A319371 Numbers k such that the characteristic polynomial of a wheel graph of k nodes has exactly one monomial with vanishing coefficient.

1, 2, 3, 4, 6, 7, 8, 10, 11, 12, 14, 15, 16, 18, 19, 20, 22, 23, 24, 26, 27, 28, 30, 31, 32, 34, 35, 36, 38, 39, 40, 42, 43, 44, 46, 47, 48, 50, 51, 52, 54, 55, 56, 58, 59, 60, 62, 63, 64, 66, 67, 68, 70, 71, 72, 74, 75, 76, 78, 79, 80, 82, 83, 84, 86, 87, 88

Offset: 1

Author

Pierandrea Formusa, Sep 17 2018

Keywords

• nonn

Comments

For the spectrum of W_n see, e.g., the Wikipedia link. - Wolfdieter Lang, Oct 30 2018

Examples

			4 is a term as the characteristic polynomial of the wheel graph of 4 nodes is x^4 - 6*x^2 - 8*x - 3, in which the monomial of x^3 has null coefficient and no other ones, so this polynomial has exactly one monomial with vanishing coefficient.
5 is not member of this sequence because the eigenvalues of A(W_5) (the adjacency matrix of W_5) has eigenvalues 0, 0, 2, 1 + sqrt(5), 1 - sqrt(5), and the monic characteristic polynomial is x^5 - 8*x^3 - 8*x^2 with three missing monomials x^0, x^1 and x^4. - _Wolfdieter Lang_, Oct 30 2018
		

Links

• Wikipedia, Wheel graph

Crossrefs

Cf. A004772.

Programs

Sage

def how_many_zeros(v):
    t=0
    for el in v:
        if el==0: t += 1
    return t
r=""
for i in range(1,100):
        p = graphs.WheelGraph(i)
        cp=p.characteristic_polynomial()
        vcp=(cp.coefficients(sparse=False))
        if how_many_zeros(vcp)==1:
            r=r+","+str(i)
print(r)

Formula

Conjecture: a(n) = A004772(n) for n> 1. [clarified by Michel Marcus, Apr 16 2019]

Conjectures from Colin Barker, Nov 02 2020: (Start)

G.f.: x*(1 + x + x^2 + x^4) / ((1 - x)^2*(1 + x + x^2)).

a(n) = a(n-1) + a(n-3) - a(n-4) for n>5.

(End)