

A193516


T(n,k) = number of ways to place any number of 4X1 tiles of k distinguishable colors into an nX1 grid.


1



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, 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, 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
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OFFSET

1,10


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
..2...3...4...5....6....7....8....9...10...11...12....13....14....15....16
..3...5...7...9...11...13...15...17...19...21...23....25....27....29....31
..4...7..10..13...16...19...22...25...28...31...34....37....40....43....46
..5...9..13..17...21...25...29...33...37...41...45....49....53....57....61
..7..15..25..37...51...67...85..105..127..151..177...205...235...267...301
.10..25..46..73..106..145..190..241..298..361..430...505...586...673...766
.14..39..76.125..186..259..344..441..550..671..804...949..1106..1275..1456
.19..57.115.193..291..409..547..705..883.1081.1299..1537..1795..2073..2371
.26..87.190.341..546..811.1142.1545.2026.2591.3246..3997..4850..5811..6886
.36.137.328.633.1076.1681.2472.3473.4708.6201.7976.10057.12468.15233.18376


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 (nz) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n1) X 1. So T(n,k) = T(n1,k) + k*T(nz,k), with T(n,k) = 1 for n=0,1,...,z1. The solution is T(n,k) = sum_r r^(n1)/(1 + z k r^(z1)) where the sum is over the roots of the polynomial k x^z + x  1.
T(n,k) = sum {s=0..[n/4]} (binomial(n3*s,s)*k^s).
For z X 1 tiles, T(n,k,z) = sum{s=0..[n/z]} (binomial(n(z1)*s,s)*k^s).  R. H. Hardin, Jul 31 2011


EXAMPLE

Some solutions for n=9 k=3; colors=1, 2, 3; empty=0
..0....3....0....0....3....3....0....0....0....0....2....2....0....0....1....2
..1....3....0....2....3....3....3....0....0....0....2....2....1....0....1....2
..1....3....0....2....3....3....3....2....0....0....2....2....1....0....1....2
..1....3....3....2....3....3....3....2....1....0....2....2....1....0....1....2
..1....0....3....2....0....3....3....2....1....0....2....0....1....0....0....0
..2....3....3....2....0....3....3....2....1....3....2....2....0....0....0....3
..2....3....3....2....0....3....3....0....1....3....2....2....0....0....0....3
..2....3....0....2....0....3....3....0....0....3....2....2....0....0....0....3
..2....3....0....2....0....0....3....0....0....3....0....2....0....0....0....3


MAPLE

T:= proc(n, k) option remember;
`if`(n<0, 0,
`if`(n<4 or k=0, 1, k*T(n4, k) +T(n1, k)))
end:
seq(seq(T(n, d+1n), 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 < 4  k == 0, 1, k*T[n4, k]+T[n1, k]]]; Table[Table[T[n, d+1n], {n, 1, d}], {d, 1, 13}] // Flatten (* JeanFrançois Alcover, Mar 04 2014, after Alois P. Heinz *)


CROSSREFS

Column 1 is A003269(n+1),
Column 2 is A052942,
Column 3 is A143454(n3),
Row 8 is A082111,
Row 9 is A100536(n+1),
Row 10 is A051866(n+1).
Sequence in context: A136043 A254055 A096815 * A124445 A124279 A135225
Adjacent sequences: A193513 A193514 A193515 * A193517 A193518 A193519


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



