login
A366837
Expansion of the unique real solution y(t) to the equation y*(1+y)^2/((1-y)*(1-y^3)) = t/(1+3t) with initial value y(0)=0.
0
0, 1, -6, 41, -314, 2630, -23532, 220869, -2147042, 21429482, -218310836, 2260625674, -23722964740, 251724659212, -2696226815960, 29112495523245, -316541381682162, 3462842349049666, -38086918804472292, 420921653789328030, -4671874553638732300
OFFSET
0,3
COMMENTS
The solution y(t) is used to express the Segre series defined in (1) in the case alpha is a rank 2 K-theory class of K-nontrivial surfaces in the reference below (Theorem 2 in Section 1.5).
LINKS
Alina Marian, Dragos Oprea, and Rahul Pandharipande, Higher rank Segre integrals over the Hilbert scheme of points, arXiv:1712.02382 [math.AG], 2017-2021.
FORMULA
G.f.: series reversion of x*(1 + x)^2/(1 - 4*x - 6*x^2 - 4*x^3 + x^4). - Andrew Howroyd, Oct 26 2023
PROG
(SageMath)
# compute y(t) satisfies y*(1+y)^2/((1-y)*(1-y^3))=t/(1+3t) with y(0)=0
# it is the same as (1+3t)*(y+2y^2+y^3)=t*(1-y-y^3+y^4)
# further, t*y^4-(1+4*t)*y^3-(2+6*t)*y^2-(1+4*t)*y+t=0
# Suppose y(t) = p_d(t)+O(t^{d+1}) where p_d is a polynomial of deg d. The above equation will become
# t*p^4-(1+4*t)*p^3-(2+6*t)*p^2-(1+4*t)*p+t+O(t^{d+1})=0
Y.<t, c> = PolynomialRing(QQ)
#YY.<t> = PolynomialRing(Y)
# initialization of y(t). Y_list are coefficients.
Y_list = [0]
# compute y(t) up to N-th degree.
N = 20
for ind in range(N):
# compute one degree higher
p = sum([Y_list[i]*t^i for i in range(0, len(Y_list))])
# compute one degree higher
itera = c*t^(len(Y_list)) + p
q = t*itera^4-(1+4*t)*itera^3-(2+6*t)*itera^2-(1+4*t)*itera+t
l = q.coefficient(t^len(Y_list)).coefficients()
new_coeff = -l[1]/l[0]
Y_list.append(new_coeff)
(PARI) seq(n)={Vec(serreverse(x*(1 + x)^2/(1 - 4*x - 6*x^2 - 4*x^3 + x^4) + O(x*x^n)), -n-1)} \\ Andrew Howroyd, Oct 26 2023
CROSSREFS
Sequence in context: A077147 A100770 A084397 * A024078 A095177 A199553
KEYWORD
sign
AUTHOR
Zhongyi Shi, Oct 25 2023
STATUS
approved