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A319371
Numbers k such that the characteristic polynomial of a wheel graph of k nodes has exactly one monomial with vanishing coefficient.
0
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
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 A376455 A029597
KEYWORD
nonn
AUTHOR
Pierandrea Formusa, Sep 17 2018
STATUS
approved