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Any finite sequences can be expressed as a polynomial, but there are many infinite sequences for which we have found no closed form. Is it possible that no closed form exists? Are there sequences in which we've been able to PROVE no closed form exists?

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    Old question, I know, but Komolgorov complexity suggests that there must be, but we probably can't produce a constructive proof.2017-01-27

2 Answers 2

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There are infinite sequences without closed forms, for any reasonable definition of "closed form". There are only countably many closed form expressions and uncountably many sequences.

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    Yep, I missed that one it's not $\alef_0^c=2^c$, it is $c^{\alef_0}=c$.But then the above argument doesn't hold. And your answer is the closed form for any sequence, of course after taking $[\arctan(x)+\frac{\pi}{2}]/\pi$2011-10-18
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As has been noted, no answer can be given until you define "closed form". You might, however, be interested in the sequence of busy beaver numbers, which is known to grow faster than any computable function (and thus can't be given by any computable function). I'm sure some people would happily accept that as an example for which there is, provably, no closed form.

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    I am one of the people you talk about, so I happily accepted this answer.2011-10-19