Let me outline an approach for computing permanents in these conjectures. For the sake of concreteness, I consider Conjecture 1 for an odd $n$. The matrix here is the sum of the following two 0-1 matrices (using [Iverson bracket](https://en.wikipedia.org/wiki/Iverson_bracket) notation):
$$A:=\big([2j-k \geq 1]\big)_{j,k=1}^n$$
and
$$B:=\big([2j-k \geq n+1]\big)_{j,k=1}^n$$
(notice that I intentionally redefine matrices $A$ and $B$). For example, for $n=5$, we have
$$A=\begin{bmatrix} 1&0&0&0&0\\ 1&1&1&0&0
\\ 1&1&1&1&1\\ 1&1&1&1&1
\\ 1&1&1&1&1\end{bmatrix}
\quad\text{and}\quad
B=\begin{bmatrix} 0&0&0&0&0\\ 0&0&0&0&0
\\ 0&0&0&0&0\\ 1&1&0&0&0
\\ 1&1&1&1&0\end{bmatrix}
$$
Our goal is to compute $\mathrm{per}(A+B)$.

The crucial observation is that 0-1 matrices can be viewed as boards on which permanent enumerates [non-attacking rook placements](https://en.wikipedia.org/wiki/Rook_polynomial). Furthermore, our matrices have the shape of [Ferrers boards](https://en.wikipedia.org/wiki/Partition_(number_theory)#Ferrers_diagram), and the one for $B$ is a sub-board for that of $A$. From now on, I will not distinguish matrices $A$ and $B$ from the corresponding Ferrers boards. 

I will use the notation and machinery from my [other answer](https://mathoverflow.net/q/386804), which computes the number of non-attacking rook placements (i.e., the permanent) for the *difference* of a Ferrers board with its sub-board. In the current problem, we need to compute the number of placements of $n$ non-attacking rooks in $A$, where each placement comes with multiplicity $2^t$, where $t$ in the number of rooks in $B\subset A$.

Board $A$ has row lengths 
$$a:=(1,3,5,\dots,n-2,\underbrace{n,n,\dots,n}_{(n+1)/2}),$$
while board $B$ has row lengths
$$b:=(\underbrace{0,0,\dots,0}_{(n+1)/2},2,4,\dots,n-1).$$

By inclusion-exclusion here, we have
$$\mathrm{per}(A+B) = \sum_{T\subseteq[n]} r_n(A_{\bar T}\| B_T),$$
where $\bar T := [n] \setminus T$ is the complement of $T$. The analog of formula $(\star)$ here gives the following expression:
$$\mathrm{per}(A+B) = \sum_{p\in\{0,1\}^n} \prod_{i=1}^n \big(p_i(a_i-\sum_{j=1}^{\tau_A(i)-1} \delta_j) + q_i(b_i-\sum_{j=1}^{\tau_B(i)-1} \delta_j)\big),$$
where $q_i:=1-p_i$ and
$$\sigma:=\big(\underbrace{0,0,\dots,0}_{(n+1)/2},1,2,\dots,n-1,\underbrace{n,n,\dots,n}_{(n+1)/2}\big),$$
$$\delta:=\big(q_1,q_2,\dots,q_{\frac{n+1}2},p_1,q_{\frac{n+1}2+1},p_2,q_{\frac{n+1}2+2},\dots,p_{\frac{n-1}2},q_n,p_{\frac{n+1}2},p_{\frac{n+1}2+1},\dots,p_n\big),$$
$$\tau_A:=\big( \frac{n+3}2,\frac{n+7}2, \dots, \frac{3n+1}2, \frac{3n+3}2,\frac{3n+5}2,\dots,2n\big),$$
$$\tau_B:=\big(1,2,\dots,\frac{n+1}2,\frac{n+1}2+2,\frac{n+1}2+4,\dots,\frac{3n-1}2\big).$$

Correspondingly, we have
$$\sum_{j=1}^{\tau_A(i)-1} \delta_j =
\begin{cases} 
i-1 + \sum_{j=i}^{\frac{n-1}2+i}q_j, & \text{if}\ i\leq\frac{n-1}2;\\
n - \sum_{j=i}^{n}p_j, & \text{if}\ i\geq\frac{n+1}2.
\end{cases}$$
and
$$\sum_{j=1}^{\tau_B(i)-1} \delta_j =
\begin{cases} 
\sum_{j=1}^{i-1} q_j & \text{if}\ i\leq\frac{n-1}2;\\
i-1 - \sum_{j=i-\frac{n-1}2}^{i-1}p_j, & \text{if}\ i\geq\frac{n+1}2.
\end{cases}$$


The formula then becomes
$$\mathrm{per}(A+B) = \sum_{p\in\{0,1\}^n} \prod_{i=1}^{(n-1)/2} \big(p_i(i - \sum_{j=i}^{\frac{n-1}2+i}q_j) - q_i \sum_{j=1}^{i-1} q_j)\big)
\prod_{i=(n+1)/2}^n \big(p_i\sum_{j=i}^{n}p_j + q_i(i-n + \sum_{j=i-\frac{n-1}2}^{i-1}p_j)\big).$$

This can surely be simplified further and linked to sequence $a(n)$ - will do that when I have time.