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Does the recurrence relation

$$ a(n+1) = a(n)^2 + 1,\quad a(1)=1, $$

have a closed form solution?

I have tried hard to find it, but failed. Any ideas ?

I am particular interested in prime factors of a(n), the least n for which I did not find a prime factor, is 89. With trial division, I checked up to about 10^9 and did not find a prime factor. Is there a better method to find prime factors for such large numbers than the time-consuming trial division ?

Peter
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    It's usually a good idea to check the OEIS for sequences like this, it doesn't appear there is a closed form for this recurrence: http://oeis.org/search?q=1%2C2%2C5%2C26%2C677%2C458330&language=english&go=Search – Thomas Russell Dec 26 '13 at 20:09
  • similar: http://math.stackexchange.com/questions/580069/recurrence-a-n-sum-k-1n-1a2-k-a-1-1 – Alex Dec 26 '13 at 20:30

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