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%I #23 Nov 14 2014 10:23:09
%S 1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,3,3,1,1,1,1,4,5,4,1,1,1,1,5,7,
%T 7,5,1,1,1,1,6,9,10,9,6,1,1,1,1,7,11,13,13,11,8,1,1,1,1,8,13,16,17,16,
%U 17,11,1,1,1,1,9,15,19,21,21,28,27,15,1,1,1,1,10,17,22,25,26,41,49,41,20,1,1,1
%N T(n,k) = number of ways to place any number of 5X1 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
%C ..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
%C ..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
%C ..3..5...7...9..11..13..15...17...19...21...23...25...27...29...31...33...35
%C ..4..7..10..13..16..19..22...25...28...31...34...37...40...43...46...49...52
%C ..5..9..13..17..21..25..29...33...37...41...45...49...53...57...61...65...69
%C ..6.11..16..21..26..31..36...41...46...51...56...61...66...71...76...81...86
%C ..8.17..28..41..56..73..92..113..136..161..188..217..248..281..316..353..392
%C .11.27..49..77.111.151.197..249..307..371..441..517..599..687..781..881..987
%C .15.41..79.129.191.265.351..449..559..681..815..961.1119.1289.1471.1665.1871
%C .20.59.118.197.296.415.554..713..892.1091.1310.1549.1808.2087.2386.2705.3044
%C .26.81.166.281.426.601.806.1041.1306.1601.1926.2281.2666.3081.3526.4001.4506
%H R. H. Hardin, <a href="/A193517/b193517.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/5]} (binomial(n-4*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=11 k=3; colors=1, 2, 3; empty=0
%e ..0....2....2....0....0....1....0....3....3....0....0....0....0....3....1....0
%e ..0....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
%e ..0....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
%e ..3....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
%e ..3....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
%e ..3....1....0....2....1....0....3....3....0....3....2....3....1....0....0....1
%e ..3....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
%e ..3....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
%e ..0....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
%e ..0....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
%e ..0....0....3....0....1....1....2....0....2....0....0....3....0....3....0....2
%p T:= proc(n, k) option remember;
%p `if`(n<0, 0,
%p `if`(n<5 or k=0, 1, k*T(n-5, 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 < 5 || k == 0, 1, k*T[n-5, 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 A003520,
%Y Column 2 is A143447(n-4),
%Y Column 3 is A143455(n-4),
%Y Row 10 is A028884.
%K nonn,tabl
%O 1,15
%A _R. H. Hardin_, with proof and formula from _Robert Israel_ in the Sequence Fans Mailing List, Jul 29 2011
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