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A209344
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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.
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13
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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, 1, 8, 24, 128, 400, 770, 514, 51, 1, 9, 31, 205, 883, 2698, 4029, 2032, 102, 1, 10, 40, 311, 1724, 7358, 18646, 22739, 8076, 249, 1, 11, 49, 448, 3045, 16968, 62853, 136000
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OFFSET
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1,3
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COMMENTS
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Table starts
..1....1.....1......1......1.......1.......1........1........1........1
..2....3.....4......5......6.......7.......8........9.......10.......11
..1....4.....7.....12.....17......24......31.......40.......49.......60
..4...15....35.....72....128.....205.....311......448......618......829
..5...40...145....400....883....1724....3045.....5026.....7827....11684
.14..146...770...2698...7358...16968...34720....64942...113288...186906
.21..514..4029..18646..62853..172610..409199...870122..1699831..3104474
.51.2032.22739.136000.563109.1830872.5016681.12099880.26438711.53392286
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LINKS
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FORMULA
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Empirical for row n:
n=2: a(k) = 2*a(k-1) - a(k-2).
n=3: a(k) = 2*a(k-1) - 2*a(k-3) + a(k-4).
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).
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).
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).
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EXAMPLE
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Some solutions for n=6, k=8:
.-4...-4...-4...-8...-7...-6...-6...-8...-7...-8...-7...-7...-8...-8...-8...-4
.-3...-3...-3...-3....0....1....1....0...-2....0....1...-2....3...-8...-4...-4
..5...-1...-4....4...-4...-1....1....1....8....3....0....8...-4...-4....0...-2
.-2....3...-3....1....2....8....6....4...-5....5...-6....1....0....6....7....5
.-1...-1....6....3....3...-5...-6....0....5...-4....8...-7....6....7....3....7
..5....6....8....3....6....3....4....3....1....4....4....7....3....7....2...-2
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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