A sequence is defined $P_n(x)=P_{n-1}(x-1)P_{n-1}(x+1)$for $n\ge1$ and $P_0(x)=x$. Find the largest $k$ such that $P_{2014}(x)$ is divisible by $x^k$.

Question

I started by finding some of the polynomials and got $$P_1(x)=x^2-1$$$$ P_2(x)=x^2(x^2-4)$$$$ P_3(x)=(x^2-1)^3(x^2-9)$$$$P_4(x)=x^6(x^2-4)^4(x^2-16)$$$$P_5(x)=(x^2-1)^{10}(x^2-9)^5(x^2-25)$$ but I’m not able to observe a general pattern for $P_n(x)$. Moreover just taking the writing the powers of $x$ when $n$ is even, I got the series $ 1,2,6,20…$. Please help to proceed further.

Thanks

Answer 1

Hint: We notice that the degree of $P_n(x)$ is $2n$ and $P_n(x)$ have $n+1$ roots. if we denote $$P_n(x):=\prod_{i=0}^n (x-a_i)^{p_i}$$ then $(a_i)_{i}\in\{-n, -n+2,-n+4,...,n-2,n\}$ and $(p_i)_i$ are the coefficients of Pascal's triangle.

It's not difficult to prove this statement by induction.

The power of $x$ in $P_{2n}(x)$ corresponds to the coefficient in the 2n-th row and n-th column of the Pascal's triangle: $$\color{red}{p_{n}=\pmatrix{2n \\n} = \frac{(2n)!}{(n!)^2}}$$