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In the article Analytic solutions and integrability for bilinear recurrences of order six Andrew Hone considered general Somos-6 recurrence $$\tau_{n+6}\tau_{n}=\alpha\tau_{n+5}\tau_{n+1}+\beta\tau_{n+5}\tau_{n+2}+\gamma\tau_{n+3}^2$$ with arbitrary coefficients $\alpha$, $\beta$, $\gamma$. He gave explicit analytic solution in the form $$\tau_n=AB^n\dfrac{\sigma(\mathbf{v}_0+n\mathbf{v})}{\sigma(\mathbf{v})^{n^2}},$$ where $\mathbf{v}$, $\mathbf{v}_0\in\mathbb{C}^2$ and $\sigma$ is a Kleinian sigma function associated with some genus $2$ curve $\mu^2=4\nu^5+c_3\nu^3+c_2\nu^2+c_1\nu+c_0.$ From Baker's addition formula (see Baker's An introduction to the theory of multiply periodic functions (1907)) $$\dfrac{\sigma(\mathbf{u}+\mathbf{v}) \sigma(\mathbf{u}-\mathbf{v})}{\sigma(\mathbf{u})^2\sigma(\mathbf{v})^2}= \wp_{22}(\mathbf{u})\wp_{12}(\mathbf{v})-\wp_{12}(\mathbf{u})\wp_{22}(\mathbf{v})+ \wp_{11}(\mathbf{v})-\wp_{11}(\mathbf{u})$$ follows that for some fumctions $f_k$, $g_k$ ($1\le k\le 4$) $$\tau_{m+n}\tau_{m-n}=\sum\limits_{k=1}^{4}f_k(m)g_k(n).$$ It means that infinite matrix consisting from $A_{mn}=\tau_{m+n}\tau_{m-n}$ has rank at most $4$ and every minor of order $5$ vanishes.

Let's apply this theory to the given sequence. (From this point I'll follow David Speyer's solution.)

Taking two $5$-tuples of $m$'s and $n$'s $(m,22,21,20,19)$ and $(24,18,12,6,0)$ from $$\left| \begin{array}{ccccc} \tau_{m-24} \tau_{m+24} & \tau_{m-18} \tau_{m+18} & \tau_{m-12} \tau_{m+12} & \tau_{m-6} \tau_{m+6} & \tau_m^2 \\ \tau_{22-24} \tau_{22+24} & \tau_{22-18} \tau_{22+18} & \tau_{22-12} \tau_{22+12} & \tau_{22-6} \tau_{22+6} & \tau_{22}^2 \\ \tau_{21-24} \tau_{21+24} & \tau_{21-18} \tau_{21+18} & \tau_{21-12} \tau_{21+12} & \tau_{21-6} \tau_{21+6} & \tau_{21}^2 \\ \tau_{20-24} \tau_{20+24} & \tau_{20-18} \tau_{20+18} & \tau_{20-12} \tau_{20+12} & \tau_{20-6} \tau_{20+6} & \tau_{20}^2 \\ \tau_{19-24} \tau_{19+24} & \tau_{19-18} \tau_{19+18} & \tau_{19-12} \tau_{19+12} & \tau_{19-6} \tau_{19+6} & \tau_{19}^2 \\ \end{array} \right|=0$$ we get his first recurrence which proves that $\tau_{6n}$, $\tau_{6n+1}$, $\tau_{6n+2}$, $\tau_{6n+3}$ are always odd. Taking two $5$-tuples of $m$'s and $n$'s as $(m,45,44,43,42)$ and $(48,36,24,12,0)$ we get his second formula which proves that $2$-adic valuation has period $24$.

(Thanks to David Speyer who found a gap in first version of this answer.)

Andrew Hone in the articles Analytic solutions and integrability for bilinear recurrences of order six and Sigma-function solution to the general Somos-6 recurrence via hyperelliptic Prym varieties (with Yuri N. Fedorov) considered general Somos-6 recurrence $$\tau_{n+6}\tau_{n}=\alpha\tau_{n+5}\tau_{n+1}+\beta\tau_{n+5}\tau_{n+2}+\gamma\tau_{n+3}^2$$ with arbitrary coefficients $\alpha$, $\beta$, $\gamma$. He gave explicit analytic solution in the form $$\tau_n=AB^n\dfrac{\sigma(\mathbf{v}_0+n\mathbf{v})}{\sigma(\mathbf{v})^{n^2}},$$ where $\mathbf{v}$, $\mathbf{v}_0\in\mathbb{C}^2$ and $\sigma$ is a Kleinian sigma function associated with some genus $2$ curve $\mu^2=4\nu^5+c_3\nu^3+c_2\nu^2+c_1\nu+c_0.$ From Baker's addition formula (see Baker's An introduction to the theory of multiply periodic functions (1907)) $$\dfrac{\sigma(\mathbf{u}+\mathbf{v}) \sigma(\mathbf{u}-\mathbf{v})}{\sigma(\mathbf{u})^2\sigma(\mathbf{v})^2}= \wp_{22}(\mathbf{u})\wp_{12}(\mathbf{v})-\wp_{12}(\mathbf{u})\wp_{22}(\mathbf{v})+ \wp_{11}(\mathbf{v})-\wp_{11}(\mathbf{u})$$ follows that for some fumctions $f_k$, $g_k$ ($1\le k\le 4$) $$\tau_{m+n}\tau_{m-n}=\sum\limits_{k=1}^{4}f_k(m)g_k(n).$$ It means that infinite matrix consisting from $A_{mn}=\tau_{m+n}\tau_{m-n}$ has rank at most $4$ and every minor of order $5$ vanishes.

Let's apply this theory to the given sequence. (From this point I'll follow David Speyer's solution.)

Taking two $5$-tuples of $m$'s and $n$'s $(m,22,21,20,19)$ and $(24,18,12,6,0)$ from $$\left| \begin{array}{ccccc} \tau_{m-24} \tau_{m+24} & \tau_{m-18} \tau_{m+18} & \tau_{m-12} \tau_{m+12} & \tau_{m-6} \tau_{m+6} & \tau_m^2 \\ \tau_{22-24} \tau_{22+24} & \tau_{22-18} \tau_{22+18} & \tau_{22-12} \tau_{22+12} & \tau_{22-6} \tau_{22+6} & \tau_{22}^2 \\ \tau_{21-24} \tau_{21+24} & \tau_{21-18} \tau_{21+18} & \tau_{21-12} \tau_{21+12} & \tau_{21-6} \tau_{21+6} & \tau_{21}^2 \\ \tau_{20-24} \tau_{20+24} & \tau_{20-18} \tau_{20+18} & \tau_{20-12} \tau_{20+12} & \tau_{20-6} \tau_{20+6} & \tau_{20}^2 \\ \tau_{19-24} \tau_{19+24} & \tau_{19-18} \tau_{19+18} & \tau_{19-12} \tau_{19+12} & \tau_{19-6} \tau_{19+6} & \tau_{19}^2 \\ \end{array} \right|=0$$ we get his first recurrence which proves that $\tau_{6n}$, $\tau_{6n+1}$, $\tau_{6n+2}$, $\tau_{6n+3}$ are always odd. Taking two $5$-tuples of $m$'s and $n$'s as $(m,45,44,43,42)$ and $(48,36,24,12,0)$ we get his second formula which proves that $2$-adic valuation has period $24$.

(Thanks to David Speyer who found a gap in first version of this answer.)

(David Speyer found a gap in this answer. So far it is not correct, but probably can be fixed up.) In the articleAnalytic solutions and integrability for bilinear recurrences of order six Andrew Hone considered general Somos-6 recurrence $$\tau_{n+6}\tau_{n}=\alpha\tau_{n+5}\tau_{n+1}+\beta\tau_{n+5}\tau_{n+2}+\gamma\tau_{n+3}^2$$ with arbitrary coefficients $\alpha$, $\beta$, $\gamma$. He gave explicit analytic solution in the form $$\tau_n=AB^n\dfrac{\sigma(\mathbf{v}_0+n\mathbf{v})}{\sigma(\mathbf{v})^{n^2}},$$ where $\mathbf{v}$, $\mathbf{v}_0\in\mathbb{C}^2$ and $\sigma$ is a Kleinian sigma function associated with some genus $2$ curve $\mu^2=4\nu^5+c_3\nu^3+c_2\nu^2+c_1\nu+c_0.$ From Baker's addition formula (see Baker's An introduction to the theory of multiply periodic functions (1907)) $$\dfrac{\sigma(\mathbf{u}+\mathbf{v}) \sigma(\mathbf{u}-\mathbf{v})}{\sigma(\mathbf{u})^2\sigma(\mathbf{v})^2}= \wp_{22}(\mathbf{u})\wp_{12}(\mathbf{v})-\wp_{12}(\mathbf{u})\wp_{22}(\mathbf{v})+ \wp_{11}(\mathbf{v})-\wp_{11}(\mathbf{u})$$ follows that for some fumctions $f_k$, $g_k$ ($1\le k\le 4$) $$\tau_{m+n}\tau_{m-n}=\sum\limits_{k=1}^{4}f_k(m)g_k(n).$$ It means that infinite matrix consisting from $A_{mn}=\tau_{m+n}\tau_{m-n}$ has rank at most $4$ and every minor of order $5$ vanishes.

Let's apply this theory to the given sequence. This minor corresponds to two $5$-tuples of $m$'s and $n$'s $(m,5,4,3,2)$ and $(n,4,2,1,0)$ $$\Delta=\left| \begin{array}{ccccc} \tau_{m-n} \tau_{m+n} & \tau_{m-4} \tau_{m+4} & \tau_{m-2} \tau_{m+2} & \tau_{m-1} \tau_{m+1} & \tau_m^2 \\ \tau_{5-n} \tau_{n+5} & 97 & 11 & 10 & 4 \\ \tau_{4-n} \tau_{n+4} & 25 & 5 & 2 & 4 \\ \tau_{3-n} \tau_{n+3} & 22 & 2 & 2 & 1 \\ \tau_{2-n} \tau_{n+2} & 10 & 2 & 1 & 1 \\ \end{array} \right|=0.$$ We can take $m=n$ if we want to find $\tau_{2n}$ and $m=n+1$ for $\tau_{2n+1}$. But it is necessary to divide by $$\left| \begin{array}{cccc} 97 & 11 & 10 & 4 \\ 25 & 5 & 2 & 4 \\ 22 & 2 & 2 & 1 \\ 10 & 2 & 1 & 1 \\ \end{array} \right|=2\cdot 9.$$ We know from David Speyer's answer that $9$ is not important for us. The only problem is $2$. ($\tau_0=\tau_1=1$ are not a problem too.) But $$\left| \begin{array}{ccccc} \tau_{m-n} \tau_{m+n} & \tau_{m-4} \tau_{m+4} & \tau_{m-2} \tau_{m+2} & \tau_{m-1} \tau_{m+1} & \tau_m^2 \\ \tau_{5-n} \tau_{n+5} & 97 & 11 & 10 & 4 \\ \tau_{4-n} \tau_{n+4} & 25 & 5 & 2 & 4 \\ \tau_{3-n} \tau_{n+3} & 22 & 2 & 2 & 1 \\ \tau_{2-n} \tau_{n+2} & 10 & 2 & 1 & 1 \\ \end{array} \right|\equiv (\tau_{m+2}\tau_{m-2}+\tau_{m+4}\tau_{m-4})(\tau_{4+n}\tau_{4-n}+\tau_{5+n}\tau_{5-n})\pmod{2}$$ (the gap is here) must be $0\pmod{2}$. So we have the expression $$0=\Delta=18\tau_{m+n}\tau_{m-n}+(\tau_{m+2}\tau_{m-2}+\tau_{m+4}\tau_{m-4})(\tau_{4+n}\tau_{4-n}+\tau_{5+n}\tau_{5-n})+2\cdot (\mathrm{Some Polynomial})$$ which can be divided by $2\tau_{m-n}=2$ (because $m=n$ or $m=n+1$) in order to get $\tau_{m+n}$.

In the article  Analytic solutions and integrability for bilinear recurrences of order six Andrew Hone considered general Somos-6 recurrence $$\tau_{n+6}\tau_{n}=\alpha\tau_{n+5}\tau_{n+1}+\beta\tau_{n+5}\tau_{n+2}+\gamma\tau_{n+3}^2$$ with arbitrary coefficients $\alpha$, $\beta$, $\gamma$. He gave explicit analytic solution in the form $$\tau_n=AB^n\dfrac{\sigma(\mathbf{v}_0+n\mathbf{v})}{\sigma(\mathbf{v})^{n^2}},$$ where $\mathbf{v}$, $\mathbf{v}_0\in\mathbb{C}^2$ and $\sigma$ is a Kleinian sigma function associated with some genus $2$ curve $\mu^2=4\nu^5+c_3\nu^3+c_2\nu^2+c_1\nu+c_0.$ From Baker's addition formula (see Baker's An introduction to the theory of multiply periodic functions (1907)) $$\dfrac{\sigma(\mathbf{u}+\mathbf{v}) \sigma(\mathbf{u}-\mathbf{v})}{\sigma(\mathbf{u})^2\sigma(\mathbf{v})^2}= \wp_{22}(\mathbf{u})\wp_{12}(\mathbf{v})-\wp_{12}(\mathbf{u})\wp_{22}(\mathbf{v})+ \wp_{11}(\mathbf{v})-\wp_{11}(\mathbf{u})$$ follows that for some fumctions $f_k$, $g_k$ ($1\le k\le 4$) $$\tau_{m+n}\tau_{m-n}=\sum\limits_{k=1}^{4}f_k(m)g_k(n).$$ It means that infinite matrix consisting from $A_{mn}=\tau_{m+n}\tau_{m-n}$ has rank at most $4$ and every minor of order $5$ vanishes.

Let's apply this theory to the given sequence. (From this point I'll follow David Speyer's solution.)

Taking two $5$-tuples of $m$'s and $n$'s $(m,22,21,20,19)$ and $(24,18,12,6,0)$ from $$\left| \begin{array}{ccccc} \tau_{m-24} \tau_{m+24} & \tau_{m-18} \tau_{m+18} & \tau_{m-12} \tau_{m+12} & \tau_{m-6} \tau_{m+6} & \tau_m^2 \\ \tau_{22-24} \tau_{22+24} & \tau_{22-18} \tau_{22+18} & \tau_{22-12} \tau_{22+12} & \tau_{22-6} \tau_{22+6} & \tau_{22}^2 \\ \tau_{21-24} \tau_{21+24} & \tau_{21-18} \tau_{21+18} & \tau_{21-12} \tau_{21+12} & \tau_{21-6} \tau_{21+6} & \tau_{21}^2 \\ \tau_{20-24} \tau_{20+24} & \tau_{20-18} \tau_{20+18} & \tau_{20-12} \tau_{20+12} & \tau_{20-6} \tau_{20+6} & \tau_{20}^2 \\ \tau_{19-24} \tau_{19+24} & \tau_{19-18} \tau_{19+18} & \tau_{19-12} \tau_{19+12} & \tau_{19-6} \tau_{19+6} & \tau_{19}^2 \\ \end{array} \right|=0$$ we get his first recurrence which proves that $\tau_{6n}$, $\tau_{6n+1}$, $\tau_{6n+2}$, $\tau_{6n+3}$ are always odd. Taking two $5$-tuples of $m$'s and $n$'s as $(m,45,44,43,42)$ and $(48,36,24,12,0)$ we get his second formula which proves that $2$-adic valuation has period $24$.

(Thanks to David Speyer who found a gap in first version of this answer.)

In the articleAnalytic solutions and integrability for bilinear recurrences of order six Andrew Hone considered general Somos-6 recurrence $$\tau_{n+6}\tau_{n}=\alpha\tau_{n+5}\tau_{n+1}+\beta\tau_{n+5}\tau_{n+2}+\gamma\tau_{n+3}^2$$ with arbitrary coefficients $\alpha$, $\beta$, $\gamma$. He gave explicit analytic solution in the form $$\tau_n=AB^n\dfrac{\sigma(\mathbf{v}_0+n\mathbf{v})}{\sigma(\mathbf{v})^{n^2}},$$ where $\mathbf{v}$, $\mathbf{v}_0\in\mathbb{C}^2$ and $\sigma$ is a Kleinian sigma function associated with some genus $2$ curve $\mu^2=4\nu^5+c_3\nu^3+c_2\nu^2+c_1\nu+c_0.$ From Baker's addition formula (see Baker's An introduction to the theory of multiply periodic functions (1907)) $$\dfrac{\sigma(\mathbf{u}+\mathbf{v}) \sigma(\mathbf{u}-\mathbf{v})}{\sigma(\mathbf{u})^2\sigma(\mathbf{v})^2}= \wp_{22}(\mathbf{u})\wp_{12}(\mathbf{v})-\wp_{12}(\mathbf{u})\wp_{22}(\mathbf{v})+ \wp_{11}(\mathbf{v})-\wp_{11}(\mathbf{u})$$ follows that for some fumctions $f_k$, $g_k$ ($1\le k\le 4$) $$\tau_{m+n}\tau_{m-n}=\sum\limits_{k=1}^{4}f_k(m)g_k(n).$$ It means that infinite matrix consisting from $A_{mn}=\tau_{m+n}\tau_{m-n}$ has rank at most $4$ and every minor of order $5$ vanishes.

Let's apply this theory to the given sequence. This minor corresponds to two $5$-tuples of $m$'s and $n$'s $(m,5,4,3,2)$ and $(n,4,2,1,0)$ $$\Delta=\left| \begin{array}{ccccc} \tau_{m-n} \tau_{m+n} & \tau_{m-4} \tau_{m+4} & \tau_{m-2} \tau_{m+2} & \tau_{m-1} \tau_{m+1} & \tau_m^2 \\ \tau_{5-n} \tau_{n+5} & 97 & 11 & 10 & 4 \\ \tau_{4-n} \tau_{n+4} & 25 & 5 & 2 & 4 \\ \tau_{3-n} \tau_{n+3} & 22 & 2 & 2 & 1 \\ \tau_{2-n} \tau_{n+2} & 10 & 2 & 1 & 1 \\ \end{array} \right|=0.$$ We can take $m=n$ if we want to find $\tau_{2n}$ and $m=n+1$ for $\tau_{2n+1}$. But it is necessary to divide by $$\left| \begin{array}{cccc} 97 & 11 & 10 & 4 \\ 25 & 5 & 2 & 4 \\ 22 & 2 & 2 & 1 \\ 10 & 2 & 1 & 1 \\ \end{array} \right|=2\cdot 9.$$ We know from David Speyer's answer that $9$ is not important for us. The only problem is $2$. ($\tau_0=\tau_1=1$ are not a problem too.) But $$\left| \begin{array}{ccccc} \tau_{m-n} \tau_{m+n} & \tau_{m-4} \tau_{m+4} & \tau_{m-2} \tau_{m+2} & \tau_{m-1} \tau_{m+1} & \tau_m^2 \\ \tau_{5-n} \tau_{n+5} & 97 & 11 & 10 & 4 \\ \tau_{4-n} \tau_{n+4} & 25 & 5 & 2 & 4 \\ \tau_{3-n} \tau_{n+3} & 22 & 2 & 2 & 1 \\ \tau_{2-n} \tau_{n+2} & 10 & 2 & 1 & 1 \\ \end{array} \right|\equiv (\tau_{m+2}\tau_{m-2}+\tau_{m+4}\tau_{m-4})(\tau_{4+n}\tau_{4-n}+\tau_{5+n}\tau_{5-n})\pmod{2}$$ must be $0\pmod{2}$. So we have the expression $$0=\Delta=18\tau_{m+n}\tau_{m-n}+(\tau_{m+2}\tau_{m-2}+\tau_{m+4}\tau_{m-4})(\tau_{4+n}\tau_{4-n}+\tau_{5+n}\tau_{5-n})+2\cdot (\mathrm{Some Polynomial})$$ which can be divided by $2\tau_{m-n}=2$ (because $m=n$ or $m=n+1$) in order to get $\tau_{m+n}$.

(David Speyer found a gap in this answer. So far it is not correct, but probably can be fixed up.) In the articleAnalytic solutions and integrability for bilinear recurrences of order six Andrew Hone considered general Somos-6 recurrence $$\tau_{n+6}\tau_{n}=\alpha\tau_{n+5}\tau_{n+1}+\beta\tau_{n+5}\tau_{n+2}+\gamma\tau_{n+3}^2$$ with arbitrary coefficients $\alpha$, $\beta$, $\gamma$. He gave explicit analytic solution in the form $$\tau_n=AB^n\dfrac{\sigma(\mathbf{v}_0+n\mathbf{v})}{\sigma(\mathbf{v})^{n^2}},$$ where $\mathbf{v}$, $\mathbf{v}_0\in\mathbb{C}^2$ and $\sigma$ is a Kleinian sigma function associated with some genus $2$ curve $\mu^2=4\nu^5+c_3\nu^3+c_2\nu^2+c_1\nu+c_0.$ From Baker's addition formula (see Baker's An introduction to the theory of multiply periodic functions (1907)) $$\dfrac{\sigma(\mathbf{u}+\mathbf{v}) \sigma(\mathbf{u}-\mathbf{v})}{\sigma(\mathbf{u})^2\sigma(\mathbf{v})^2}= \wp_{22}(\mathbf{u})\wp_{12}(\mathbf{v})-\wp_{12}(\mathbf{u})\wp_{22}(\mathbf{v})+ \wp_{11}(\mathbf{v})-\wp_{11}(\mathbf{u})$$ follows that for some fumctions $f_k$, $g_k$ ($1\le k\le 4$) $$\tau_{m+n}\tau_{m-n}=\sum\limits_{k=1}^{4}f_k(m)g_k(n).$$ It means that infinite matrix consisting from $A_{mn}=\tau_{m+n}\tau_{m-n}$ has rank at most $4$ and every minor of order $5$ vanishes.

Let's apply this theory to the given sequence. This minor corresponds to two $5$-tuples of $m$'s and $n$'s $(m,5,4,3,2)$ and $(n,4,2,1,0)$ $$\Delta=\left| \begin{array}{ccccc} \tau_{m-n} \tau_{m+n} & \tau_{m-4} \tau_{m+4} & \tau_{m-2} \tau_{m+2} & \tau_{m-1} \tau_{m+1} & \tau_m^2 \\ \tau_{5-n} \tau_{n+5} & 97 & 11 & 10 & 4 \\ \tau_{4-n} \tau_{n+4} & 25 & 5 & 2 & 4 \\ \tau_{3-n} \tau_{n+3} & 22 & 2 & 2 & 1 \\ \tau_{2-n} \tau_{n+2} & 10 & 2 & 1 & 1 \\ \end{array} \right|=0.$$ We can take $m=n$ if we want to find $\tau_{2n}$ and $m=n+1$ for $\tau_{2n+1}$. But it is necessary to divide by $$\left| \begin{array}{cccc} 97 & 11 & 10 & 4 \\ 25 & 5 & 2 & 4 \\ 22 & 2 & 2 & 1 \\ 10 & 2 & 1 & 1 \\ \end{array} \right|=2\cdot 9.$$ We know from David Speyer's answer that $9$ is not important for us. The only problem is $2$. ($\tau_0=\tau_1=1$ are not a problem too.) But $$\left| \begin{array}{ccccc} \tau_{m-n} \tau_{m+n} & \tau_{m-4} \tau_{m+4} & \tau_{m-2} \tau_{m+2} & \tau_{m-1} \tau_{m+1} & \tau_m^2 \\ \tau_{5-n} \tau_{n+5} & 97 & 11 & 10 & 4 \\ \tau_{4-n} \tau_{n+4} & 25 & 5 & 2 & 4 \\ \tau_{3-n} \tau_{n+3} & 22 & 2 & 2 & 1 \\ \tau_{2-n} \tau_{n+2} & 10 & 2 & 1 & 1 \\ \end{array} \right|\equiv (\tau_{m+2}\tau_{m-2}+\tau_{m+4}\tau_{m-4})(\tau_{4+n}\tau_{4-n}+\tau_{5+n}\tau_{5-n})\pmod{2}$$ (the gap is here) must be $0\pmod{2}$. So we have the expression $$0=\Delta=18\tau_{m+n}\tau_{m-n}+(\tau_{m+2}\tau_{m-2}+\tau_{m+4}\tau_{m-4})(\tau_{4+n}\tau_{4-n}+\tau_{5+n}\tau_{5-n})+2\cdot (\mathrm{Some Polynomial})$$ which can be divided by $2\tau_{m-n}=2$ (because $m=n$ or $m=n+1$) in order to get $\tau_{m+n}$.