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I have the recurrence relation $a_{n}=2a_{n-1}^{2}-1$ with the initial term $a_{0}=3$. The first few terms in this sequence are $3,17,577,665857,\dots$

Does anyone know how to find an explicit formula for this relation?

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closed as off-topic by heropup, abiessu, José Carlos Santos, John B, rtybase Jan 5 '18 at 22:56

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    $\begingroup$ If $a=\cosh t$ then $2a^2-1=\cosh2t$ etc. $\endgroup$ – Lord Shark the Unknown Jan 5 '18 at 15:46
  • $\begingroup$ One thing I see is that $3=2+1,17=2^4+1$, and we have $$2(2^k+1)^2-1=2^{2k+1}+2^{k+2}+1$$ but this doesn't immediately lend itself to a nice formula... $\endgroup$ – abiessu Jan 5 '18 at 15:48
  • $\begingroup$ Lord Shark of the Unknown's comment above is the key to solving this. Set $a_n = \cosh t_n$. $\endgroup$ – John Barber Jan 5 '18 at 16:04
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$a_n = \frac{1}{2}((1 + \sqrt2)^{2^n} + (1 - \sqrt2)^{2^n})$

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    $\begingroup$ The question asks: "Does anyone know how to find an explicit formula for this relation?" $\endgroup$ – John Barber Jan 5 '18 at 15:58
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    $\begingroup$ you can find an explicit formula by looking the sequence up in the OEIS $\endgroup$ – Dan Jan 5 '18 at 16:02
  • $\begingroup$ This is correct, but I really am interested how you got this. $\endgroup$ – Krypto14 Jan 5 '18 at 16:03

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