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A245964
Triangle read by rows: T(n,k) is the number of maximal hypercubes Q(p) in the Lucas cube Lambda(n) (i.e., Q(p) is an induced subgraph of Lambda(n) that is not a subgraph of a subgraph of Lambda(n) that is isomorphic to the hypercube Q(p+1)).
1
1, 1, 0, 2, 0, 3, 0, 0, 2, 0, 0, 5, 0, 0, 3, 2, 0, 0, 0, 7, 0, 0, 0, 8, 2, 0, 0, 0, 3, 9, 0, 0, 0, 0, 15, 2, 0, 0, 0, 0, 11, 11, 0, 0, 0, 0, 3, 24, 2, 0, 0, 0, 0, 0, 26, 13, 0, 0, 0, 0, 0, 14, 35, 2, 0, 0, 0, 0, 0, 3, 50, 15, 0, 0, 0, 0, 0, 0, 40, 48, 2, 0, 0, 0, 0, 0, 0, 17, 85, 17
OFFSET
0,4
COMMENTS
Row n contains 1 + floor(n/2) entries.
Sum of entries in row n = A001608(n) (n>=2) (the Perrin sequence).
LINKS
S. Klavzar, Structure of Fibonacci cubes: a survey, J. Comb. Optim., 25, 2013, 505-522.
M. Mollard, Maximal hypercubes in Fibonacci and Lucas cubes, arXiv:1201.1494 [math.CO], 2012.
M. Mollard, Maximal hypercubes in Fibonacci and Lucas cubes, Discrete Appl. Math., 160, 2012, 2479-2483.
FORMULA
T(n,k) = (n/k)*binomial(k,n-2k) (k>=1).
G.f.: (1+z+t*z^2+t*z^3-t*z^4)/(1-t*(1+z)*z^2).
EXAMPLE
Row 4 is 0,0,2. Indeed, the Lucas cube Lambda(4) is the graph <><> obtained by identifying a vertex of a square with a vertex of another square; each square is a maximal Q(2).
Triangle starts:
1;
1;
0, 2;
0, 3;
0, 0, 2;
0, 0, 5;
0, 0, 3, 2;
0, 0, 0, 7;
0, 0, 0, 8, 2;
MAPLE
T := proc (n, k) if n = 0 and k = 0 then 1 elif n = 1 and k = 0 then 1 elif k = 0 then 0 else n*binomial(k, n-2*k)/k end if end proc: for n from 0 to 20 do seq(T(n, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
MATHEMATICA
T[0|1, 0] = 1; T[_, 0] = 0; T[n_, k_] := (n/k)*Binomial[k, n - 2k];
Table[T[n, k], {n, 0, 20}, {k, 0, n/2}] // Flatten (* Jean-François Alcover, Dec 01 2017 *)
CROSSREFS
Sequence in context: A268726 A035182 A280720 * A141700 A345228 A357887
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Aug 13 2014
STATUS
approved