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

1,2

Comments

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

Links

Table of n, a(n) for n=1..67.

Wikipedia, Wheel graph

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)

Example

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

Prog

(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)

Crossrefs

Cf. A004772.

Sequence in context: A184478 A188188 A230319 * A004772 A029597 A188262

Adjacent sequences: A319368 A319369 A319370 * A319372 A319373 A319374

Keyword

nonn

Author

Pierandrea Formusa, Sep 17 2018

Status

approved