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%I #11 Feb 27 2020 15:39:01
%S 1,1,3,1,1,6,10,6,1,1,9,27,38,27,9,1,1,12,52,116,150,116,52,12,1,1,15,
%T 85,260,490,602,490,260,85,15,1,1,18,126,490,1215,2052,2436,2052,1215,
%U 490,126,18,1,1,21,175,826,2541,5467,8547,9900,8547,5467,2541,826,175,21,1
%N Irregular triangle read by rows: T(n,k) = number of independent vertex subsets of size k of the graph g_n obtained by attaching two pendant edges to each vertex of the complete graph K_n (0 <= k <= 2n).
%C Sum of entries in row n = 2^{2n-4}*(4 + n) = A079028(n).
%C In the Maple program P[n] gives the independence polynomial of the graph g_n.
%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="https://draganstevanovic.files.wordpress.com/2012/09/gtn-341998-31-36-acro6.pdf">Graphs with palindromic independence polynomial</a>, Graph Theory Notes of New York, 34, 1998, 31-36.
%F Generating polynomial of row n (n>=0) is (1+x)^{2n-2}*((1+x)^2 + nx) (it is palindromic).
%F Bivariate generating polynomial: G(x,z) = (1-z-xz-x^2*z)/(1-z-2xz-x^2*z)^2.
%F G(1/x, x^2*z) = G(x,z) (this implies the above mentioned palindromicity).
%e Row 1 is 1,3,1; indeed, K_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}.
%e Triangle begins:
%e 1;
%e 1,3,1;
%e 1,6,10,6,1;
%e 1,9,27,38,27,9,1;
%e 1,12,52,116,150,116,52,12,1;
%p G := (1-z-x*z-x^2*z)/(1-z-2*x*z-x^2*z)^2: Gser := simplify(series(G, z = 0, 10)): for n from 0 to 9 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 9 do seq(coeff(P[n], x, i), i = 0 .. 2*n) end do;# yields sequence in triangular form
%Y Cf. A079028
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Jan 13 2014
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