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Triangle read by rows: T(n,k) is the number of maximal hypercubes Q(p) in the Fibonacci cube Gamma(n) (i.e., Q(p) is an induced subgraph of Gamma(n) that is not a subgraph of a subgraph of Gamma(n) that is isomorphic to the hypercube Q(p+1)).
3

%I #19 Jul 17 2017 01:42:18

%S 1,0,1,0,2,0,1,1,0,0,3,0,0,3,1,0,0,1,4,0,0,0,6,1,0,0,0,4,5,0,0,0,1,10,

%T 1,0,0,0,0,10,6,0,0,0,0,5,15,1,0,0,0,0,1,20,7,0,0,0,0,0,15,21,1,0,0,0,

%U 0,0,6,35,8,0,0,0,0,0,1,35,28,1,0,0,0,0,0,0,21,56,9,0,0,0,0,0,0,7,70,36,1

%N Triangle read by rows: T(n,k) is the number of maximal hypercubes Q(p) in the Fibonacci cube Gamma(n) (i.e., Q(p) is an induced subgraph of Gamma(n) that is not a subgraph of a subgraph of Gamma(n) that is isomorphic to the hypercube Q(p+1)).

%C The nonzero entries in columns 0,1,2,... are rows 0,2,3,... of the Pascal triangle.

%C Row n contains 1+ceiling(n/2) entries.

%C Sum of entries in row n = A000931(n+6) (the Padovan sequence).

%C Sum_{k>=0}k*T(n,k) = A228364(n+1).

%H Michael De Vlieger, <a href="/A245963/b245963.txt">Table of n, a(n) for n = 0..10300</a> (Rows 1 <= n <= 200).

%H S. Klavzar, <a href="http://dx.doi.org/10.1007/s10878-011-9433-z">Structure of Fibonacci cubes: a survey</a>, J. Comb. Optim., 25, 2013, 505-522.

%H M. Mollard, <a href="http://arxiv.org/abs/1201.1494">Maximal hypercubes in Fibonacci and Lucas cubes</a>, arXiv:1201.1494 [math.CO], 2012.

%H M. Mollard, <a href="http://dx.doi.org/10.1016/j.dam.2012.06.003">Maximal hypercubes in Fibonacci and Lucas cubes</a>, Discrete Appl. Math., 160, 2012, 2479-2483.

%F T(n,k) = binomial(k+1,n-2*k+1).

%F G.f.: (1+t*z*(1+z))/(1-t*(1+z)*z^2).

%e Row 3 is 0,1,1. Indeed, the Fibonacci cube Gamma(3) is a square with an additional pendant edge attached to one of its vertices; the pendant edge is a maximal Q(1) and the square is a maximal Q(2).

%e Triangle starts:

%e 1;

%e 0, 1;

%e 0, 2;

%e 0, 1, 1;

%e 0, 0, 3;

%e 0, 0, 3, 1;

%e 0, 0, 1, 4;

%e 0, 0, 0, 6, 1;

%p T := proc (n, k) options operator, arrow: binomial(1+k, n-2*k+1) end proc: for n from 0 to 20 do seq(T(n, k), k = 0 .. (n+1)*(1/2)) end do; # yields sequence in triangular form

%t Table[Binomial[k + 1, n - 2 k + 1], {n, 0, 17}, {k, 0, Ceiling[n/2]}] // Flatten (* _Michael De Vlieger_, Jul 16 2017 *)

%Y Cf. A000931, A228364, A245964.

%K nonn,tabf

%O 0,5

%A _Emeric Deutsch_, Aug 13 2014