. - _David Wasserman_, Apr 15 2008
%C Solutions do not always exist for table sizes less than a(n). For example, for n = 5 there is no solution for a table of size 20. - _David Wasserman_, Apr 15 2008
%C Equivalently, largest value of S such that in some cyclic array of positive integers of length n, every positive integer <= S is the sum of consecutive terms. For example, the numbers 1..21 can be written as the sum of consecutive terms in the cyclic array [10,3,1,5,2]. So a(5) = 21. - _Phil Scovis_, Jan 29 2016
%H Don Reble, <a href="/A102508/a102508.txt">C++ Program</a>
%e a(5)=21 because if we have 21 chairs, 5 persons can sit down on chairs 1, 4, 5, 10 and 12. 1 == 5-4 (mod 21). 2 == 12-10 (mod 21). 3 == 4-1 (mod 21). 4 == 5-1 (mod 21). 5 == 10-5 (mod 21). 6 == 10-4 (mod 21). 7 == 12-5 (mod 21). 8 == 12-4 (mod 21). 9 == 10-1 (mod 21). 10 == 1-12 (mod 21). It is impossible to do the same with 22 or more chairs.
%Y Cf. A004137, A058241.
%K nonn,more,hard
%O 1,2
%A Ard Van Moer (ard.van.moer(AT)vub.ac.be), Mar 15 2005
%E 3 more terms from _David Wasserman_, Apr 15 2008
%E Edited by _Max Alekseyev_, Apr 29 2010, Mar 01 2015
%E a(11) = 95 from _Don Reble_, Feb 25 2015. - _N. J. A. Sloane_, Mar 01 2015
%E a(12) from _Max Alekseyev_, Mar 01 2015
|
