A178523 The path length of the Fibonacci tree of order n.
0, 0, 2, 6, 16, 36, 76, 152, 294, 554, 1024, 1864, 3352, 5968, 10538, 18478, 32208, 55852, 96420, 165800, 284110, 485330, 826752, 1404816, 2381616, 4029216, 6803666, 11468502, 19300624, 32433204, 54426364, 91216184, 152691702, 255313658, 426460288, 711634648
Offset: 0
Examples
a(2)=2 because the Fibonacci tree of order 2 is /\ with path length 1 + 1. - _Emeric Deutsch_, Sep 13 2010
References
- Ralph P. Grimaldi, Properties of Fibonacci trees, In Proceedings of the Twenty-second Southeastern Conference on Combinatorics, Graph Theory, and Computing (Baton Rouge, LA, 1991); Congressus Numerantium 84 (1991), 21-32. - Emeric Deutsch, Sep 13 2010
- D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.
Links
- Muniru A Asiru, Table of n, a(n) for n = 0..4500
- Carlos Alirio Rico Acevedo, and Ana Paula Chaves, Double-Recurrence Fibonacci Numbers and Generalizations, arXiv:1903.07490 [math.NT], 2019.
- Y. Horibe, An entropy view of Fibonacci trees, Fibonacci Quarterly, 20, No. 2, 1982, 168-178.
- W. Kuszmaul, Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations, arXiv preprint arXiv:1509.08216 [cs.DM], 2015-2017.
- D. Young, The Number of Domino Matchings in the Game of Memory, Journal of Integer Sequences, Vol. 21 (2018), Article 18.8.1.
- Donovan Young, Generating Functions for Domino Matchings in the 2 * k Game of Memory, arXiv:1905.13165 [math.CO], 2019. Also in J. Int. Seq., Vol. 22 (2019), Article 19.8.7.
- Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,1,1).
Programs
-
GAP
a:=[0,2];; for n in [3..35] do a[n]:=a[n-1]+a[n-2]+ 2*Fibonacci(n +1) -2; od; Concatenation([0],a); # Muniru A Asiru, Oct 23 2018
-
Magma
[2+(2/5)*(4*n-9)*Fibonacci(n)+(2/5)*(3*n-5)*Fibonacci(n-1): n in [0..40]]; // Vincenzo Librandi, Oct 24 2018
-
Maple
with(combinat): a := proc (n) options operator, arrow: 2+((8/5)*n-18/5)*fibonacci(n)+((6/5)*n-2)*fibonacci(n-1) end proc: seq(a(n), n = 0 .. 35); G := 2*z^2/((1-z)*(1-z-z^2)^2): Gser := series(G, z = 0, 40): seq(coeff(Gser, z, n), n = 0 .. 35);
-
Mathematica
Table[2 +2/5 (4n-9) Fibonacci[n] +2/5 (3n -5) Fibonacci[n-1], {n, 0, 40}] (* or *) LinearRecurrence[{3, -1, -3, 1, 1}, {0, 0, 2, 6, 16}, 40] (* Harvey P. Dale, Oct 02 2016 *)
-
PARI
vector(40, n, n--; (10+(8*n-18)*fibonacci(n)+(6*n-10)*fibonacci(n-1))/5) \\ G. C. Greubel, Jan 31 2019
-
Sage
[(10+(8*n-18)*fibonacci(n)+(6*n-10)*fibonacci(n-1))/5 for n in range(40)] # G. C. Greubel, Jan 31 2019
Formula
a(n) = 2 + (2/5)*(4n-9)*F(n) + (2/5)*(3n-5)*F(n-1), where F(n) = A000045(n) (Fibonacci numbers).
a(n) = 2*A006478(n+1).
a(n) = Sum_{k=0..n-1} k*A178522(n,k).
G.f.: 2*z^2/((1-z)*(1-z-z^2)^2).
From Emeric Deutsch, Sep 13 2010: (Start)
a(0)=a(1)=0, a(n) = a(n-1)+a(n-2)+2F(n+1)-2 if n>=2; here F(j)=A000045(j) are the Fibonacci numbers (see the Grimaldi reference, Eq. (**) on p. 23).
An explicit formula for a(n) is given in the Grimaldi reference (Theorem 2).
(End)
E.g.f.: 2*exp(x) + 2*exp(x/2)*(5*(4*x - 5)*cosh(sqrt(5)*x/2) + sqrt(5)*(10*x - 13)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Dec 04 2023
Comments