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%I #10 Mar 13 2017 04:25:22
%S 1,1,2,1,3,1,1,4,4,4,1,5,7,15,5,1,6,12,35,40,14,1,7,17,72,145,146,21,
%T 1,8,24,128,400,770,514,51,1,9,31,205,883,2698,4029,2032,102,1,10,40,
%U 311,1724,7358,18646,22739,8076,249,1,11,49,448,3045,16968,62853,136000
%N T(n,k) is the number of n-bead necklaces labeled with numbers -k..k allowing reversal, with sum zero with no three beads in a row equal.
%C Table starts
%C ..1....1.....1......1......1.......1.......1........1........1........1
%C ..2....3.....4......5......6.......7.......8........9.......10.......11
%C ..1....4.....7.....12.....17......24......31.......40.......49.......60
%C ..4...15....35.....72....128.....205.....311......448......618......829
%C ..5...40...145....400....883....1724....3045.....5026.....7827....11684
%C .14..146...770...2698...7358...16968...34720....64942...113288...186906
%C .21..514..4029..18646..62853..172610..409199...870122..1699831..3104474
%C .51.2032.22739.136000.563109.1830872.5016681.12099880.26438711.53392286
%H R. H. Hardin, <a href="/A209344/b209344.txt">Table of n, a(n) for n = 1..165</a>
%F Empirical for row n:
%F n=2: a(k) = 2*a(k-1) - a(k-2).
%F n=3: a(k) = 2*a(k-1) - 2*a(k-3) + a(k-4).
%F n=4: a(k) = 3*a(k-1) - 3*a(k-2) + 2*a(k-3) - 3*a(k-4) + 3*a(k-5) - a(k-6).
%F n=5: a(k) = 2*a(k-1) + a(k-2) - 3*a(k-3) - a(k-4) + a(k-5) + 3*a(k-6) - a(k-7) - 2*a(k-8) + a(k-9).
%F n=6: a(k) = 4*a(k-1) - 5*a(k-2) + a(k-3) + a(k-4) + a(k-5) + a(k-6) - 5*a(k-7) + 4*a(k-8) - a(k-9).
%e Some solutions for n=6, k=8:
%e .-4...-4...-4...-8...-7...-6...-6...-8...-7...-8...-7...-7...-8...-8...-8...-4
%e .-3...-3...-3...-3....0....1....1....0...-2....0....1...-2....3...-8...-4...-4
%e ..5...-1...-4....4...-4...-1....1....1....8....3....0....8...-4...-4....0...-2
%e .-2....3...-3....1....2....8....6....4...-5....5...-6....1....0....6....7....5
%e .-1...-1....6....3....3...-5...-6....0....5...-4....8...-7....6....7....3....7
%e ..5....6....8....3....6....3....4....3....1....4....4....7....3....7....2...-2
%Y Row 3 is A074148.
%K nonn,tabl
%O 1,3
%A _R. H. Hardin_, Mar 06 2012
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