The square Frobenius number conjecture for consecutive generators differing by two

Let $a\geqslant 3$ be an odd integer, and let $\langle a,a+2\rangle$ denote the numerical semigroup generated by $a$ and $a+2$. Let $(u_n)_{n\geqslant 1}$ be the recursive sequence given in the paper's defining recurrence, and let ${}^{2\!}r(\langle a,a+2\rangle)$ denote the least square integer that does not belong to this semigroup. Square Frobenius number conjecture. If $a=(2b+1)^2$ for an integer $b\geqslant 1$, then

$${}^{2\!}r\!\left(\left\langle a,a+2 \right\rangle\right) = \left\{ \begin{array}{cl} \left(a-2\left\lfloor\frac{(2b+1)\sqrt{2}}{2}\right\rfloor\right)^2 & \text{if } (2b+1)\not\in\displaystyle\bigcup_{n\geqslant1}\left\{u_{4n+1}\right\}, \\ \\ \left(a-\left\lfloor (2b+1)\sqrt{3} \right\rfloor\right)^2 & \text{if } (2b+1)\in\displaystyle\bigcup_{n\geqslant2}\left\{u_{4n+1}\right\}, \\ \\ 38^2 & \text{if } 2b+1=u_5=7. \end{array} \right.$$

If $a+2=(2b+1)^2$ for an integer $b\geqslant 1$, then

$${}^{2\!}r\!\left(\left\langle a,a+2 \right\rangle\right) = \left\{ \begin{array}{ll} \left(a-2\left\lfloor\frac{(2b+1)\sqrt{2}}{2}\right\rfloor\right)^2 & \text{if } (2b+1)\not\in\displaystyle\bigcup_{n\geqslant0}\left\{u_{4n+3}\right\}, \\ \\ \left(a-\left\lfloor (2b+1)\sqrt{3} \right\rfloor\right)^2 & \text{if } (2b+1)\in\displaystyle\bigcup_{n\geqslant0}\left\{u_{4n+3}\right\}. \end{array} \right.$$

The preceding proposition establishes the corresponding formula when neither generator is a square, while the displayed cases concern the remaining square-generator situations. The formulas of the preceding conjecture were verified computationally for all integers $a\geqslant 2$ up to $10^6$, but the conjecture itself is not established in the supplied text.