A166106 a(n) = a(n-1) + a(n-2) + F(n), with a(0) = 0, a(1) = 1, a(2) = a(1) + a(0), a(3) = a(2) + a(1), a(4) = a(3) + a(2) + 2.

0, 1, 1, 2, 5, 12, 25, 50, 96, 180, 331, 600, 1075, 1908, 3360, 5878, 10225, 17700, 30509, 52390, 89664, 153000, 260375, 442032, 748775, 1265832, 2136000, 3598250, 6052061, 10164540, 17048641, 28559450, 47786400, 79870428, 133359715, 222457608, 370747675

Offset: 0

Author

Gerald McGarvey, Oct 06 2009

Keywords

• nonn
• easy

Comments

Consider the recursive call tree for Fibonacci numbers shown in the Wilson, Abelson et al., and Bloch links. This type of tree is a variant of Fibonacci trees shown/defined elsewhere. Here, let us refer it as a recursive Fibonacci tree. A Fibonacci number F(n) is the weight of the root of a weighted recursive Fibonacci tree of order n in which the leafs have a weight of 1, and the weight of a node equals the sum of the weights of its two children. If instead we weight each leaf by the number of nodes above it (considering the root as the topmost node), then for n > 2, a(n) is the weight of the root of such a weighted tree of order n. For example, a(5) = 2+2+2+3+3.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Harold Abelson and Gerald Jay Sussman with Julie Sussman, Tree Recursion from "Structure and Interpretation of Computer Programs", MIT Press, 1996, LaTeX2HTML translation by Ryan Bender.
• Laurent Bloch, Analyse de l'algorithme de Fibonacci
• Bill Wilson, The Prolog Dictionary - memoisation (shows Recursive call tree for Fibonacci number f_6).
• Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Crossrefs

Cf. A000045.

Programs

Mathematica

a[n_] := a[n] = a[n-1] + a[n-2] + Fibonacci[n]; a[0] = 0; a[1] = 1; a[2] = 1; a[3] = 2; a[4] = 5; Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Oct 03 2011 *)

PARI

s = 33; a = concat([0,1,1,2,5],vector(s-5)); for(n=6,s,a[n]=a[n-1]+a[n-2]+fibonacci(n)); for(n=1,s,print1(a[n],", "))

PARI

concat(0, Vec(x*(x^5+3*x^4+2*x^3-x^2-x+1)/(x^2+x-1)^2 + O(x^100))) \\ Colin Barker, May 25 2014

Formula

For n > 1, a(n) = A067331(n-2).

From Colin Barker, May 25 2014: (Start)

a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) for n > 6.

G.f.: x*(x^5 + 3*x^4 + 2*x^3 - x^2 - x + 1) / (x^2+x-1)^2. (End)

a(n) = (1/25)*2^(-n-1)*(5*((1 - sqrt(5))^(n+1) + (1 + sqrt(5))^(n+1))*n - (25 + sqrt(5))*(1 + sqrt(5))^n + (sqrt(5) - 25)*(1 - sqrt(5))^n), n > 2. - Ilya Gutkovskiy, Apr 26 2016

Extensions

More terms from Colin Barker, May 25 2014