cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A209344 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.

Original entry on oeis.org

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
Offset: 1

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Author

R. H. Hardin, Mar 06 2012

Keywords

Examples

			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
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
		

Crossrefs

Row 3 is A074148.

Formula

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).