Degree recursion conjecture for squared Schubert varieties

Let $1\leq i<j\leq n$ with $j<n$, and let $\mathcal{S}_{ij}^2$ be the square image of the Schubert variety $\mathcal{S}_{ij}$. Degree recursion conjecture. The degrees satisfy

$$\deg(\mathcal{S}_{i,n-1}^2)=4(2^{n-i-1}-1),$$

and

$$\deg(\mathcal{S}_{ij}^2)=\deg(\mathcal{S}_{i+1,j}^2)+\deg(\mathcal{S}_{i,j+1}^2),\qquad \deg(\mathcal{S}_{jj}^2)=0.$$

Consequently, the proposed closed form is

$$\deg(\mathcal{S}_{ij}^2)=\sum_{k=n-j+1}^{n-i-1}\binom{2n-i-j}{k}+2\binom{2n-i-j-1}{n-j}.$$

The formula is inferred from computed values and is not resolved in the supplied text.