A235114 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 star graph S_n (having n vertices).
1, 1, 3, 1, 1, 6, 10, 6, 1, 1, 9, 28, 40, 28, 9, 1, 1, 12, 55, 129, 170, 129, 55, 12, 1, 1, 15, 91, 300, 597, 748, 597, 300, 91, 15, 1, 1, 18, 136, 580, 1560, 2783, 3368, 2783, 1560, 580, 136, 18, 1, 1, 21, 190, 996, 3391, 7923, 13067, 15418, 13067, 7923, 3391, 996, 190, 21, 1
Offset: 0
Examples
Row 1 is 1,3,1; 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}.
Triangle begins:
1;
1,3,1;
1,6,10,6,1;
1,9,28,40,28,9,1;
Links
- E. Mandrescu, Unimodality of some independence polynomials via their palindromicity, Australasian J. of Combinatorics, 53, 2012, 77-82.
- D. Stevanovic, Graphs with palindromic independence polynomial, Graph Theory Notes of New York, 34, 1998, 31-36.
Crossrefs
Cf. A235115.
Programs
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Maple
G := (1-z-2*x*z-x^2*z-x^2*z^2)/((1-z-2*x*z-x^2*z)*(1-z-3*x*z-x^2*z)): 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
Formula
Generating polynomial of row n (n>=1) is x(1 + x)^{2n-2} + (1 + x)^2*(1 + 3*x + x^2)^{n-1} (it is palindromic).
Bivariate generating polynomial: G(x,z) = (1 - z - 2xz - x^2*z - x^2*z^2)/((1 - z - 2xz - x^2*z)(1 - z - 3xz - x^2*z)).
G(1/x, x^2*z) = G(x,z) (this implies the above mentioned palindromicity).
Comments