A209485 T(n,k) is the number of n-bead necklaces labeled with numbers -k..k allowing reversal, with sum zero and avoiding the patterns z z+1 z+2 and z z-1 z-2.
1, 1, 2, 1, 3, 1, 1, 4, 4, 4, 1, 5, 7, 15, 4, 1, 6, 12, 35, 38, 11, 1, 7, 17, 72, 140, 136, 15, 1, 8, 24, 128, 390, 731, 458, 43, 1, 9, 31, 205, 866, 2606, 3740, 1781, 77, 1, 10, 40, 311, 1702, 7179, 17771, 20888, 6912, 199, 1, 11, 49, 448, 3014, 16660, 60778, 128598, 118137
Offset: 1
Examples
Some solutions for n=6, k=6: .-5...-4...-5...-6...-6...-5...-6...-4...-3...-6...-6...-3...-5...-5...-6...-4 ..0....0...-2...-3...-2...-4...-5...-3...-1....5....2...-2....0....2...-2....2 .-2...-2....2....4....0...-1....4....0...-1...-5....0...-2...-3...-4....6...-4 ..2....2...-2....3....1....5....3....4....1....0...-4....5....2....1...-4....4 ..5....0....5...-2....1....0....4....3...-2....0....6....0....5....0....0...-4 ..0....4....2....4....6....5....0....0....6....6....2....2....1....6....6....6
Links
- R. H. Hardin, Table of n, a(n) for n = 1..148
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-3) - 2*a(k-5) + 2*a(k-6) + a(k-8) - 2*a(k-10) + a(k-11).
n=6: a(k) = 5*a(k-1) - 10*a(k-2) + 11*a(k-3) - 10*a(k-4) + 11*a(k-5) - 10*a(k-6) + 5*a(k-7) - a(k-8) for k > 9.
Comments