OFFSET
0,2
COMMENTS
Number of entries in row n is 1 + ceiling(n/2).
T(n,0) = F(n+2) = A000045(n+2) (Fibonacci); number of vertices of Gamma(n).
Sum of entries in row n is A278137(n).
T(n,1) = floor(F(n+2)/2) (see Lemma 2.1 in the Gravier et al. paper).
The generating function of column k is x^{2k-1}/((1-x^3)^k*(1-x-x^2)) (k>=0) (see Corollary 2.5 in the Gravier et al. paper).
LINKS
Indranil Ghosh, Rows 0..100, flattened
S. Gravier, M. Mollard, S. Spacapan, S. S. Zemljic, On disjoint hypercubes in Fibonacci cubes, Discrete Appl. Math., 190-191, 2015, 50-55.
S. Klavzar, Structure of Fibonacci cubes: a survey, J. Comb. Optim., 25, 2013, 505-522.
M. Mollard, Maximal hypercubes in Fibonacci and Lucas cubes, Discrete Appl. Math., 160, 2012, 2479-2483.
FORMULA
T(n,k) = Sum_{i=k-1..floor((n+k-2)/3)} binomial(i,k-1)*F(n+k-3i-1), where F(j) = A000045(j) (Fibonacci); (see Corollary 2.4 in the Gravier et al. paper).
T(n,k) = T(n-2,k-1) + T(n-3,k) (n>=3, k>=1) (see Theorem 2.2 in the Gravier et al. paper).
EXAMPLE
Row 3 is 5,2,1. Indeed, the Fibonacci cube Gamma(3) has 5 vertices A, B, C, D, E and edges AB, BC, CD, DA, DE and so it has at most 2 disjoint edges and it has one square.
Triangle starts:
1;
2, 1;
3, 1;
5, 2, 1;
8, 4, 1;
13, 6, 2, 1;
MAPLE
with(combinat): F := proc (k) options operator, arrow: fibonacci(k) end proc; T := proc (n, k) options operator, arrow: sum(binomial(i, k-1)*F(n+k-3*i-1), i = k-1 .. floor((1/3)*n+(1/3)*k-2/3)) end proc: for n from 0 to 20 do seq(T(n, k), k = 0 .. ceil((1/2)*n)) end do; # yields sequence in triangular form
MATHEMATICA
Flatten[Table[Sum[Binomial[i, k-1] Fibonacci[n+k-3i-1], {i, k-1, Floor[(n+k-2)/3]}], {n, 0, 14}, {k, 0, Ceiling[n/2]}]] (* Indranil Ghosh, Mar 05 2017 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Feb 26 2017
STATUS
approved