Leading term of difference of two polynomials defined recursively

Question

Define two polynomial sequences by $$ \cases{P_0(x)=x\\ Q_0(x)=1}\\ \cases{P_{n+1}(x)=P_n(x-1)Q_n(x+1)\\ Q_{n+1}(x)=P_n(x+1)Q_n(x-1)} $$ Is it true that the leading term of $ Q_n(x) - P_n(x) $ is $ 2^n \cdot (n-1)! \cdot x^{2^{n-1}-n} $ for all $ n \geq 1 $?

(From this post“the leading term of $g_n$ is $2^{n-1}(n-1)!x^{2^{n-1}-n}$.” I set $x_1=\dots=x_n=1$ to simplify the setting.)

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Verification for $ n = 1 $ to $ 10 $.

var('x')
# P_0 and Q_0
P = [x]
Q = [1]
Compute P_n and Q_n for n=1 to 10
for n in range(10):
    P_next = P[n].subs(x=x-1) * Q[n].subs(x=x+1)
    Q_next = P[n].subs(x=x+1) * Q[n].subs(x=x-1)
    P.append(P_next)
    Q.append(Q_next)
for n in range(1, 11):
    diff = (Q[n] - P[n]).expand()
    LHS = diff.leading_coefficient(x) * x^diff.degree(x)
    RHS = 2^n * factorial(n-1) * x^(2^(n-1) - n)
    print(f"n={n}: {bool(LHS == RHS)}")

Answer 1

By induction, we can write

$$P_n(x) = \prod_{i=0}^{\lfloor n/2\rfloor}(x - n + 4i)^{\binom{n}{2i}}, \quad Q_n(x) = \prod_{i=0}^{\lfloor(n-1)/2\rfloor} (x - n + 4i + 2)^{\binom{n}{2i + 1}}.$$

So for $n ≥ 1$, as $x → ∞$, we have

\begin{align*} Q_n(x) - P_n(x) &= \left(\frac{Q_n(x)}{P_n(x)} - 1\right)P_n(x) \\ &= \left(\exp\left(\ln Q_n(x) - \ln P_n(x)\right) - 1\right)P_n(x) \\ &= \left(\exp\left(-\sum_{i=0}^n (-1)^i \binom ni \ln (x-n+2i)\right) - 1\right)P_n(x) \\ &= \left(\exp\left(-\sum_{i=0}^n (-1)^i \binom ni \ln \left(1 + \left(\frac n2-i\right)⋅-\frac2x\right)\right) - 1\right)P_n(x) \\ &\sim -\ln^{(n)} 1⋅\left(-\frac2x\right)^n x^{2^{n-1}} \\ &= 2^n(n-1)!x^{2^{n-1} - n}. \end{align*}

Here we have used the facts that

$$\sum_{i=0}^n (-1)^i \binom ni \ln x = 0, \\ \sum_{i=0}^n (-1)^i \binom ni f(a + (b - i)h) \sim f^{(n)}(a)h^n \quad \text{as $h → 0$}, \\ \exp u - 1 \sim u \quad \text{as $u → 0$}, \\ P_n(x) → x^{2^{n-1}}. $$