

A193515


T(n,k) = number of ways to place any number of 3X1 tiles of k distinguishable colors into an nX1 grid.


1



1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 7, 7, 6, 1, 1, 6, 9, 10, 13, 9, 1, 1, 7, 11, 13, 22, 23, 13, 1, 1, 8, 13, 16, 33, 43, 37, 19, 1, 1, 9, 15, 19, 46, 69, 73, 63, 28, 1, 1, 10, 17, 22, 61, 101, 121, 139, 109, 41, 1, 1, 11, 19, 25, 78, 139, 181, 253, 268, 183, 60, 1, 1, 12
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OFFSET

1,6


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
..2...3...4...5...6....7....8....9...10...11...12...13....14....15....16....17
..3...5...7...9..11...13...15...17...19...21...23...25....27....29....31....33
..4...7..10..13..16...19...22...25...28...31...34...37....40....43....46....49
..6..13..22..33..46...61...78...97..118..141..166..193...222...253...286...321
..9..23..43..69.101..139..183..233..289..351..419..493...573...659...751...849
.13..37..73.121.181..253..337..433..541..661..793..937..1093..1261..1441..1633
.19..63.139.253.411..619..883.1209.1603.2071.2619.3253..3979..4803..5731..6769
.28.109.268.529.916.1453.2164.3073.4204.5581.7228.9169.11428.14029.16996.20353


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/3]} (binomial(n2*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=7 k=3; colors=1,2,3 and empty=0
..3....0....0....2....0....1....3....0....0....0....1....0....3....1....0....0
..3....0....0....2....2....1....3....2....1....0....1....3....3....1....0....0
..3....1....0....2....2....1....3....2....1....2....1....3....3....1....0....3
..1....1....3....0....2....0....0....2....1....2....3....3....0....2....0....3
..1....1....3....0....0....2....2....2....1....2....3....2....1....2....1....3
..1....0....3....0....0....2....2....2....1....0....3....2....1....2....1....0
..0....0....0....0....0....2....2....2....1....0....0....2....1....0....1....0


MAPLE

T:= proc(n, k) option remember;
`if`(n<0, 0,
`if`(n<3 or k=0, 1, k*T(n3, k) +T(n1, k)))
end:
seq(seq(T(n, d+1n), n=1..d), d=1..13); # Alois P. Heinz, Jul 29 2011


MATHEMATICA

nmax = 13; t[_?Negative, _] = 0; t[n_, k_] /; (n < 3  k == 0) = 1; t[n_, k_] := t[n, k] = k*t[n3, k] + t[n1, k]; Flatten[ Table[ t[nk+1, k], {n , 1, nmax}, {k, n, 1, 1}]](* JeanFrançois Alcover, Nov 28 2011, after Maple *)


CROSSREFS

Column 1 is A000930,
Column 2 is A003229(n1),
Column 3 is A084386,
Column 4 is A089977,
Column 10 is A178205,
Row 6 is A028872(n+2),
Row 7 is A144390(n+1),
Row 8 is A003154(n+1).
Sequence in context: A174448 A077028 A114225 * A259874 A256141 A072704
Adjacent sequences: A193512 A193513 A193514 * A193516 A193517 A193518


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



