Let $\{f_n\}_{n=1}^\infty\in \mathbb{C}[x,y]$ be a sequence of polynomials given by the following expressions $$ f_n(x,y)=\sum_{i=0}^{[\dfrac{n}{2}]}(-1)^{n-i}{{n-i}\choose i}x^{n-2i}y^i. $$

Let $(f_n,f_{n+1})$ denote the ideal generated by $f_n,f_{n+1}$.

Let $h(n)$ be the smallest positive integer such that $x^{h(n)}\in (f_n,f_{n+1})$.

I want to prove $h(n)=2n-1$.

The result is tested by computer "sagemath". I do not know how to prove it?

My attempt:

Step~1. By Euclid Algorithm, I have proved $x^{2n-1} \in (f_n,f_{n+1})$. Hence $h(n)\leq 2n-1$.

Step~2. Let $k$ be an arbitrary positive integer such that $x^k\in (f_n,f_{n+1})$. I want to prove $k\geq 2n-1$.

Ideals: $(x^k)\subseteq (f_n,f_{n+1})$.

Varieties: $V(f_n)\cap V(f_{n+1})=V(f_n,f_{n+1})\subseteq V(x^k)$.

Transform into $\mathbb{C}P^2$.

Multiplicities of $0$: $m_{(0,0)}+m_\infty=n(n+1)\leq M_{(0,0)}+M_\infty=k+M_\infty$.

Hence $k\geq m_{(0,0)}+m_\infty-M_\infty$.

Does $m_{(0,0)}+m_\infty-M_\infty=2n-1$? How to compute $m_{(0,0)},m_\infty,M_\infty$?

Let $\{f_n\}_{n=1}^\infty\in \mathbb{C}[x,y]$ be a sequence of polynomials given by the following expressions $$ f_n(x,y)=\sum_{i=0}^{[\dfrac{n}{2}]}(-1)^{n-i}{{n-i}\choose i}x^{n-2i}y^i. $$

Let $(f_n,f_{n+1})$ denote the ideal generated by $f_n,f_{n+1}$.

Let $h(n)$ be the smallest positive integer such that $x^{h(n)}\in (f_n,f_{n+1})$.

I want to prove $h(n)=2n-1$.

The result is tested by computer "sagemath". I do not know how to prove it?

My attempt:

Step~1. By Euclid Algorithm, I have proved $x^{2n-1} \in (f_n,f_{n+1})$. Hence $h(n)\leq 2n-1$.

Step~2. Let $k$ be an arbitrary positive integer such that $x^k\in (f_n,f_{n+1})$. I want to prove $k\geq 2n-1$.