Divisor-count conjecture for classes of upper k-gap balancing numbers

Let $k\geq1$ and put $N=2k^2-1$. By the seed characterization following Proposition 4.1, the number of classes equals the number of pairs $(x,y)$ satisfying

$$0\leq x<k,\qquad y>0,\qquad y^2=8x^2+8(1-k)x+(2k-1)^2.$$

With $z=2x+1-k$, this becomes

$$y^2-2z^2=N. \tag{1}$$

Conversely, every integer solution of (1) with $y>0$ and $|z|<k$ yields a unique admissible seed. Reduction modulo $8$ shows that $z$ is odd when $k$ is even and even when $k$ is odd. Hence $z\equiv1-k\pmod2$, so

$$x=\frac{z+k-1}{2}\in\mathbb Z,$$

and $|z|<k$ gives $0\leq x<k$.

Consider the Euclidean domain $R=\mathbb Z[\sqrt2]$, with norm

$$\operatorname{Nm}(u+v\sqrt2)=u^2-2v^2.$$

Every prime $p\mid N$ is odd and satisfies $2k^2\equiv1\pmod p$. Therefore $2$ is a quadratic residue modulo $p$, and $p$ splits in $R$ as

$$p=\pi_p\overline{\pi_p}.$$

Because $R$ is a principal ideal domain and $1+\sqrt2$ has norm $-1$, the factors can be chosen with

$$\operatorname{Nm}(\pi_p)=\operatorname{Nm}(\overline{\pi_p})=p.$$

Writing $N=\prod_p p^{e_p}$, unique factorization shows that every element of norm $N$ has the form

$$\alpha=u\prod_{p\mid N}\pi_p^{j_p} \overline{\pi_p}^{\,e_p-j_p}, \qquad 0\leq j_p\leq e_p,$$

where $u$ has norm $1$. Distinct exponent vectors give distinct unit orbits. The norm-one units are

$$\pm(3+2\sqrt2)^m,\qquad m\in\mathbb Z.$$

Restricting to positive elements removes the sign. Thus the positive elements of norm $N$ split into exactly

$$\prod_{p\mid N}(e_p+1)=\tau(N)$$

orbits under multiplication by $\varepsilon=3+2\sqrt2$.

Each orbit contains exactly one admissible seed. Indeed, since $\varepsilon=(1+\sqrt2)^2$, there is a unique integer $m$ such that

$$\frac{\sqrt N}{1+\sqrt2} < \alpha\varepsilon^m < \sqrt N(1+\sqrt2). \tag{2}$$

Neither endpoint can occur, since equality would imply $z^2=N/2$, impossible for odd $N$. Writing $\alpha\varepsilon^m=y+z\sqrt2$, positivity of this element and its conjugate $N/(\alpha\varepsilon^m)$ gives $y>0$, and

$$z=\frac{\alpha\varepsilon^m-N/(\alpha\varepsilon^m)}{2\sqrt2}.$$

Therefore (2) is equivalent to

$$|z|<\sqrt{N/2}=\sqrt{k^2-\tfrac12},$$

which for integral $z$ is exactly $|z|<k$. Hence each positive norm-$N$ orbit produces exactly one class seed.

The number of classes is consequently

$$\boxed{\tau(2k^2-1)}.$$

Finally, $k=0$ is the ordinary cobalancing case with one class, agreeing with the single positive divisor of $-1$. Thus the formula holds for every $k\geq0$.