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%I #21 Jan 14 2022 10:15:52
%S 4,6,6,20,118,282,1540,7092,36128,206848,1439500,10066876,74931690,
%T 613296028,5263250382,47965088850
%N Number of distinct vertex-magic total labelings on cycle C_n.
%H Andrew Baker and Joe Sawada, <a href="http://dx.doi.org/10.1007/978-3-540-85097-7_34">Magic Labelings on Cycles and Wheels</a>, Lecture Notes in Computer Science 5165 (Combinatorial Optimization and Applications. Second International Conference, COCOA 2008). pp. 361-373.
%H Mukkai S. Krishnamoorthy, Allen Lavoie, Ali Dasdan, Bharath Santosh, <a href="http://arxiv.org/abs/1402.2878">Number of unique Edge-magic total labelings on Path P_n</a>, arXiv:1402.2878 [math.CO], 2014.
%e From _Gheorghe Coserea_, May 23 2018: (Start)
%e For n=4 the a(4)=6 solutions are:
%e [1, 4, 8, 3, 2, 6, 5, 7]
%e [1, 5, 6, 4, 2, 7, 3, 8]
%e [1, 5, 8, 2, 4, 3, 7, 6]
%e [1, 7, 5, 2, 6, 3, 4, 8]
%e [3, 4, 8, 1, 6, 2, 7, 5]
%e [3, 6, 5, 1, 8, 2, 4, 7]
%e The solution [1, 4, 8, 3, 2, 6, 5, 7] is an encoding of the following vertex-magic labeling on C_4:
%e 7 1 4 8 3 2 6
%e o-----o-----o-----o
%e | |
%e +-----------------+
%e 5
%e In this labeling vertices are labeled 7, 4, 3, 6 while edges are labeled 1, 8, 2, 5 respectively. The vertex-magic constant of labeling k is 13 since k = 5+7+1 = 1+4+8 = 8+3+2 = 2+6+5.
%e In general, for C_n the magic constant of labeling k satisfies 3*n+1 - floor((n-1)/2) <= k <= 3*n+2 + floor((n-1)/2) and this bounds are tight for n>=6.
%e The solutions for n=4 have been generated using the MiniZinc model (e.g. $ minizinc -a -D"n=4;" magiccn.mzn | sort).
%e (End)
%o (MiniZinc)
%o % filename: magiccn.mzn : generate solution of size n
%o % usage: minizinc -a --soln-sep "" --search-complete-msg "" -D"n=5;" magiccn.mzn
%o include "globals.mzn";
%o int: n;
%o int: lo = 3*n+1 - (n-1) div 2;
%o int: hi = 3*n+2 + (n-1) div 2;
%o array[1..2*n] of var 1..2*n: x;
%o var lo..hi: h;
%o constraint alldifferent(x);
%o constraint forall([h = x[2*i-1] + x[2*i] + x[2*i+1] | i in 1..n-1]);
%o constraint h = x[2*n-1] + x[2*n] + x[1];
%o constraint forall([x[1] < x[2*i+1] | i in 1..n-1]); % break rotations
%o constraint x[2] < x[2*n]; % break reflection symmetry
%o solve satisfy;
%o output [show(x)];
%o % _Gheorghe Coserea_, May 22 2018
%Y Cf. A237426.
%K nonn,more
%O 3,1
%A Andrew Baker (abaker04(AT)uoguelph.ca), Oct 16 2008
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