

A193376


T(n,k) = number of ways to place any number of 2 X 1 tiles of k distinguishable colors into an n X 1 grid.


5



1, 1, 2, 1, 3, 3, 1, 4, 5, 5, 1, 5, 7, 11, 8, 1, 6, 9, 19, 21, 13, 1, 7, 11, 29, 40, 43, 21, 1, 8, 13, 41, 65, 97, 85, 34, 1, 9, 15, 55, 96, 181, 217, 171, 55, 1, 10, 17, 71, 133, 301, 441, 508, 341, 89, 1, 11, 19, 89, 176, 463, 781, 1165, 1159, 683, 144, 1, 12, 21, 109, 225, 673
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OFFSET

1,3


COMMENTS

Table starts:
..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
..3...5....7....9....11....13....15.....17.....19.....21.....23.....25.....27
..5..11...19...29....41....55....71.....89....109....131....155....181....209
..8..21...40...65....96...133...176....225....280....341....408....481....560
.13..43...97..181...301...463...673....937...1261...1651...2113...2653...3277
.21..85..217..441...781..1261..1905...2737...3781...5061...6601...8425..10557
.34.171..508.1165..2286..4039..6616..10233..15130..21571..29844..40261..53158
.55.341.1159.2929..6191.11605.19951..32129..49159..72181.102455.141361.190399
.89.683.2683.7589.17621.35839.66263.113993.185329.287891.430739.624493.881453
Transposed variant of A083856.  R. J. Mathar, Aug 23 2011
As to the sequences by columns beginning (1, N,...), let m = (N1). The g.f. for the sequence (1, N,...) is 1/(1  x  m*x^2). Alternatively, the corresponding matrix generator is [(1,1},(m,0}]. Another equivalency is simply: The sequence beginning (1, N,...) is the INVERT transform of (1, m, 0, 0, 0,...). Convergents to the sequences a(n)/a(n1) are (1 + sqrt(4*m+1))/2.  Gary W. Adamson, Feb 25 2014


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.
For z = 2, T(n,k) = ((2 k/(sqrt(1+4 k)  1))^(n+1)  (2 k/(sqrt(1+4 k) + 1))^(n+1))/sqrt(1+4 k).
T(n,k) = sum {s=0..[n/2]} (binomial(ns,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=5 k=3; colors=1, 2, 3; empty=0
..0....2....3....2....0....1....0....0....2....0....0....2....3....0....0....0
..0....2....3....2....2....1....2....3....2....1....0....2....3....1....1....1
..1....0....0....0....2....0....2....3....2....1....0....1....0....1....1....1
..1....2....2....0....3....2....2....3....2....0....3....1....3....3....2....1
..0....2....2....0....3....2....2....3....0....0....3....0....3....3....2....1


MAPLE

T:= proc(n, k) option remember; `if`(n<0, 0,
`if`(n<2 or k=0, 1, k*T(n2, k) +T(n1, k)))
end;
seq(seq(T(n, d+1n), n=1..d), d=1..12); # Alois P. Heinz, Jul 29 2011


MATHEMATICA

T[n_, k_] := T[n, k] = If[n < 0, 0, If[n < 2  k == 0, 1, k*T[n2, k]+T[n1, k]]]; Table[Table[T[n, d+1n], {n, 1, d}], {d, 1, 12}] // Flatten (* JeanFrançois Alcover, Mar 04 2014, after Alois P. Heinz *)


CROSSREFS

Column 1 is A000045(n+1),
Column 2 is A001045(n+1),
Column 3 is A006130,
Column 4 is A006131,
Column 5 is A015440,
Column 6 is A015441(n+1),
Column 7 is A015442(n+1),
Column 8 is A015443,
Column 9 is A015445,
Column 10 is A015446,
Column 11 is A015447,
Column 12 is A053404,
Diagonal is A171180,
Row 2 is A000027(n+1),
Row 3 is A004273(n+1),
Row 4 is A028387,
Row 5 is A000567(n+1),
Row 6 is A106734(n+2),
Superdiagonal 1 is A083859(n+1),
Superdiagonal 2 is A083860(n+1).
Sequence in context: A210489 A125175 A210552 * A185095 A177888 A073020
Adjacent sequences: A193373 A193374 A193375 * A193377 A193378 A193379


KEYWORD

nonn,tabl


AUTHOR

R. H. Hardin, Jul 24 2011


EXTENSIONS

Formula and proof from Robert Israel in the Sequence Fans mailing list.


STATUS

approved



