Let $v_{i,j}$ be the value of $\nu_{2,j} - \nu_{1,j}$ after round $i$, with initial values $v_{0,j} = j-1$ for all $j\geq 1$. For $i\geq 1$, we have $v_{i,j} = v_{i-1,j}$ for $j\leq i$, while for $j\geq i+1$, 
$$v_{i,j} = v_{i-1,j} - v_{i,j-1}.$$
Correspondingly, the generating function
$$F(x,y) := \sum_{i\geq 0} x^i \sum_{j\geq i+1} y^{j-i-1} v_{i,j}$$
satisfies
$$(x-y-y^2)F(x,y) = x(1+y)F(x,0) - \frac{y^2(1+y)}{(1-y)^2}.$$
Plugging $y=\frac{-1+\sqrt{1+4x}}{2}$ (zero of $x-y-y^2$) in, we get
$$F(x,0) = \frac1{x} \bigg(\frac{-1+\sqrt{1+4x}}{-3+\sqrt{1+4x}}\bigg)^2,$$
and correspondingly
$$F(-x,0) = -\frac1{x} \bigg(\frac{1-\sqrt{1-4x}}{3-\sqrt{1-4x}}\bigg)^2,$$
matching the negated o.g.f. for `A065601`. Hence, $$v_{i,i} = v_{i-1,i} = (-1)^{i-1} [x^{i-1}]\ F(-x,0) = (-1)^i a(i-1).$$ QED

---

PS. The same result can also be obtained from [this earlier question](https://mathoverflow.net/q/477098) by taking $f(i)=-1$ for all $i$.