I want to know if there is an explicit solution to the following recurrence relation:
$f(n+1) = \frac{1}{2} (f(n)^2 + 1)$
with the intial condition $f(1) = 3$.
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Ohhh now you edit it that makes much more sense (I'll fix my solution) – Eric Lee May 07 '17 at 04:59
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According to http://oeis.org/A053630, it would seem unlikely.
Marcus Andrews
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As Amarnath Murthy commented in the OEIS page, using the numbers of the sequence, $\log(\log(f_n))$ is very close to $a+bn$ – Claude Leibovici May 07 '17 at 06:04
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@ClaudeLeibovici Amarnath Murthy's comment was
a(1) = 3, a(n+1) = (1/4)[{a(n)-1}^2 + {a(n)+1}^2]while $a(n) = \text{floor}\left[2 \times 1.597910218031873...^{2^n}\right]$ was by Michael Somos – Henry Dec 08 '19 at 17:40