Find the infinite product

CLAIM 1 $a_k \; = \; 2^{2^k} + 2^{-2^k}$ for all $k \ge 0$

This is proved by induction on $k$ .

CLAIM 2 $\prod_{k=0}^K a_k \; = \; \tfrac23\big(2^{2^{K+1}} - 2^{-2^{K+1}}\big)$ for all $k \ge 0$ .

This is proved by induction on $K$ .

CLAIM 3 $\prod_{k=0}^K (a_k-1) \; = \; \tfrac27\big(2^{2^{K+1}} + 1 +2^{-2^{K+1}}\big)$ for all $k \ge 0$ .

Note that $a_{k+1}+1 \,=\, a_k^2-1 \,=\, (a_k-1)(a_k+1)$ , and hence that $a_k - 1 \; = \; \frac{a_{k+1} +1}{a_k + 1}$ for all $k \ge 0$ , so that $\prod_{k=0}^K (a_k-1) \; =\; \frac{a_{K+1}+1}{a_0+1} \; = \; \tfrac27\big(2^{2^{K+1}} + 1 + 2^{-2^{K+1}}\big)$ for all $K \ge 0$

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Putting this all together, we see that $\prod_{k=0}^K \left(1 - \frac{1}{a_k}\right) \; = \; \frac{\displaystyle \prod_{k=0}^K (a_k-1)}{\displaystyle\prod_{k=0}^K a_k} \; = \; \frac37 \times \frac{2^{2^{K+1}} + 1 + 2^{-2^{K+1}}}{2^{2^{K+1}} - 2^{-2^{K+1}}}$ and hence, letting $K \to \infty$ , $\prod_{k=0}^\infty \left(1 - \frac{1}{a_k}\right) \; = \; \frac37$ making the answer $3+7 = \boxed{10}$ .