A178524 Triangle read by rows: T(n,k) is the number of leaves at level k in the Fibonacci tree of order n (n>=0, 0<=k<=n-1).

1, 1, 0, 2, 0, 1, 2, 0, 0, 3, 2, 0, 0, 1, 5, 2, 0, 0, 0, 4, 7, 2, 0, 0, 0, 1, 9, 9, 2, 0, 0, 0, 0, 5, 16, 11, 2, 0, 0, 0, 0, 1, 14, 25, 13, 2, 0, 0, 0, 0, 0, 6, 30, 36, 15, 2, 0, 0, 0, 0, 0, 1, 20, 55, 49, 17, 2, 0, 0, 0, 0, 0, 0, 7, 50, 91, 64, 19, 2, 0, 0, 0, 0, 0, 0, 1, 27, 105, 140, 81, 21, 2

Offset: 0

Author

Emeric Deutsch, Jun 15 2010

Keywords

• nonn
• tabf

Comments

A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node.

Sum of entries in row n is the Fibonacci number F(n+1) (A000045(n+1)).

Sum(k*T(n,k),k>=0)=A067331(n-2).

T(n,k) is the number of vertices in the Fibonacci cube G_{n-1} that have eccentricity k (see Klavzar and Mollard reference). - Michel Mollard, Aug 20 2014

Examples

			Triangle starts:
1,
1,
0, 2,
0, 1, 2,
0, 0, 3, 2,
0, 0, 1, 5, 2,
0, 0, 0, 4, 7, 2,
0, 0, 0, 1, 9, 9, 2,
0, 0, 0, 0, 5, 16, 11, 2,
0, 0, 0, 0, 1, 14, 25, 13, 2,
0, 0, 0, 0, 0, 6, 30, 36, 15, 2,
		

References

• D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.

Links

• A. Castro and M. Mollard, The eccentricity sequences of Fibonacci and Lucas cubes, Discrete Math., 312 (2012), 1025-1037. See Table 1. [From _N. J. A. Sloane_, Mar 22 2012]
• Y. Horibe, An entropy view of Fibonacci trees, Fibonacci Quarterly, 20, No. 2, 1982, 168-178.
• S. Klavzar, M. Mollard, Asymptotic Properties of Fibonacci Cubes and Lucas Cubes, Annals of Combinatorics, 18, 2014, 447-457.

Crossrefs

Cf. A000045, A067331, A178522, A004070

Programs

Maple

G := (1+z-t*z)/(1-t*z-t*z^2): Gser := simplify(series(G, z = 0, 17)): for n from 0 to 15 do P[n] := sort(coeff(Gser, z, n)) end do: 1; for n to 13 do seq(coeff(P[n], t, k), k = 0 .. n-1) end do; # yields sequence in triangular form

Formula

G.f.: G(t,z) = (1+z-t*z) / (1-t*z-t*z^2).