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 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 (list; table; graph; refs; listen; history; text; internal format)
 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 (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. T(n,k) = sum {s=0..[n/4]} (binomial(n-3*s,s)*k^s). 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 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(n-4, k) +T(n-1, k)))     end: seq(seq(T(n, d+1-n), 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[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 *) CROSSREFS Column 1 is A003269(n+1), Column 2 is A052942, Column 3 is A143454(n-3), 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

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