Rank of a combinatorial matrix For any $2n$ with $n\in\mathbb{N}$, we conjecture the matrix $A\in\mathcal{M}_{3n\times 2n}$ to have rank $\lceil{\frac{5n-1}{3}}\rceil$ where $$A=\left(\begin{array}{cccccc}
0 &  &  & & &   \\
1 & 2 &  & \\
1 & 2 & 0 &  &   \\
\vdots & \vdots  & & \;\ddots   \\
1 & \binom{2n-2}{1} & \cdots & \binom{2n-2}{2n-3} & 0 & \\
1 & \binom{2n-1}{1} & \cdots & \binom{2n-1}{2n-3} & \binom{2n-1}{2n-2} & 2\\
\hline\\
1 & & & & & -1\\
 & \ddots & & & \cdot^{\large{\cdot^{\Large{\cdot}}}} &\\ 
 & & 1 & -1 & &
\end{array}
\right).$$
That is, the kernel of $A$ is the solution to the following set of equations. For $k=1,\dots,2n$, we set, for $k$ odd, $$x_{1}+\binom{k-1}{1}x_2+\dots+\binom{k-1}{k-2}x_{k-1}=0$$ and, for $k$ even, $$x_{1}+\binom{k-1}{1}x_2+\dots+\binom{k-1}{k-2}x_{k-1}+2x_k=0.$$
Additionally, for $i=1,\dots,n$, we take $$x_i-x_{2n+1-i}=0.$$ 
Computationally (with SageMath), we have found this to be true for up to at least $n=150$. We have obtained these equations while studying tensor valuations on lattice polygons.
 A: If $(c_0,\dots,c_{2n-1})$ is a row of $A$, we look at a polynomial $c_0-c_1s+\dots-c_{2n-1}s^{2n-1}$. We have to find the dimension of the subspace $\Sigma$ formed by such polynomials in the space $\Pi$ of all polynomials of degree at most $2n-1$. The polynomials are $(1-s)^{k-1}-s^{k-1}$ for $k=1,\dots,2n$ and $\pm(s^{i-1}+s^{2n-i})$, $i=1,\dots,n$. Denote $$A={\text{span}}\,\left((1-s)^{m}-s^{m},0\leqslant m\leqslant 2n-1\right),\\B={\text{span}}\,\left(s^{i-1}+s^{2n-i},1\leqslant i\leqslant n\right).$$
We have to find $\dim(A+ B)=\dim A+\dim B-\dim A\cap B$. Note that $A$ is the subspace of polynomials $f(s)\in \Pi$ satisfying $f(s)=-f(1-s):=\alpha(f)$ (this is clear if we denote $s=1/2+x$: then $A$ consists of all polynomials in $\Pi$ which are odd in the variable $x$.) Thus $\dim A=n$. Obviously $\dim B=n$, and $B$ is the space of all polynomials in $\Pi$ satisfying $f(s)=s^{2n-1}f(1/s):=\beta(f)$. So, $A\cap B$ is a set of polynomials in $\Pi$ which are invariant under both involutions $\alpha(f),\beta(f)$. It is straightforward (but crucial) that $(\alpha \beta)^3=\rm id$, so the group generated by $\alpha$, $\beta$ consists of 6 elements $\rm{id},\alpha,\beta,\alpha\beta,\alpha\beta\alpha,\beta\alpha$. 
So, we have to prove that the space $C=A\cap B$ of the polynomials in $\Pi$ invariant under the action of this group $G\cong S_3$ has a dimension $2n-\lceil(5n-1)/3\rceil=\lfloor(n+1)/3\rfloor$. In other words, the multiplicity of the trivial 1-dimensional representation in our representation $\lambda$ of $S_3$ must be equal to $\lfloor(n+1)/3\rfloor$. This multiplicity is a scalar product of the characters of $\lambda$ and the trivial representation, that is, the sum of traces of all the six above linear operators divided by 6. The trace of $\rm id$ equals $2n$, the traces of involutions $\alpha,\beta,\alpha\beta\alpha$ are equal to $0$ (for $\beta$ or $\alpha$ this is clear in the standard basis, and $\alpha\beta\alpha$ is conjugate to $\beta$), and it remains to prove that the trace $t$ of $\alpha\beta$ (and of $\beta\alpha$ too) equals $3\lfloor(n+1)/3\rfloor-n$. There should be some clever way, but in any case we may do it in the standard basis $\{1,s,\dots,s^{2n-1}\}$. Indeed, $$(\beta\alpha)s^i=-s^{2n-1-i}(s-1)^i,$$
so we should sum up the coefficients of $s^i$ in $-s^{2n-1-i}(s-1)^i$, or, the same, sum up the constant terms of $-s^{2n-1-2i}(s-1)^i$, or, the same, to take the constant term of $$-s^{2n-1}\sum_{i=0}^{2n-1}\left(\frac{s-1}{s^2}\right)^i=-s^{1-2n}\frac{s^{4n}-(s-1)^{2n}}{s^2-s+1},$$
or, the same, to take the coefficient of $s^{2n-1}$ in $\frac{(s-1)^{2n}}{1-s+s^2}$. For doing this we note that $(s-1)^{2n}-(-s)^n$ is divisible by $1-s+s^2$, and the ratio has coefficient 0 of $s^{2n-1}$. While the coefficient of $s^n/(1-s+s^2)=(s^n+s^{n+1})/(1+s^3)=(s^n+s^{n+1})(1-s^3+s^6-\dots)$ is easy to find. Checking separately remainders of $n$ modulo 3 we verify that it is indeed $(-1)^n(3\lfloor(n+1)/3\rfloor-n)$.
