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EDIT: By scaling, we may as well assume $\Delta=1$. I don't know if this leads to a "closed form", but the eigenvalue $\lambda_0$ that is analytic in a neighbourhood of $\beta=0$ has some interesting regularities in its Maclaurin series:

$$\matrix{n=2 & -\beta^{2}+2 \beta^{3}-5 \beta^{4}+14 \beta^{5}-42 \beta^{6}\cr&+132 \beta^{7}-429 \beta^{8}+1430 \beta^{9}-4862 \beta^{10}\crn=3 & -\beta^{3}+2 \beta^{4}-\beta^{5}-6 \beta^{6}+20 \beta^{7}\cr&-22 \beta^{8}-49 \beta^{9}+260 \beta^{10}-441 \beta^{11}\cr&-320 \beta^{12}+3652 \beta^{13}\crn=4 & -\beta^{4}+2 \beta^{5}-\beta^{6}-8 \beta^{8}+26 \beta^{9}\cr&-28 \beta^{10}+10 \beta^{11}-100 \beta^{12}+442 \beta^{13}\cr&-729 \beta^{14}+532 \beta^{15}-1641 \beta^{16}\crn=5 & -\beta^{5}+2 \beta^{6}-\beta^{7}-10 \beta^{10}+32 \beta^{11}\cr&-34 \beta^{12}+12 \beta^{13}-155 \beta^{15}+672 \beta^{16}\cr&-1089 \beta^{17}+782 \beta^{18}-210 \beta^{19}\crn=6 & -\beta^{6}+2 \beta^{7}-\beta^{8}-12 \beta^{12}+38 \beta^{13}\cr&-40 \beta^{14}+14 \beta^{15}-222 \beta^{18}+950 \beta^{19}\cr&-1521 \beta^{20}+1080 \beta^{21}-287 \beta^{22}\crn=7 & -\beta^{7}+2 \beta^{8}-\beta^{9}-14 \beta^{14}+44 \beta^{15}\cr&-46 \beta^{16}+16 \beta^{17}-301 \beta^{21}+1276 \beta^{22}\cr&-2025 \beta^{23}+1426 \beta^{24}-376 \beta^{25}\crn=8 & -\beta^{8}+2 \beta^{9}-\beta^{10}-16 \beta^{16}+50 \beta^{17}\cr&-52 \beta^{18}+18 \beta^{19}-392 \beta^{24}+1650 \beta^{25}\cr&-2601 \beta^{26}+1820 \beta^{27}-477 \beta^{28}\crn=9 & -\beta^{9}+2 \beta^{10}-\beta^{11}-18 \beta^{18}+56 \beta^{19}\cr&-58 \beta^{20}+20 \beta^{21}-495 \beta^{27}+2072 \beta^{28}\cr&-3249 \beta^{29}+2262 \beta^{30}-590 \beta^{31}\crn=10 & -\beta^{10}+2 \beta^{11}-\beta^{12}-20 \beta^{20}+62 \beta^{21}\cr&-64 \beta^{22}+22 \beta^{23}-610 \beta^{30}+2542 \beta^{31}\cr&-3969 \beta^{32}+2752 \beta^{33}-715 \beta^{34}\cr}$$

All coefficients are integers, and it starts with$-\beta^n + 2 \beta^{n+1} - \beta^{n+2} - 2 n \beta^{2n} + (6n+2) \beta^{2n+1} - (6n+4) \beta^{2n+2} + (2n+2) \beta^{2n+3} - (6n+1)n \beta^{3n} \ldots$

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As Denis remarked, the characteristic polynomial $P_n(\lambda)$ of the $n \times n$ matrix does satisfy a two-term three-term linear recurrence $$P_{n+2}(\lambda) = (\lambda + \beta + \Delta) P_{n+1}(\lambda) - \beta \Delta P_n(\lambda)$$ with initial conditions $P_1(\lambda) = \lambda+\beta$, $P_2(\lambda) = \lambda^2+(2 \beta+\Delta)\lambda +\beta^2$, and then $$P_n(\lambda) = \frac{(P_2 - r_2 P_1)}{r_1 (r_1 - r_2)} r_1^n + \frac{(P_2 - r_1 P_1)}{r_2 (r_2 - r_1)} r_2^n$$ where $r_1$ and $r_2$ are the roots of $r^2 - (\lambda + \beta + \Delta) r + \beta \Delta$. However, I don't see how this implies a "closed form" for the eigenvalues.

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As Denis remarked, the characteristic polynomial $P_n(\lambda)$ of the $n \times n$ matrix does satisfy a two-term linear recurrence $$P_{n+2}(\lambda) = (\lambda + \beta + \Delta) P_{n+1}(\lambda) - \beta \Delta P_n(\lambda)$$ with initial conditions $P_1(\lambda) = \lambda+\beta$, $P_2(\lambda) = \lambda^2+(2 \beta+\Delta)\lambda +\beta^2$, and then $$P_n(\lambda) = \frac{(P_2 - r_1 r_2 P_1)}{r_1 (r_1 - r_2)} r_1^n + \frac{(P_2 - r_2 r_1 P_1)}{r_2 (r_2 - r_1)} r_2^n$$ where $r_1$ and $r_2$ are the roots of $r^2 - (\lambda + \beta + \Delta) r + \beta \Delta$. However, I don't see how this implies a "closed form" for the eigenvalues.

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