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%I #10 Mar 13 2017 04:25:03
%S 1,1,1,1,2,1,1,2,3,2,1,3,5,7,3,1,3,7,16,17,5,1,4,11,33,59,51,9,1,4,15,
%T 58,159,253,155,18,1,5,19,95,351,874,1111,508,35,1,5,25,144,683,2354,
%U 4935,5067,1683,73,1,6,31,209,1207,5404,16307,28816,23483,5709,151,1,6,37,290
%N T(n,k) is the number of n-bead necklaces labeled with numbers -k..k not allowing reversal, with sum zero and first differences in -k..k.
%C Table starts
%C ..1...1....1.....1......1......1.......1.......1.......1........1........1
%C ..1...2....2.....3......3......4.......4.......5.......5........6........6
%C ..1...3....5.....7.....11.....15......19......25......31.......37.......45
%C ..2...7...16....33.....58.....95.....144.....209.....290......391......512
%C ..3..17...59...159....351....683....1207....1989....3099.....4623.....6647
%C ..5..51..253...874...2354...5404...11010...20553...35781....58971....92851
%C ..9.155.1111..4935..16307..44209..104001..219929..428031...779487..1344353
%C .18.508.5067.28816.116183.371893.1008880.2416967.5255962.10577092.19976373
%H R. H. Hardin, <a href="/A208993/b208993.txt">Table of n, a(n) for n = 1..203</a>
%F Empirical for row n:
%F n=2: a(k) = a(k-1) + a(k-2) - a(k-3).
%F n=3: a(k) = 2*a(k-1) - a(k-2) + a(k-3) - 2*a(k-4) + a(k-5).
%F n=4: a(k) = 3*a(k-1) - 2*a(k-2) - 2*a(k-3) + 3*a(k-4) - a(k-5).
%F n=5: a(k) = 2*a(k-1) + a(k-2) - 4*a(k-3) + a(k-4) + 3*a(k-5) - 3*a(k-6) - a(k-7) + 4*a(k-8) - a(k-9) - 2*a(k-10) + a(k-11).
%e Some solutions for n=6, k=6:
%e .-3...-4...-5...-5...-6...-5...-4...-2...-4...-5...-4...-4...-3...-6...-5...-5
%e ..0...-3...-4...-3...-1....1....2...-2...-2....1...-1....1....3....0...-1....1
%e ..0....3....1....3....2....5....5....1....3....1....4....3....0....4....0...-2
%e .-1....5....5....2....3....2...-1....2....2....1....0....4....0....2....5....0
%e ..1....1....4....2....3...-4....0...-2...-1....2....2...-1...-2....2....2....6
%e ..3...-2...-1....1...-1....1...-2....3....2....0...-1...-3....2...-2...-1....0
%Y Row 2 is A004526(n+2).
%K nonn,tabl
%O 1,5
%A _R. H. Hardin_, Mar 04 2012
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