Computation of the pfaffian of a particular matrix 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
$$0<\mathrm{pf}(A(n))<\mathrm{pf}(A(n-2))<1,$$ for an even $n.$
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?
Edit: I guess the following should be true for each $n$:
$\mathrm{pf}(A(n))=\alpha^{m(n)_1}-\alpha^{m(n)_2}+\alpha^{m(n)_3}-\alpha^{m(n)_4}+\cdots +\alpha^{m(n)_r},$ for a strictly increasing sequence of numbers  $m(n)_1,m(n)_2,\cdots, m(n)_r,$ where $r$ is odd.
If we can show this, we are through.
 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: A track, not a solution:
Let us first recall the Schur's complement lemma (that one finds in the "Properties" part in this reference):
Lemma: For a block-partitioned square matrix:
$$M=\begin{pmatrix}A&B\\C&D\end{pmatrix}$$
where $A$ and $D$ are themselves square matrices (not necessarily the same size) with $A$ invertible, we have:
$$\det M= \det(A) \det(S) \ \ \text{where} \ \ S:=D-CA^{-1}B$$
($S$ is called the Schur's complement of $A$ with respect to matrix $M$).
*** End of recall ***
The triple inequality you want to establish is equivalent to:
$$0<|\det(A(n))|<|\det(A(n-2))|<1 \tag{1}$$
But, as $A(n-2)$ is the upper-left block of a certain block partition of $A(n)$, the Schur determinant lemma can be used:
$$\det(A(n))=\det(A(n-2)) \cdot \det(S(n)) \tag{2}$$
where Schur's complement $S(n)$ is in fact a $2 \times 2$ matrix.
Therefore, relationships (1) and (2) bring back your issue to prove that $$|\det(S(n))|<1.$$
The remaining (hopefully not easy) part is to be able to find a workable expression for $A(n-2)^{-1}$...
