A208970 T(n,k) is the number of n-bead necklaces labeled with numbers -k..k allowing reversal, with sum zero and first and second differences in -k..k.
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 1, 1, 2, 2, 4, 3, 2, 1, 2, 2, 8, 9, 8, 2, 1, 2, 5, 11, 19, 29, 15, 4, 1, 3, 5, 18, 40, 90, 87, 42, 4, 1, 3, 5, 24, 77, 221, 371, 325, 94, 7, 1, 3, 8, 35, 130, 495, 1185, 1755, 1148, 246, 7, 1, 3, 8, 45, 213, 967, 3186, 6883, 8092, 4168, 613, 14, 1
Offset: 1
Examples
Some solutions for n=5, k=5: .-2...-2...-1...-3...-2...-1...-2...-2...-2...-1...-2...-1...-3...-1...-2...-2 .-2...-1....0...-2...-1...-1...-1....0....0....0....0....0...-1....0....0...-2 ..0....1....1....2....2....0....2...-1....2...-1....1....0....2....0....0...-1 ..2....2...-1....2....1....2....0....2...-1....0....1....0....3....1....2....3 ..2....0....1....1....0....0....1....1....1....2....0....1...-1....0....0....2
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-4) - a(k-5).
n=3: a(k) = a(k-1) + a(k-3) - a(k-4) + a(k-6) - a(k-7) - a(k-9) + a(k-10).
n=4: a(k) = 2*a(k-1) + a(k-2) - 4*a(k-3) + a(k-4) + 2*a(k-5) - a(k-6).
Comments