cp's OEIS Frontend

A192019 The Wiener index of the binary Fibonacci tree of order n.

%I A192019 #15 Dec 10 2016 03:00:11
%S A192019 1,10,50,214,802,2802,9275,29580,91668,277924,828092,2433140,7067885,
%T A192019 20337318,58054534,164602410,463990190,1301338150,3633753815,
%U A192019 10107239160,28016346216,77419909800,213349801560,586471432104,1608485221177,4402406713762
%N A192019 The Wiener index of the binary Fibonacci tree of order n.
%C A192019 The binary Fibonacci trees f(k) of order k are rooted binary trees defined as follows: 1. f(0) has no nodes and f(1) consists of a single node. 2. For k>=2, f(k) is constructed from a root with f(k-1) as its left subtree and f(k-2) as its right subtree. See the Iyer & Reddy references.
%D A192019 K. Viswanathan Iyer and K. R. Udaya Kumar Reddy, Wiener index of Binomial trees and Fibonacci trees, Int'l. J. Math. Engin. with Comp., Accepted for publication, Sept. 2009.
%H A192019 B. E. Sagan, Y-N. Yeh and P. Zhang, <a href="http://dx.doi.org/10.1002/(SICI)1097-461X(1996)60:5&lt;959::AID-QUA2&gt;3.0.CO;2-W">The Wiener Polynomial of a Graph</a>, Internat. J. of Quantum Chem., 60, 1996, 959-969.
%H A192019 K. Viswanathan Iyer and K. R. Udaya Kumar Reddy, <a href="http://arxiv.org/abs/0910.4432">Wiener index of binomial trees and Fibonacci trees</a>, arXiv:0910.4432 [cs.DM], 2009.
%F A192019 a(n) = Sum_{k>=1} k*A192018(n,k).
%F A192019 The Wiener index of a connected graph is the derivative of the Wiener polynomial W(t) of the graph, evaluated at t=1. The Wiener polynomial w(n,t) of the binary Fibonacci tree of order n satisfies the recurrence relation w(n,t) = w(n-1,t) + w(n-2,t) + t*r(n-1,t) + t*r(n-2) + t^2*r(n-1,t)*r(n-2,t), w(1,t)=0, w(2,t)=t, where r(n,t) is the generating polynomial of the nodes of the binary Fibonacci tree f(n) with respect to the level of the nodes (for example, r(2,t) = 1 + t for the tree / ; see A004070 and the Maple program).
%F A192019 Empirical G.f.: x^2*(x^4-3*x^2+4*x+1)/((x+1)^2*(x^2-3*x+1)^2*(x^2+x-1)^2). [_Colin Barker_, Nov 17 2012]
%e A192019 a(3)=10 because the binary Fibonacci tree of order 3 is basically the path graph A - B - R - C and we have 3 distances equal to 1 (AB, BR, RC), 2 distances equal to 2 (AR and BC) and 1 distance equal to 3 (AC); 3*1 + 2*2 + 1*3 = 10.
%p A192019 G := z/((1-z)*(1-t*z-t*z^2)): Gser := simplify(series(G, z = 0, 30)): for n to 27 do r[n] := sort(coeff(Gser, z, n)) end do: w[1] := 0: w[2] := t: for n from 3 to 27 do w[n] := sort(expand(w[n-1]+w[n-2]+t*r[n-1]+t*r[n-2]+t^2*r[n-1]*r[n-2])) end do: seq(subs(t = 1, diff(w[n], t)), n = 2 .. 27);
%Y A192019 Cf. A192018.
%K A192019 nonn
%O A192019 2,2
%A A192019 _Emeric Deutsch_, Jun 21 2011