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The Wiener index of the binomial tree of order n.
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%I #25 Sep 15 2019 11:05:37

%S 0,1,10,68,392,2064,10272,49216,229504,1048832,4719104,20972544,

%T 92276736,402657280,1744838656,7516209152,32212287488,137439019008,

%U 584115683328,2473901424640,10445360988160,43980466159616,184717955563520,774056190148608

%N The Wiener index of the binomial tree of order n.

%C The binomial trees b(k) of order k are ordered trees defined as follows: 1. b(0) consists of a single node. 2. For k>=1, b(k) is obtained from two copies of b(k-1) by linking them in such a way that the root of one is the leftmost child of the root of the other. See the Iyer & Reddy references.

%D 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.

%D T. H. Cormen, C. E. Leiserson and R. L. Rivest: Introduction to Algorithms. MIT Press / McGraw-Hill (1990)

%H 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 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.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (10,-32,32).

%F a(n) = Sum_{k>=1} k*A192020(n,k).

%F From _Colin Barker_, Jul 07 2012: (Start)

%F a(n) = 2^(n-1)*(1+2^n*(n-1)).

%F a(n) = 10*a(n-1) - 32*a(n-2) + 32*a(n-3).

%F G.f.: x/((1-2*x)*(1-4*x)^2). (End)

%p a := proc(n) (n-1)*2^(2*n-1)+2^(n-1) end proc: seq(a(n), n = 0 .. 23);

%t LinearRecurrence[{10, -32, 32}, {0, 1, 10}, 23] (* _Jean-François Alcover_, Sep 23 2017 *)

%Y Cf. A192020.

%K nonn,easy

%O 0,3

%A _Emeric Deutsch_, Jun 22 2011

%E Initial 0 in the sample values which is Wiener index of singleton tree b(0), and consequent amendments to formulas. - _Kevin Ryde_, Sep 12 2019