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%I #16 Jun 22 2015 19:45:08
%S 1,0,2,0,2,1,0,1,3,1,0,0,5,2,1,0,0,3,7,2,1,0,0,1,10,7,2,1,0,0,0,9,14,
%T 8,2,1,0,0,0,4,23,16,9,2,1,0,0,0,1,22,34,19,10,2,1,0,0,0,0,14,50,44,
%U 22,11,2,1,0,0,0,0,5,55,77,56,25,12,2,1,0,0,0,0,1,40,117,106,69,28,13,2,1,0,0,0,0,0,20,131,188,140,83,31,14,2,1
%N Triangle read by rows: T(n,k) is the number of the vertices of the Fibonacci cube G_n that have degree k (0<=k<=n).
%C The Fibonacci cube G_n is obtained from the n-cube Q_n by removing all the vertices that contain two consecutive 1s.
%C Sum of entries in row n is the Fibonacci number F_{n+2}.
%C Sum of entries in column k (k>=1) is the Fibonacci number F_{2k+3}. - _Emeric Deutsch_, Jun 22 2015
%C Sum(k*T(n,k), k=0..n) = 2*sum(F(k)*F(n+1-k),k=0..n+1) = 2*A001629(n+1).
%H S. Klavzar, M. Mollard, M. Petkovsek, <a href="http://dx.doi.org/10.1016/j.disc.2011.03.019">The degree sequence of Fibonacci and Lucas cubes</a>, Discrete Math., 311, 2011, 1310-1322.
%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
%F T(n,k) = sum(binomial(n-2i, k-i)*binomial(i+1,n-k-i+1), i=0..k).
%F G.f.: (1 + t*z + (1 - t)*t*z^2)/((1 - t*z)*(1 - t*z^2) - t*z^3).
%e Row 2 is 0,2,1 because the Fibonacci cube G_2 is the path-tree P_3 having 2 vertices of degree 1 and 1 vertex of degree 2.
%e Triangle starts:
%e 1;
%e 0,2;
%e 0,2,1;
%e 0,1,3,1;
%e 0,0,5,2,1;
%e 0,0,3,7,2,1;
%e 0,0,1,10,7,2,1;
%p T := proc (n, k) options operator, arrow: sum(binomial(n-2*i, k-i)*binomial(i+1, n-k-i+1), i = 0 .. k) end proc: seq(seq(T(n, k), k = 0 .. n), n = 0 .. 13);
%Y Cf. A000045, A001519, A001629.
%K nonn,tabl
%O 0,3
%A _Emeric Deutsch_, Aug 03 2014
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