A193517 T(n,k) = number of ways to place any number of 5X1 tiles of k distinguishable colors into an nX1 grid.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 4, 5, 4, 1, 1, 1, 1, 5, 7, 7, 5, 1, 1, 1, 1, 6, 9, 10, 9, 6, 1, 1, 1, 1, 7, 11, 13, 13, 11, 8, 1, 1, 1, 1, 8, 13, 16, 17, 16, 17, 11, 1, 1, 1, 1, 9, 15, 19, 21, 21, 28, 27, 15, 1, 1, 1, 1, 10, 17, 22, 25, 26, 41, 49, 41, 20, 1, 1, 1

Offset

1,15

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

..2..3...4...5...6...7...8....9...10...11...12...13...14...15...16...17...18

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

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

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

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

..8.17..28..41..56..73..92..113..136..161..188..217..248..281..316..353..392

.11.27..49..77.111.151.197..249..307..371..441..517..599..687..781..881..987

.15.41..79.129.191.265.351..449..559..681..815..961.1119.1289.1471.1665.1871

.20.59.118.197.296.415.554..713..892.1091.1310.1549.1808.2087.2386.2705.3044

.26.81.166.281.426.601.806.1041.1306.1601.1926.2281.2666.3081.3526.4001.4506

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/5]} (binomial(n-4*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=11 k=3; colors=1, 2, 3; empty=0
..0....2....2....0....0....1....0....3....3....0....0....0....0....3....1....0
..0....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..0....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..3....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..3....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..3....1....0....2....1....0....3....3....0....3....2....3....1....0....0....1
..3....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..3....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..0....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..0....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..0....0....3....0....1....1....2....0....2....0....0....3....0....3....0....2

Maple

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<5 or k=0, 1, k*T(n-5, 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 < 5 || k == 0, 1, k*T[n-5, 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 A003520,

Column 2 is A143447(n-4),

Column 3 is A143455(n-4),

Row 10 is A028884.

Sequence in context: A363096 A260222 A181386 * A296554 A327482 A189006

Adjacent sequences: A193514 A193515 A193516 * A193518 A193519 A193520

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