

A235113


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).


2



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, 85, 260, 490, 602, 490, 260, 85, 15, 1, 1, 18, 126, 490, 1215, 2052, 2436, 2052, 1215, 490, 126, 18, 1, 1, 21, 175, 826, 2541, 5467, 8547, 9900, 8547, 5467, 2541, 826, 175, 21, 1
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OFFSET

0,3


COMMENTS

Sum of entries in row n = 2^{2n4}*(4 + n) = A079028(n).
In the Maple program P[n] gives the independence polynomial of the graph g_n.


REFERENCES

E. Mandrescu, Unimodality of some independence polynomials via their palindromicity, Australasian J. of Combinatorics, 53, 2012, 7782.
D. Stevanovic, Graphs with palindromic independence polynomial, Graph Theory Notes of New York, 34, 1998, 3136.


LINKS

Table of n, a(n) for n=0..63.


FORMULA

Generating polynomial of row n (n>=0) is (1+x)^{2n2}*((1+x)^2 + nx) (it is palindromic).
Bivariate generating polynomial: G(x,z) = (1zxzx^2*z)/(1z2xzx^2*z)^2.
G(1/x, x^2*z) = G(x,z) (this implies the above mentioned palindromicity).


EXAMPLE

Row 1 is 1,3,1; indeed, K_1 is the onevertex 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}.
Triangle begins:
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;


MAPLE

G := (1zx*zx^2*z)/(1z2*x*zx^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


CROSSREFS

Cf. A079028
Sequence in context: A298636 A123354 A120247 * A235116 A235114 A272866
Adjacent sequences: A235110 A235111 A235112 * A235114 A235115 A235116


KEYWORD

nonn,tabf


AUTHOR

Emeric Deutsch, Jan 13 2014


STATUS

approved



