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%I #27 Nov 14 2014 10:22:34
%S 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,
%T 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,
%U 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
%N T(n,k) = number of ways to place any number of 4X1 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
%C ..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
%C ..2...3...4...5....6....7....8....9...10...11...12....13....14....15....16
%C ..3...5...7...9...11...13...15...17...19...21...23....25....27....29....31
%C ..4...7..10..13...16...19...22...25...28...31...34....37....40....43....46
%C ..5...9..13..17...21...25...29...33...37...41...45....49....53....57....61
%C ..7..15..25..37...51...67...85..105..127..151..177...205...235...267...301
%C .10..25..46..73..106..145..190..241..298..361..430...505...586...673...766
%C .14..39..76.125..186..259..344..441..550..671..804...949..1106..1275..1456
%C .19..57.115.193..291..409..547..705..883.1081.1299..1537..1795..2073..2371
%C .26..87.190.341..546..811.1142.1545.2026.2591.3246..3997..4850..5811..6886
%C .36.137.328.633.1076.1681.2472.3473.4708.6201.7976.10057.12468.15233.18376
%H R. H. Hardin, <a href="/A193516/b193516.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/4]} (binomial(n-3*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=9 k=3; colors=1, 2, 3; empty=0
%e ..0....3....0....0....3....3....0....0....0....0....2....2....0....0....1....2
%e ..1....3....0....2....3....3....3....0....0....0....2....2....1....0....1....2
%e ..1....3....0....2....3....3....3....2....0....0....2....2....1....0....1....2
%e ..1....3....3....2....3....3....3....2....1....0....2....2....1....0....1....2
%e ..1....0....3....2....0....3....3....2....1....0....2....0....1....0....0....0
%e ..2....3....3....2....0....3....3....2....1....3....2....2....0....0....0....3
%e ..2....3....3....2....0....3....3....0....1....3....2....2....0....0....0....3
%e ..2....3....0....2....0....3....3....0....0....3....2....2....0....0....0....3
%e ..2....3....0....2....0....0....3....0....0....3....0....2....0....0....0....3
%p T:= proc(n, k) option remember;
%p `if`(n<0, 0,
%p `if`(n<4 or k=0, 1, k*T(n-4, 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 < 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_ *)
%Y Column 1 is A003269(n+1),
%Y Column 2 is A052942,
%Y Column 3 is A143454(n-3),
%Y Row 8 is A082111,
%Y Row 9 is A100536(n+1),
%Y Row 10 is A051866(n+1).
%K nonn,tabl
%O 1,10
%A _R. H. Hardin_, with proof and formula from _Robert Israel_ in the Sequence Fans Mailing List, Jul 29 2011
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