Vanishing of a product of cyclotomic polynomials in characteristic 2 Let $n$ be a positive integer and $1\leq j\leq n$. Consider the following polynomial:
$$p_{n,j}(x)=\frac{\prod\limits_{i=1}^{n+1}\frac{x^{i}+1}{x+1}}{\prod\limits_{i=1}^{j}\frac{x^{i}+1}{x+1}\prod\limits_{i=1}^{n-j+1}\frac{x^{i}+1}{x+1}}\in\mathbb{F}[x]$$
This polynomial can be computed for each $j,n$ given. What I want to do is to understand in which cases $p(1)=0$ and in which cases $p(1)=1$ when working over a field $\mathbb{F}$ of characteristic $2$. I checked some cases by hand and this is what I got:
$$p_{2,1}(1)=p_{2,2}(1)=1$$
$$p_{3,1}(1)=p_{3,2}(1)=p_{3,3}(1)=0$$
$$p_{4,1}(1)=p_{4,4}(1)=1~~p_{4,2}(1)=p_{4,3}(1)=0$$
$$p_{5,2}(1)=p_{5,4}(1)=1~~p_{5,1}(1)=p_{5,3}(1)=p_{5,5}=0$$
$$p_{6,1}(1)=p_{6,2}(1)=p_{6,3}(1)=p_{6,4}(1)=p_{6,5}(1)=p_{6,6}(1)=1$$
So apparently there is no easy pattern.  It is also easily expressible as a product of cyclotomic polynomials, but I'm not sure if this is of any help. I would like (if possible) to have a way to know if $p_{n,j}(1)=0$ or $p_{n,j}(1)=1$ just in terms of $n$ and $j$.
If you are courious, these are the coefficients that has appeared in the computation of the homology groups of a certain family of Artin groups. I want to understand the behaviour of these polynomials over a field of characteristic $2$ in $x=1$ because this is the situation where I can assure that the homology groups are infinite dimensional.
Thanks for your help.

Claim: $p_{n,j}(1)$ vanishes modulo $2$ if and only there is a carry when adding $j$ and $n+1-j$ in base-$2$.
Proof: Let us write $x^n+1 \equiv x^n-1= \prod_{d \mid n} \Phi_d(x)$ where $\Phi_d$ is the $d$th cyclotomic polynomial. Then, since there are $\lfloor m/d\rfloor$ multiplies of $d$ in $\{1,2,\ldots,m\}$,
$$p_{n,j}(x) = \prod_{d=2}^{n} \Phi_d(x)^{\lfloor \frac{n+1}{d} \rfloor- \lfloor \frac{j}{d}\rfloor - \lfloor \frac{n-j+1}{d}\rfloor }.$$
Note that $\lfloor a+b \rfloor \ge \lfloor a\rfloor + \lfloor b \rfloor$ so your expression is necessarily a polynomial (and not a strictly rational function).
The value of $\Phi_d(x)$ at $x=1$ is well-understood. One can use the property $x^n-1= \prod_{d \mid n} \Phi_d(x)$ to show $\Phi_d(1)$ is $1$ unless $d$ is a prime power $p^k$ in which case $\Phi_d(1)=p$ (this is classical, see here for proofs). This implies
$$p_{n,j}(1) \equiv \prod_{p^k \le n} p^{\lfloor \frac{n+1}{p^k} \rfloor- \lfloor \frac{j}{p^k}\rfloor - \lfloor \frac{n-j+1}{p^k}\rfloor }.$$
Since $2$ is the only even prime,
$$p_{n,j}(1) \equiv \prod_{2^k \le n} 2^{\lfloor \frac{n+1}{2^k} \rfloor- \lfloor \frac{j}{2^k}\rfloor - \lfloor \frac{n-j+1}{2^k}\rfloor }.$$
So your expression vanishes modulo $2$ if and only if
$$ \sum_{2 \le 2^k \le n} \left\lfloor \frac{n+1}{2^k} \right\rfloor- \left\lfloor \frac{j}{2^k}\right\rfloor- \left\lfloor \frac{n-j+1}{2^k}\right\rfloor\ge 1.$$
Let us express this criterion using base-$2$ representation. If $j = \sum_{i \ge 0} a_i 2^i$, $n-j+1 = \sum_{i \ge 0} b_i 2^i$ and $n+1 = \sum_{i \ge 0} c_i 2^i$ then
$$ \left\lfloor \frac{n+1}{2^k} \right\rfloor- \left\lfloor \frac{j}{2^k}\right\rfloor- \left\lfloor \frac{n-j+1}{2^k}\right\rfloor =\sum_{i \ge k} (c_i-a_i-b_i) 2^{i-k} $$
so that
$$\begin{align} 
\sum_{2 \le 2^k \le n} \left\lfloor \frac{n+1}{2^k} \right\rfloor- \left\lfloor \frac{j}{2^k}\right\rfloor- \left\lfloor \frac{n-j+1}{2^k}\right\rfloor &= \sum_{i \ge 0} (c_i-a_i-b_i)(2^{i-1}+2^{i-2}+\ldots+1) \\
&= \sum_{i \ge 0} (c_i-a_i-b_i)(2^{i}-1)\\
&= n+1 - (j + (n+1-j)) - \sum_{i \ge 0} (c_i-a_i-b_i)\\
&= \sum_{i \ge 0} (a_i+b_i-c_i).
\end{align}$$
Let $s_2(m)$ be the sum of digits of $m$ in base-$2$. We see that $p_{n,j}(1)\equiv 0$ if and only if $s_2(j) + s_2(n+1-j) > s_2(n+1)$.
In general, $s_2(a)+s_2(b)-s_2(a+b)$ is the number of carries when adding $a$ and $b$ in base-$2$ (can be proved by induction, goes back to Kummer).
