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%I #60 Jan 02 2023 12:30:48
%S 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,
%T 13,41,65,97,85,34,1,9,15,55,96,181,217,171,55,1,10,17,71,133,301,441,
%U 508,341,89,1,11,19,89,176,463,781,1165,1159,683,144,1,12,21,109,225,673
%N 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; array read by descending antidiagonals, with n, k >= 1.
%C Transposed variant of A083856. - _R. J. Mathar_, Aug 23 2011
%C As to the sequences by columns beginning (1, N, ...), let m = (N-1). 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(n-1) are (1 + sqrt(4*m+1))/2. - _Gary W. Adamson_, Feb 25 2014
%H R. H. Hardin, <a href="/A193376/b193376.txt">Table of n, a(n) for n = 1..9999</a>
%H Ron H. Hardin, <a href="http://list.seqfan.eu/oldermail/seqfan/2011-July/007739.html">Re: A193376 Tabl = 20 existing sequences</a>, Sequence Fans mailing list, 2011.
%H Robert Israel, <a href="http://list.seqfan.eu/oldermail/seqfan/2011-July/007738.html">Re: A193376 Tabl = 20 existing sequences</a>, Sequence Fans mailing list, 2011.
%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. Thus, T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n = 0, 1, ..., z-1.
%F The solution is T(n,k) = Sum_r r^(-n-1)/(1 + z*k*r^(z-1)), where the sum is over the roots r of the polynomial k*x^z + x - 1.
%F 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).
%F T(n,k) = Sum_{s=0..[n/2]} binomial(n-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 Array T(n,k) (with rows n >= 1 and column k >= 1) begins as follows:
%e ..1...1....1....1.....1.....1.....1......1......1......1......1......1...
%e ..2...3....4....5.....6.....7.....8......9.....10.....11.....12.....13...
%e ..3...5....7....9....11....13....15.....17.....19.....21.....23.....25...
%e ..5..11...19...29....41....55....71.....89....109....131....155....181...
%e ..8..21...40...65....96...133...176....225....280....341....408....481...
%e .13..43...97..181...301...463...673....937...1261...1651...2113...2653...
%e .21..85..217..441...781..1261..1905...2737...3781...5061...6601...8425...
%e .34.171..508.1165..2286..4039..6616..10233..15130..21571..29844..40261...
%e .55.341.1159.2929..6191.11605.19951..32129..49159..72181.102455.141361...
%e .89.683.2683.7589.17621.35839.66263.113993.185329.287891.430739.624493...
%e ...
%e Some solutions for n = 5 and k = 3 with colors = 1, 2, 3 and empty = 0:
%e ..0....2....3....2....0....1....0....0....2....0....0....2....3....0....0....0
%e ..0....2....3....2....2....1....2....3....2....1....0....2....3....1....1....1
%e ..1....0....0....0....2....0....2....3....2....1....0....1....0....1....1....1
%e ..1....2....2....0....3....2....2....3....2....0....3....1....3....3....2....1
%e ..0....2....2....0....3....2....2....3....0....0....3....0....3....3....2....1
%p T:= proc(n,k) option remember; `if`(n<0, 0,
%p `if`(n<2 or k=0, 1, k*T(n-2, k) +T(n-1, k)))
%p end;
%p seq(seq(T(n, d+1-n), n=1..d), d=1..12); # _Alois P. Heinz_, Jul 29 2011
%t T[n_, k_] := T[n, k] = If[n < 0, 0, If[n < 2 || k == 0, 1, k*T[n-2, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 12}] // Flatten (* _Jean-François Alcover_, Mar 04 2014, after _Alois P. Heinz_ *)
%Y 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, and column 12 is A053404,
%Y Row 2 is A000027(n+1), row 3 is A004273(n+1), row 4 is A028387, row 5 is A000567(n+1), and row 6 is A106734(n+2).
%Y Diagonal is A171180, superdiagonal 1 is A083859(n+1), and superdiagonal 2 is A083860(n+1).
%K nonn,tabl
%O 1,3
%A _R. H. Hardin_, Jul 24 2011
%E Formula and proof from _Robert Israel_ in the Sequence Fans mailing list.
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