

A334057


Triangle read by rows: T(n,k) is the number of configurations with exactly k polyomino matchings in a generalized game of memory played on the path of length 4n.


6



1, 0, 1, 31, 3, 1, 5474, 288, 12, 1, 2554091, 72026, 1476, 31, 1, 2502018819, 43635625, 508610, 5505, 65, 1, 4456194509950, 52673302074, 394246455, 2559565, 16710, 120, 1, 13077453070386914, 111562882654972, 580589062179, 2504572910, 10288390, 43806, 203, 1
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OFFSET

0,4


COMMENTS

In this generalized game of memory n indistinguishable quadruples of matched cards are placed on the vertices of the path of length 4n. A polyomino is a quadruple on four adjacent vertices.
T(n,k) is the number of set partitions of {1..4n} into n sets of 4 with k of the sets being a contiguous set of elements.  Andrew Howroyd, Apr 16 2020


LINKS



FORMULA

G.f.: Sum_{j>=0} (4*j)! * y^j / (j! * 24^j * (1+(1z)*y)^(4*j+1)).
T(n,k) = Sum_{j=0..nk} (1)^(njk)*(n+3*j)!/(24^j*j!*(njk)!*k!).  Andrew Howroyd, Apr 16 2020


EXAMPLE

The first few rows of T(n,k) are:
1;
0, 1;
31, 3, 1;
5474, 288, 12, 1;
2554091, 72026, 1476, 31, 1;
...
For n=2 and k=1 the polyomino must start either on the second vertex of the path, the third, or the fourth, otherwise the remaining quadruple will also form a polyomino; thus T(2,1) = 3.


MATHEMATICA

CoefficientList[Normal[Series[Sum[y^j*(4*j)!/24^j/j!/(1+y*(1z))^(4*j+1), {j, 0, 20}], {y, 0, 20}]], {y, z}]


PROG

(PARI) T(n, k)={sum(j=0, nk, (1)^(njk)*(n+3*j)!/(24^j*j!*(njk)!*k!))} \\ Andrew Howroyd, Apr 16 2020


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



