A193518 T(n,k) = number of ways to place any number of 6X1 tiles of k distinguishable colors into an nX1 grid.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 4, 5, 4, 1, 1, 1, 1, 1, 5, 7, 7, 5, 1, 1, 1, 1, 1, 6, 9, 10, 9, 6, 1, 1, 1, 1, 1, 7, 11, 13, 13, 11, 7, 1, 1, 1, 1, 1, 8, 13, 16, 17, 16, 13, 9, 1, 1, 1, 1, 1, 9, 15, 19, 21, 21, 19, 19, 12, 1, 1, 1, 1, 1, 10, 17

Offset

1,21

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....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...17...18...19

..3..5...7...9..11..13..15..17..19...21...23...25...27...29...31...33...35...37

..4..7..10..13..16..19..22..25..28...31...34...37...40...43...46...49...52...55

..5..9..13..17..21..25..29..33..37...41...45...49...53...57...61...65...69...73

..6.11..16..21..26..31..36..41..46...51...56...61...66...71...76...81...86...91

..7.13..19..25..31..37..43..49..55...61...67...73...79...85...91...97..103..109

..9.19..31..45..61..79..99.121.145..171..199..229..261..295..331..369..409..451

.12.29..52..81.116.157.204.257.316..381..452..529..612..701..796..897.1004.1117

.16.43..82.133.196.271.358.457.568..691..826..973.1132.1303.1486.1681.1888.2107

.21.61.121.201.301.421.561.721.901.1101.1321.1561.1821.2101.2401.2721.3061.3421

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/6]} (binomial(n-5*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=13 k=3; colors=1, 2, 3; empty=0
..0....0....0....0....0....3....0....0....0....0....2....0....0....2....2....1
..3....0....1....2....1....3....0....0....0....2....2....0....0....2....2....1
..3....0....1....2....1....3....0....0....0....2....2....0....0....2....2....1
..3....0....1....2....1....3....0....0....0....2....2....2....0....2....2....1
..3....0....1....2....1....3....0....0....3....2....2....2....0....2....2....1
..3....1....1....2....1....3....0....0....3....2....2....2....0....2....2....1
..3....1....1....2....1....0....3....0....3....2....3....2....0....0....0....2
..1....1....1....0....0....1....3....3....3....3....3....2....2....2....0....2
..1....1....1....0....0....1....3....3....3....3....3....2....2....2....0....2
..1....1....1....0....0....1....3....3....3....3....3....0....2....2....0....2
..1....1....1....0....0....1....3....3....0....3....3....0....2....2....0....2
..1....0....1....0....0....1....3....3....0....3....3....0....2....2....0....2
..1....0....1....0....0....1....0....3....0....3....0....0....2....2....0....0

Maple

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<6 or k=0, 1, k*T(n-6, 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 < 6 || k == 0, 1, k*T[n-6, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 14}] // Flatten (* Jean-François Alcover, Mar 04 2014, after Alois P. Heinz *)

Crossrefs

Column 1 is A005708,

Column 2 is A143448(n-5),

Column 3 is A143456(n-5),

Row 12 is A190576(n+1),

Row 15 is A069133(n+1).

Sequence in context: A268372 A361754 A298848 * A060176 A305297 A218220

Adjacent sequences: A193515 A193516 A193517 * A193519 A193520 A193521

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