A193516 T(n,k) = number of ways to place any number of 4X1 tiles of k distinguishable colors into an nX1 grid.

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 5, 4, 1, 1, 1, 5, 7, 7, 5, 1, 1, 1, 6, 9, 10, 9, 7, 1, 1, 1, 7, 11, 13, 13, 15, 10, 1, 1, 1, 8, 13, 16, 17, 25, 25, 14, 1, 1, 1, 9, 15, 19, 21, 37, 46, 39, 19, 1, 1, 1, 10, 17, 22, 25, 51, 73, 76, 57, 26, 1, 1, 1, 11, 19, 25, 29, 67, 106, 125

Offset

1,10

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...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

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

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

..5...9..13..17...21...25...29...33...37...41...45....49....53....57....61

..7..15..25..37...51...67...85..105..127..151..177...205...235...267...301

.10..25..46..73..106..145..190..241..298..361..430...505...586...673...766

.14..39..76.125..186..259..344..441..550..671..804...949..1106..1275..1456

.19..57.115.193..291..409..547..705..883.1081.1299..1537..1795..2073..2371

.26..87.190.341..546..811.1142.1545.2026.2591.3246..3997..4850..5811..6886

.36.137.328.633.1076.1681.2472.3473.4708.6201.7976.10057.12468.15233.18376

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/4]} (binomial(n-3*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=9 k=3; colors=1, 2, 3; empty=0
..0....3....0....0....3....3....0....0....0....0....2....2....0....0....1....2
..1....3....0....2....3....3....3....0....0....0....2....2....1....0....1....2
..1....3....0....2....3....3....3....2....0....0....2....2....1....0....1....2
..1....3....3....2....3....3....3....2....1....0....2....2....1....0....1....2
..1....0....3....2....0....3....3....2....1....0....2....0....1....0....0....0
..2....3....3....2....0....3....3....2....1....3....2....2....0....0....0....3
..2....3....3....2....0....3....3....0....1....3....2....2....0....0....0....3
..2....3....0....2....0....3....3....0....0....3....2....2....0....0....0....3
..2....3....0....2....0....0....3....0....0....3....0....2....0....0....0....3

Maple

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<4 or k=0, 1, k*T(n-4, 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

T[n_, k_] := T[n, k] = If[n < 0, 0, If[n < 4 || k == 0, 1, k*T[n-4, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 13}] // Flatten (* Jean-François Alcover, Mar 04 2014, after Alois P. Heinz *)

Crossrefs

Column 1 is A003269(n+1),

Column 2 is A052942,

Column 3 is A143454(n-3),

Row 8 is A082111,

Row 9 is A100536(n+1),

Row 10 is A051866(n+1).

Sequence in context: A336420 A254055 A096815 * A124445 A124279 A297299

Adjacent sequences: A193513 A193514 A193515 * A193517 A193518 A193519

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