A209032 T(n,k) is the number of n-bead necklaces labeled with numbers -k..k allowing reversal, with sum zero and first differences in -k..k.
1, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 3, 4, 6, 2, 1, 3, 5, 12, 11, 4, 1, 4, 7, 23, 34, 33, 6, 1, 4, 10, 38, 88, 144, 86, 13, 1, 5, 12, 60, 187, 471, 576, 278, 21, 1, 5, 15, 88, 358, 1237, 2517, 2613, 873, 45, 1, 6, 19, 125, 625, 2798, 8235, 14611, 11841, 2938, 83, 1, 6, 22, 170, 1023
Offset: 1
Examples
Some solutions for n=6, k=6: .-4...-3...-2...-4...-2...-5...-3...-2...-5...-6...-2...-3...-3...-3...-4...-3 .-2....1...-1....2...-1...-5...-1...-1...-1...-2...-1...-1...-3...-3...-4....1 ..2...-2...-1...-3....0...-1...-1....0....5....3....0....3...-3...-1...-1...-2 .-1....1....2....0...-1....5....1....3....2....5...-1...-1....1....2....4....3 ..3....1....3....3....0....6....5...-1...-1....0....4....3....5....4....4....0 ..2....2...-1....2....4....0...-1....1....0....0....0...-1....3....1....1....1
Links
- R. H. Hardin, Table of n, a(n) for n = 1..184
Formula
Empirical for row n:
n=2: a(k) = a(k-1) + a(k-2) - a(k-3).
n=3: a(k) = 2*a(k-1) - a(k-2) + a(k-3) - 2*a(k-4) + a(k-5).
n=4: a(k) = 3*a(k-1) - 2*a(k-2) - 2*a(k-3) + 3*a(k-4) - a(k-5).
n=5: a(k) = 2*a(k-1) - 2*a(k-3) + 2*a(k-4) - a(k-5) - 2*a(k-6) + 2*a(k-7) + a(k-8) - 2*a(k-9) + 2*a(k-10) - 2*a(k-12) + a(k-13).
Comments