1,21

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

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..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

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

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

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

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

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

6 11 16 21 26 31 36 41 46 51 56 61 66 71

7 13 19 25 31 37 43 49 55 61 67 73 79 85

9 19 31 45 61 79 99 121 145 171 199 229 261 295

12 29 52 81 116 157 204 257 316 381 452 529 612 701

16 43 82 133 196 271 358 457 568 691 826 973 1132 1303

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

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

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 *)

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

nonn,tabl

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

approved