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).
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
Offset: 0
Examples
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}.
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;
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. A079028
Programs
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Maple
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
Formula
Generating polynomial of row n (n>=0) is (1+x)^{2n-2}*((1+x)^2 + nx) (it is palindromic).
Bivariate generating polynomial: G(x,z) = (1-z-xz-x^2*z)/(1-z-2xz-x^2*z)^2.
G(1/x, x^2*z) = G(x,z) (this implies the above mentioned palindromicity).
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