A245825 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).
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, 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, 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
Offset: 0
Examples
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. Triangle starts: 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;
Links
- S. Klavzar, M. Mollard, M. Petkovsek, The degree sequence of Fibonacci and Lucas cubes, Discrete Math., 311, 2011, 1310-1322.
- S. Klavzar, Structure of Fibonacci cubes: a survey, J. Comb. Optim., 25, 2013, 505-522
Programs
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Maple
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);
Formula
T(n,k) = sum(binomial(n-2i, k-i)*binomial(i+1,n-k-i+1), i=0..k).
G.f.: (1 + t*z + (1 - t)*t*z^2)/((1 - t*z)*(1 - t*z^2) - t*z^3).
Comments