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%I #30 Nov 14 2014 10:21:57
%S 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,
%T 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,
%U 10,17,22,61,101,121,139,109,41,1,1,11,19,25,78,139,181,253,268,183,60,1,1,12
%N T(n,k) = number of ways to place any number of 3X1 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
%C ..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
%C ..3...5...7...9..11...13...15...17...19...21...23...25....27....29....31....33
%C ..4...7..10..13..16...19...22...25...28...31...34...37....40....43....46....49
%C ..6..13..22..33..46...61...78...97..118..141..166..193...222...253...286...321
%C ..9..23..43..69.101..139..183..233..289..351..419..493...573...659...751...849
%C .13..37..73.121.181..253..337..433..541..661..793..937..1093..1261..1441..1633
%C .19..63.139.253.411..619..883.1209.1603.2071.2619.3253..3979..4803..5731..6769
%C .28.109.268.529.916.1453.2164.3073.4204.5581.7228.9169.11428.14029.16996.20353
%H R. H. Hardin, <a href="/A193515/b193515.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/3]} (binomial(n-2*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=7 k=3; colors=1,2,3 and empty=0
%e ..3....0....0....2....0....1....3....0....0....0....1....0....3....1....0....0
%e ..3....0....0....2....2....1....3....2....1....0....1....3....3....1....0....0
%e ..3....1....0....2....2....1....3....2....1....2....1....3....3....1....0....3
%e ..1....1....3....0....2....0....0....2....1....2....3....3....0....2....0....3
%e ..1....1....3....0....0....2....2....2....1....2....3....2....1....2....1....3
%e ..1....0....3....0....0....2....2....2....1....0....3....2....1....2....1....0
%e ..0....0....0....0....0....2....2....2....1....0....0....2....1....0....1....0
%p T:= proc(n, k) option remember;
%p `if`(n<0, 0,
%p `if`(n<3 or k=0, 1, k*T(n-3, 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 nmax = 13; t[_?Negative, _] = 0; t[n_, k_] /; (n < 3 || k == 0) = 1; t[n_, k_] := t[n, k] = k*t[n-3, k] + t[n-1, k]; Flatten[ Table[ t[n-k+1, k], {n , 1, nmax}, {k, n, 1, -1}]](* _Jean-François Alcover_, Nov 28 2011, after Maple *)
%Y Column 1 is A000930,
%Y Column 2 is A003229(n-1),
%Y Column 3 is A084386,
%Y Column 4 is A089977,
%Y Column 10 is A178205,
%Y Row 6 is A028872(n+2),
%Y Row 7 is A144390(n+1),
%Y Row 8 is A003154(n+1).
%K nonn,tabl
%O 1,6
%A _R. H. Hardin_, with proof and formula from Robert Israel in the Sequence Fans Mailing List, Jul 29 2011
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