OFFSET
0,5
COMMENTS
Equivalently, T(n,k) is the number of k-matchings in the ladder graph L_n = P_2 X P_n. - Emeric Deutsch, Dec 25 2004
In other words, triangle of number of monomer-dimer tilings on (2,n)-block with k dimers. If z marks the size of the block and t marks the dimers, then it is easy to see that the g.f. for indecomposable tilings, i.e., those that cannot be split vertically into smaller tilings, is g = (1+t)*z + t^2*z^2 + 2*t*z^2 + 2*t^2*z^3 + 2*t^3*z^4 + ... = (1+t)*z + t^2*z^2 + 2*t*z^2/(1-t*z); then the g.f. is 1/(1-g) = (1-t*z)/(1 - z - 2*t*z - t*z^2 + t^3*z^3) (see eq. (4) of the Grimson reference). From this the recurrence of the McQuistan & Lichtman reference follows at once. - Emeric Deutsch, Oct 16 2006
LINKS
Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
R. C. Grimson, Exact formulas for 2 x n arrays of dumbbells, J. Math. Phys., 15.2 (1974), 214-216. (Annotated scanned copy)
R. C. Grimson, Exact formulas for 2 x n arrays of dumbbells, J. Math. Phys., 15 (1974), 214-216.
H. Hosoya and A. Motoyama, An effective algorithm for obtaining polynomials for dimer statistics. Application of operator technique on the topological index to two- and three-dimensional rectangular and torus lattices, J. Math. Physics 26 (1985) 157-167.
R. B. McQuistan and S. J. Lichtman, Exact recursion relation for 2 x N arrays of dumbbells, J. Math. Phys., 11 (1970), 3095-3099.
D. G. Rogers, An application of renewal sequences to the dimer problem, pp. 142-153 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979.
Eric Weisstein's World of Mathematics, Ladder Graph
Eric Weisstein's World of Mathematics, Matching-Generating Polynomial
Donovan Young, The Number of Domino Matchings in the Game of Memory, J. Int. Seq., Vol. 21 (2018), Article 18.8.1.
Donovan Young, Generating Functions for Domino Matchings in the 2 * k Game of Memory, arXiv:1905.13165 [math.CO], 2019. Also in J. Int. Seq., Vol. 22 (2019), Article 19.8.7.
FORMULA
From Emeric Deutsch, Dec 25 2004: (Start)
The row generating polynomials P[n] satisfy P[n] = (1 + 2*t)*P[n-1] + t*P[n-2] - t^3*P[n-3] with P[0] = 1, P[1] = 1+t, P[2] = 1 + 4*t + 2*t^2.
G.f.: (1-t*z)/(1 - z - 2*t*z - t*z^2 + t^3*z^3). (End)
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) + T(n-2,k-1) - T(n-3,k-3).
EXAMPLE
T(3, 2)=11 because in the 2 X 3 grid with vertex set {O(0, 0), A(1, 0), B(2, 0), C(2, 1), D(1, 1), E(0, 1)} and edge set {OA, AB, ED, DC, UE, AD, BC} we have the following eleven 2-matchings: {OA, BC}, {OA, DC}, {OA, ED}, {AB, DC}, {AB, ED}, {AB, OE}, {BC, AD}, {BC, ED}, {BC, OA}, {BC, OE} and {DC, OE}. - Emeric Deutsch, Dec 25 2004
Triangle starts:
1;
1, 1;
1, 4, 2;
1, 7, 11, 3;
1, 10, 29, 26, 5;
MAPLE
F[0]:=1:F[1]:=1+t:F[2]:=1+4*t+2*t^2:for n from 3 to 10 do F[n]:=sort(expand((1+2*t)*F[n-1]+t*F[n-2]-t^3*F[n-3])) od: for n from 0 to 10 do seq(coeff(t*F[n], t^k), k=1..n+1) od; # yields sequence in triangular form - Emeric Deutsch
MATHEMATICA
p[n_] := p[n] = (1 + 2t) p[n-1] + t*p[n-2] - t^3*p[n-3]; p[0] = 1; p[1] = 1+t; p[2] = 1 + 4t + 2t^2; Flatten[Table[CoefficientList[Series[p[n], {t, 0, n}], t], {n, 0, 10}]][[;; 62]] (* Jean-François Alcover, Jul 13 2011, after Emeric Deutsch *)
CoefficientList[LinearRecurrence[{1 + 2 x, x, -x^3}, {1 + x, 1 + 4 x + 2 x^2, 1 + 7 x + 11 x^2 + 3 x^3}, {0, 10}], x] // Flatten (* Eric W. Weisstein, Apr 03 2018 *)
CoefficientList[CoefficientList[Series[-(1 + x z) (-1 - x + x^2 z)/(1 - z - 2 x z - x z^2 + x^3 z^3), {z, 0, 10}], z], x] // Flatten (* Eric W. Weisstein, Apr 03 2018 *)
PROG
(Haskell)
a046741 n k = a046741_tabl !! n !! k
a046741_row n = a046741_tabl !! n
a046741_tabl = [[1], [1, 1], [1, 4, 2]] ++ f [1] [1, 1] [1, 4, 2] where
f us vs ws = ys : f vs ws ys where
ys = zipWith (+) (zipWith (+) (ws ++ [0]) ([0] ++ map (* 2) ws))
(zipWith (-) ([0] ++ vs ++ [0]) ([0, 0, 0] ++ us))
-- Reinhard Zumkeller, Jan 18 2014
CROSSREFS
KEYWORD
AUTHOR
EXTENSIONS
More terms from Larry Reeves (larryr(AT)acm.org), Apr 07 2000
Formula fixed by Reinhard Zumkeller, Jan 18 2014
STATUS
approved