Computation of the pfaffian of a particular matrix This question was originally asked in (https://math.stackexchange.com/questions/4265063/computation-of-the-pfaffian-of-a-particular-matrix). I did not find any satisfactory answer there. So I am posting it here (with a bit modification). I hope this is suitable for mathoverflow.
Let $\{s_i\}_i$ be a sequence of integers such that $s_i>\sum\limits_{j=1}^{i-1}s_j$ with $s_1=1.$ For $\alpha \in (0,1)$ define the $n \times n$ antisymmetric matrix $A(n)$ by induction:
$A(2)=\left(\begin{array}{cc}0 & \alpha^{s_1} \\ -\alpha^{s_1} & 0\end{array}\right);$
$A(n)_{ij} := A(n-1)_{ij},$ for $1<i<j<n,$ and $A(n)_{in}:= \alpha^{s_{p+i}},$ for $i=1,\cdots, n-1,$ where $p=\frac{(n-1)(n-2)}{2}.$
Hence  $A(3)=\left(\begin{array}{ccc}0 & \alpha^{s_1} & \alpha^{s_2}\\ -\alpha^{s_1} & 0 & \alpha^{s_3} \\ -\alpha^{s_2} & -\alpha^{s_3} & 0 \end{array}\right)$,
$A(4)=\left(\begin{array}{cccc}0 & \alpha^{s_1} & \alpha^{s_2} & \alpha^{s_4}\\ -\alpha^{s_1} & 0 & \alpha^{s_3} & \alpha^{s_5} \\ -\alpha^{s_2} & -\alpha^{s_3} & 0 & \alpha^{s_6} \\ -\alpha^{s_4} & -\alpha^{s_5}  & -\alpha^{s_6}& 0 \end{array}\right),$ and so on.
Let us denote by $\mathrm{pf}(A)$, the pfaffian (https://en.wikipedia.org/wiki/Pfaffian) of an antisymmetric matrix $A$. I wish to prove that for each even $n$:
$$\mathrm{pf}(A(n))=\alpha^{m(n)_1}-\alpha^{m(n)_2}+\alpha^{m(n)_3}-\alpha^{m(n)_4}+\cdots +\alpha^{m(n)_r},$$ for a strictly increasing sequence of numbers  $m(n)_1,m(n)_2,\cdots, m(n)_r,$ where $r$ is odd.
I am able to prove the above statement for $n=2,4,6$ by explicit computations of the pfaffians. Can anyone help me proving the statement?
 A: I think I have a solution to the problem.
Let us first recall the following recursive definition of pfaffian from Page 116 of Schubert varieties and degeneracy loci, Fulton-Pragacz. Let $\mathrm{pf}^{i j}(A)$ denote the pfaffian of the skew-symmetric matrix obtained from $A$ by removing the $i^{t h}$ and $j^{t h}$ row and column. Let $n$ be even. Then for a fixed integer $j, 1 \leq j \leq n$, one has
$$
\operatorname{pf}(A)=\sum_{i<j}(-1)^{i+j-1} a_{i j} \operatorname{pf}^{i j}(A)+\sum_{i>j}(-1)^{i+j} a_{i j} \operatorname{pf}^{i j}(A).
$$
In particular, when $j=n,$ we have
\begin{equation}\label{eq:pfaffian_recursive}
\operatorname{pf}(A)=\sum_{i=1}^{n-1}(-1)^{i+1} a_{i n} \operatorname{pf}^{i n}(A). 
\end{equation}
Now for $A(n),$ if we write the above sum starting from $i=n-1$ to $1,$ using induction on the size $n,$ we see directly that the above sum is of the required form.
A: This is not an answer, but a reformulation. Consider all permutations $\sigma$ of the numbers $\{1,2,\dots,2n\}$ such that
$$\sigma(1)>\sigma(2),\,\sigma(3)>\sigma(4),\dots,\sigma(2n-1)>\sigma(2n)$$
and also
$$\sigma(1)>\sigma(3)>\dots>\sigma(2n-1).$$
Note that these can be identified with perfect matchings on $\{1,2,\dots,2n\}$ ( each $\sigma(2i-1)$ is paired with $\sigma(2i)$).
Write these in lexicographic order. For instance, when $2n=6$, there are $15$ such permutations and writing them from largest to smallest gives $654321$, $654231$, $654132$, $645321$, $\dots$, $615243$. Then I think your observation would follow from the statement ``Any two adjacent permutations have distinct parity".
In fact, the pfaffian can be defined as
$$\sum_{\sigma}\operatorname{sgn}(\sigma)a_{\sigma(2i-1),\sigma(2i)},$$
summed over such sigma. In your case, this sum will be a sum of $\pm\alpha^{n_{\sigma}}$ for some integers $n_{\sigma}$, and I think it is clear from your condition on $s_i$ that $n_{\sigma}>n_\tau$ if and only if $\sigma>\tau$. So if the statement above holds the signs alternate.
I suggested above that any two adjacent permutations differ by a transposition, but that is certainly false as can be seen from the example 654132,645321 above. But I did check that the claim holds for $2n=6$.
Updated after seeing SiOn's answer to their own question:
The claim above can be proved by induction on $n$. Sketch of proof: Let $\sigma>\tau$ be two lexicographically adjacent matchings. We always have $\sigma(1)=\tau(1)=2n$. If $\sigma(2)=\tau(2)=j$, then the remaining parts of $\sigma$ and $\tau$ are permutations $\sigma'$, $\tau'$ of $\{1,2,\dots,\hat j,\dots,2n-1\}$. Identifying that set with $\{1,2,\dots,2n-2\}$ it follows inductively that $\operatorname{sgn}(\sigma')=-\operatorname{sgn}(\tau')$ and hence $\operatorname{sgn}(\sigma)=-\operatorname{sgn}(\tau)$.
If $\sigma(2)\neq \tau(2)$ we can write $\sigma(2)=j$ and $\tau(2)=j-1$. Recall that the parity of perfect matchings is the crossing number. That is, if we draw $\{1,2,\dots,n\}$ as dots on a line and then connect $\sigma(2j-1)$ and $\sigma(2j)$ by arcs in the upper half-plane, then $\operatorname{sgn}(\sigma)$ is the parity of the number of crossings. In the situation at hand, $\sigma'$ is the lexicographically minimal matching on $\{1,2,\dots,\hat j,\dots,2n-1\}$, that is, $1$ is paired with $2n-1$, $2$ with $2n-2$ and so on. This matching has no crossings. The arc from $2n$ to $j$ crosses the arcs starting at $1,\dots,j-1$ and hence $\operatorname{sgn}(\sigma)=(-1)^{j+1}$. Likewise, $\tau'$ is the maximal matching on  $\{1,2,\dots,\widehat {j-1},\dots,2n-1\}$, which pairs $1$ with $2$, $3$ with $4$ and so on. Again, $\operatorname{sgn}(\tau')=1$. If $j$ is even, the arc from $2n$ to $j-1$ can be drawn without crossing any other arcs, but if $j$ is odd it must cross the arc from $j-2$ to $j$. Hence, $\operatorname{sgn}(\tau)=(-1)^j$ and we see that also in this case $\operatorname{sgn}(\sigma)=-\operatorname{sgn}(\tau)$.
