A193515 T(n,k) = number of ways to place any number of 3X1 tiles of k distinguishable colors into an nX1 grid.

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 7, 7, 6, 1, 1, 6, 9, 10, 13, 9, 1, 1, 7, 11, 13, 22, 23, 13, 1, 1, 8, 13, 16, 33, 43, 37, 19, 1, 1, 9, 15, 19, 46, 69, 73, 63, 28, 1, 1, 10, 17, 22, 61, 101, 121, 139, 109, 41, 1, 1, 11, 19, 25, 78, 139, 181, 253, 268, 183, 60, 1, 1, 12

Offset

1,6

Comments

Table starts:

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

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

..2...3...4...5...6....7....8....9...10...11...12...13....14....15....16....17

..3...5...7...9..11...13...15...17...19...21...23...25....27....29....31....33

..4...7..10..13..16...19...22...25...28...31...34...37....40....43....46....49

..6..13..22..33..46...61...78...97..118..141..166..193...222...253...286...321

..9..23..43..69.101..139..183..233..289..351..419..493...573...659...751...849

.13..37..73.121.181..253..337..433..541..661..793..937..1093..1261..1441..1633

.19..63.139.253.411..619..883.1209.1603.2071.2619.3253..3979..4803..5731..6769

.28.109.268.529.916.1453.2164.3073.4204.5581.7228.9169.11428.14029.16996.20353

Links

R. H. Hardin, Table of n, a(n) for n = 1..9999

Formula

With z X 1 tiles of k colors on an n X 1 grid (with n >= z), either there is a tile (of any of the k colors) on the first spot, followed by any configuration on the remaining (n-z) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n-1) X 1. So T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n=0,1,...,z-1. The solution is T(n,k) = sum_r r^(-n-1)/(1 + z k r^(z-1)) where the sum is over the roots of the polynomial k x^z + x - 1.

T(n,k) = sum {s=0..[n/3]} (binomial(n-2*s,s)*k^s).

For z X 1 tiles, T(n,k,z) = sum{s=0..[n/z]} (binomial(n-(z-1)*s,s)*k^s). - R. H. Hardin, Jul 31 2011

Example

Some solutions for n=7 k=3; colors=1,2,3 and empty=0
..3....0....0....2....0....1....3....0....0....0....1....0....3....1....0....0
..3....0....0....2....2....1....3....2....1....0....1....3....3....1....0....0
..3....1....0....2....2....1....3....2....1....2....1....3....3....1....0....3
..1....1....3....0....2....0....0....2....1....2....3....3....0....2....0....3
..1....1....3....0....0....2....2....2....1....2....3....2....1....2....1....3
..1....0....3....0....0....2....2....2....1....0....3....2....1....2....1....0
..0....0....0....0....0....2....2....2....1....0....0....2....1....0....1....0

Maple

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<3 or k=0, 1, k*T(n-3, k) +T(n-1, k)))
    end:
seq(seq(T(n, d+1-n), n=1..d), d=1..13); # Alois P. Heinz, Jul 29 2011

Mathematica

nmax = 13; t[_?Negative, _] = 0; t[n_, k_] /; (n < 3 || k == 0) = 1; t[n_, k_] := t[n, k] = k*t[n-3, k] + t[n-1, k]; Flatten[ Table[ t[n-k+1, k], {n , 1, nmax}, {k, n, 1, -1}]](* Jean-François Alcover, Nov 28 2011, after Maple *)

Crossrefs

Column 1 is A000930,

Column 2 is A003229(n-1),

Column 3 is A084386,

Column 4 is A089977,

Column 10 is A178205,

Row 6 is A028872(n+2),

Row 7 is A144390(n+1),

Row 8 is A003154(n+1).

Sequence in context: A174448 A077028 A114225 * A259874 A256141 A072704

Adjacent sequences: A193512 A193513 A193514 * A193516 A193517 A193518

Keyword

nonn,tabl

Author

R. H. Hardin, with proof and formula from Robert Israel in the Sequence Fans Mailing List, Jul 29 2011

Status

approved