%I #17 Oct 28 2023 20:35:31

%S 0,1,-6,41,-314,2630,-23532,220869,-2147042,21429482,-218310836,

%T 2260625674,-23722964740,251724659212,-2696226815960,29112495523245,

%U -316541381682162,3462842349049666,-38086918804472292,420921653789328030,-4671874553638732300

%N 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.

%C 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).

%H Alina Marian, Dragos Oprea, and Rahul Pandharipande, <a href="https://arxiv.org/abs/1712.02382">Higher rank Segre integrals over the Hilbert scheme of points</a>, arXiv:1712.02382 [math.AG], 2017-2021.

%F 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

%o (SageMath)

%o # compute y(t) satisfies y*(1+y)^2/((1-y)*(1-y^3))=t/(1+3t) with y(0)=0

%o # it is the same as (1+3t)*(y+2y^2+y^3)=t*(1-y-y^3+y^4)

%o # further, t*y^4-(1+4*t)*y^3-(2+6*t)*y^2-(1+4*t)*y+t=0

%o # Suppose y(t) = p_d(t)+O(t^{d+1}) where p_d is a polynomial of deg d. The above equation will become

%o # t*p^4-(1+4*t)*p^3-(2+6*t)*p^2-(1+4*t)*p+t+O(t^{d+1})=0

%o Y.<t,c> = PolynomialRing(QQ)

%o #YY.<t> = PolynomialRing(Y)

%o # initialization of y(t). Y_list are coefficients.

%o Y_list = [0]

%o # compute y(t) up to N-th degree.

%o N = 20

%o for ind in range(N):

%o # compute one degree higher

%o p = sum([Y_list[i]*t^i for i in range(0,len(Y_list))])

%o # compute one degree higher

%o itera = c*t^(len(Y_list)) + p

%o q = t*itera^4-(1+4*t)*itera^3-(2+6*t)*itera^2-(1+4*t)*itera+t

%o l = q.coefficient(t^len(Y_list)).coefficients()

%o new_coeff = -l[1]/l[0]

%o Y_list.append(new_coeff)

%o (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

%K sign

%O 0,3

%A _Zhongyi Shi_, Oct 25 2023