%I #23 Nov 14 2014 10:23:44
%S 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,
%T 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,
%U 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
%N T(n,k) = number of ways to place any number of 6X1 tiles of k distinguishable colors into an nX1 grid.
%C Table starts:
%C ..1..1...1...1...1...1...1...1...1....1....1....1....1....1....1....1....1....1
%C ..1..1...1...1...1...1...1...1...1....1....1....1....1....1....1....1....1....1
%C ..1..1...1...1...1...1...1...1...1....1....1....1....1....1....1....1....1....1
%C ..1..1...1...1...1...1...1...1...1....1....1....1....1....1....1....1....1....1
%C ..1..1...1...1...1...1...1...1...1....1....1....1....1....1....1....1....1....1
%C ..2..3...4...5...6...7...8...9..10...11...12...13...14...15...16...17...18...19
%C ..3..5...7...9..11..13..15..17..19...21...23...25...27...29...31...33...35...37
%C ..4..7..10..13..16..19..22..25..28...31...34...37...40...43...46...49...52...55
%C ..5..9..13..17..21..25..29..33..37...41...45...49...53...57...61...65...69...73
%C ..6.11..16..21..26..31..36..41..46...51...56...61...66...71...76...81...86...91
%C ..7.13..19..25..31..37..43..49..55...61...67...73...79...85...91...97..103..109
%C ..9.19..31..45..61..79..99.121.145..171..199..229..261..295..331..369..409..451
%C .12.29..52..81.116.157.204.257.316..381..452..529..612..701..796..897.1004.1117
%C .16.43..82.133.196.271.358.457.568..691..826..973.1132.1303.1486.1681.1888.2107
%C .21.61.121.201.301.421.561.721.901.1101.1321.1561.1821.2101.2401.2721.3061.3421
%H R. H. Hardin, <a href="/A193518/b193518.txt">Table of n, a(n) for n = 1..9999</a>
%F 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.
%F T(n,k) = sum {s=0..[n/6]} (binomial(n-5*s,s)*k^s).
%F 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
%e Some solutions for n=13 k=3; colors=1, 2, 3; empty=0
%e ..0....0....0....0....0....3....0....0....0....0....2....0....0....2....2....1
%e ..3....0....1....2....1....3....0....0....0....2....2....0....0....2....2....1
%e ..3....0....1....2....1....3....0....0....0....2....2....0....0....2....2....1
%e ..3....0....1....2....1....3....0....0....0....2....2....2....0....2....2....1
%e ..3....0....1....2....1....3....0....0....3....2....2....2....0....2....2....1
%e ..3....1....1....2....1....3....0....0....3....2....2....2....0....2....2....1
%e ..3....1....1....2....1....0....3....0....3....2....3....2....0....0....0....2
%e ..1....1....1....0....0....1....3....3....3....3....3....2....2....2....0....2
%e ..1....1....1....0....0....1....3....3....3....3....3....2....2....2....0....2
%e ..1....1....1....0....0....1....3....3....3....3....3....0....2....2....0....2
%e ..1....1....1....0....0....1....3....3....0....3....3....0....2....2....0....2
%e ..1....0....1....0....0....1....3....3....0....3....3....0....2....2....0....2
%e ..1....0....1....0....0....1....0....3....0....3....0....0....2....2....0....0
%p T:= proc(n, k) option remember;
%p `if`(n<0, 0,
%p `if`(n<6 or k=0, 1, k*T(n-6, k) +T(n-1, k)))
%p end:
%p seq(seq(T(n, d+1-n), n=1..d), d=1..13); # _Alois P. Heinz_, Jul 29 2011
%t 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_ *)
%Y Column 1 is A005708,
%Y Column 2 is A143448(n-5),
%Y Column 3 is A143456(n-5),
%Y Row 12 is A190576(n+1),
%Y Row 15 is A069133(n+1).
%K nonn,tabl
%O 1,21
%A _R. H. Hardin_, with proof and formula from _Robert Israel_ in the Sequence Fans Mailing List, Jul 29 2011