

A318267


a(n) is the number of configurations of n indistinguishable pairs placed on the vertices of the ladder graph P_2 X P_n such that all but two such pairs are joined by an edge.


8



0, 0, 1, 8, 39, 138, 414, 1104, 2715, 6282, 13875, 29540, 61060, 123192, 243589, 473540, 907335, 1716974, 3214066, 5959704, 10958687, 20001526, 36264579, 65359752, 117165096, 209008464, 371190217, 656540768, 1156924167, 2031676818, 3556517478
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OFFSET

0,4


COMMENTS

The generating function has been obtained using the calculus of the rook polynomial associated with A046741.
The case of all but one pair joined by an edge is given by A178523(n1). The case of all pairs joined by an edge is given by A000045(n+1), i.e., the number of perfect matchings in the ladder graph.
This is also the number of "(n2)domino" configurations in the game of memory played on a 2 X n rectangular array, see [Young].  Donovan Young, Oct 23 2018


LINKS



FORMULA

G.f.: x^2*(1 + 3*x + 6*x^2 + x^3 + 3*x^4)/((1  x)^2*(1  x  x^2)^3).


EXAMPLE

Consider the case n=3. Let the 2 X 3 grid have 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, OE, AD, BC}.
If DC represents the one pair which is joined by an edge, the remaining pairs must be placed on AE and OB; there are three other such configurations where the joined pair is placed instead on ED, OA, or AB. Our count is now at 4. If the joined pair is placed on OE then the remaining pairs must be placed on BD and AC; there is one other such configuration where the joined pair is placed on BC, bringing the count to 6. Finally, let the joined pair be placed on AD, then the remaining pairs may be placed either on OB, EC or on OC, EB, and thus we have a(3) = 8.


MAPLE

seq(coeff(series(x^2*(1+3*x+6*x^2+x^3+3*x^4)/((1x)^2*(1xx^2)^3), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 23 2018


MATHEMATICA

CoefficientList[Normal[Series[x^2(1 + 3*x + 6*x^2 + x^3 + 3*x^4)/(1  x)^2/(1  x  x^2)^3, {x, 0, 30}]], x]


PROG

(GAP) a:=[0, 0, 1, 8, 39, 138, 414, 1104];; for n in [9..35] do a[n]:=5*a[n1]7*a[n2]2*a[n3]+10*a[n4]2*a[n5]5*a[n6]+a[n7]+a[n8]; od; a; # Muniru A Asiru, Oct 23 2018


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



