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%I #21 Sep 08 2022 08:46:06
%S 5,24,116,564,2756,13524,66596,328884,1628036,8074644,40111076,
%T 199506804,993339716,4949921364,24682497956,123144054324,614646529796,
%U 3068937681684,15327508539236,76568823219444,382569238190276,1911746679323604,9554335350106916,47754084564490164,238700054078273156
%N Number of independent vertex subsets of the graph obtained by attaching two pendant edges to each vertex of the star graph S_n (having n vertices; see A235114).
%C a(n) is the sum of the entries of row n of the triangle A235114.
%H Colin Barker, <a href="/A235115/b235115.txt">Table of n, a(n) for n = 1..1000</a>
%H E. Mandrescu, <a href="http://ajc.maths.uq.edu.au/pdf/53/ajc_v53_p077.pdf">Unimodality of some independence polynomials via their palindromicity</a>, Australasian J. of Combinatorics, 53, 2012, 77-82.
%H D. Stevanovic, <a href="http://www.pmf.ni.ac.rs/pmf/licne_prezentacije/101/radovi/GTN%20-%20Palindromic%20Independence%20Polynomial/GTN.34(1998).31-36.Acro6.pdf">Graphs with palindromic independence polynomial</a>, Graph Theory Notes of New York, 34, 1998, 31-36.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (9,-20).
%F a(n) = 4*5^(n-1) + 2^(2*n-2) for n>=1.
%F G.f.: x*(5 - 21*x)/((1 - 4*x)*(1 - 5*x)).
%F a(n) = 9*a(n-1) - 20*a(n-2) for n>1. - _Colin Barker_, Jul 31 2017
%e a(1)=5; indeed, S_1 is the one-vertex graph and after attaching two pendant vertices we obtain the path graph ABC; the independent vertex subsets are: empty, {A}, {B}, {C}, and {A,C}.
%p seq(4*5^(n-1)+2^(2*n-2), n = 1 .. 27);
%t Rest@ CoefficientList[Series[x (5 - 21 x)/((1 - 4 x) (1 - 5 x)), {x, 0, 25}], x] (* or *)
%t LinearRecurrence[{9, -20}, {5, 24}, 25] (* _Michael De Vlieger_, Jul 31 2017 *)
%o (PARI) Vec(x*(5 - 21*x) / ((1 - 4*x)*(1 - 5*x)) + O(x^30)) \\ _Colin Barker_, Jul 31 2017
%o (Magma) [4*5^(n-1)+2^(2*n-2): n in [1..25]]; // _Vincenzo Librandi_, Aug 01 2017
%Y Cf. A235118.
%K nonn,easy
%O 1,1
%A _Emeric Deutsch_, Jan 13 2014
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