A208825 T(n,k) is the number of n-bead necklaces labeled with numbers -k..k allowing reversal, with sum zero.

1, 1, 2, 1, 3, 2, 1, 4, 5, 5, 1, 5, 8, 16, 7, 1, 6, 13, 38, 45, 18, 1, 7, 18, 75, 155, 167, 32, 1, 8, 25, 131, 415, 828, 609, 84, 1, 9, 32, 210, 905, 2821, 4390, 2471, 185, 1, 10, 41, 316, 1755, 7582, 19657, 25202, 10143, 486, 1, 11, 50, 453, 3085, 17339, 65134, 144871

Offset

1,3

Comments

Table starts

..1....1.....1......1......1.......1.......1........1........1........1

..2....3.....4......5......6.......7.......8........9.......10.......11

..2....5.....8.....13.....18......25......32.......41.......50.......61

..5...16....38.....75....131.....210.....316......453......625......836

..7...45...155....415....905....1755....3085.....5077.....7891....11761

.18..167...828...2821...7582...17339...35288....65769...114442...188463

.32..609..4390..19657..65134..177097..417204...883409..1720628..3135633

.84.2471.25202.144871.587682.1888153.5134796.12322101.26828152.54037203

Links

R. H. Hardin, Table of n, a(n) for n = 1..165

Formula

Empirical for row n:

n=2: a(k) = k + 1.

n=3: a(k) = 2*a(k-1) - 2*a(k-3) + a(k-4).

n=4: a(k) = (2/3)*k^3 + (3/2)*k^2 + (11/6)*k + 1.

n=5: a(k) = 3*a(k-1) - a(k-2) - 5*a(k-3) + 5*a(k-4) + a(k-5) - 3*a(k-6) + a(k-7).

n=6: a(k) = (22/15)*k^5 + (11/3)*k^4 + (14/3)*k^3 + (13/3)*k^2 + (43/15)*k + 1.

n=7: a(k) = 4*a(k-1) - 3*a(k-2) - 8*a(k-3) + 14*a(k-4) - 14*a(k-6) + 8*a(k-7) + 3*a(k-8) - 4*a(k-9) + a(k-10).

Example

All solutions for n=3, k=3:
.-2....0...-1...-1...-3...-2...-3...-2
.-1....0...-1....0....1....1....0....0
..3....0....2....1....2....1....3....2

Crossrefs

Row 3 is A000982(n+1).

Row 4 is A174723(n+1).

Sequence in context: A167237 A277813 A200154 * A344391 A089353 A136451

Adjacent sequences: A208822 A208823 A208824 * A208826 A208827 A208828

Keyword

nonn,tabl

Author

R. H. Hardin, Mar 01 2012

Status

approved