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Define a sequence $(a_n)_{n \geq 1}$ by $$na_n = 2 + \sum_{i = 1}^{n - 1} a_i^2.$$

(In particular, $a_1 = 2$.)

How can you show - preferably without using a pc! - that not all terms of the sequence are integral?

And which will be the first such term?

Motivation: nothing interesting to say, it's a random problem which I got from someone - I have no reference - and which interested me. Usually one has to prove that all terms are integral :)

Thoughts: nothing interesting. The terms are quickly getting enormous...

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    $\begingroup$ Another description: $a_1=2$ and for $n>1$, $a_n=a_{n-1}+\frac{a_{n-1}(a_{n-1}-1)}{n}$ $\endgroup$ – Jonas Meyer Feb 1 '10 at 0:11
  • $\begingroup$ Why a downvote? $\endgroup$ – Wanderer Feb 3 '10 at 15:37
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Sequences like this are sometimes called Somos sequences (and sometimes Gobel sequences) and you can find information about them at Problem E15 in Guy, Unsolved Problems in Number Theory and in the references Guy gives; also I'm sure typing Somos or Gobel into your favorite search engine will turn up something.

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    $\begingroup$ I deleted my answer (it is $k=43$, which I found by fairly routine computation) because of this excellent reference. $\endgroup$ – moonface Feb 1 '10 at 0:25
  • $\begingroup$ Yes, excellent. $\endgroup$ – Wanderer Feb 1 '10 at 0:26
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    $\begingroup$ Another reference is www-groups.dcs.st-and.ac.uk/~john/Zagier/Solution5.3.html. $\endgroup$ – Richard Stanley Feb 1 '10 at 1:48
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    $\begingroup$ Although the definition of a Somos sequence is not settled, I would deny that this is a Somos sequence. It is certainly not one of the sequences Michael Somos studied; the recursion does not involve dividing by earlier terms of the series (only by n); and there is no symmetry of the form a_n --> c^n a_n taking one solution to another. I think of those as the essential features of the Somos sequences. $\endgroup$ – David E Speyer Feb 1 '10 at 13:09
  • $\begingroup$ The Gobel reference, on the other hand, is right on target. $\endgroup$ – David E Speyer Feb 1 '10 at 13:10
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My 2 cents (ha ha): perhaps the approach is to show that the power of 2 dividing $2+\sum_{i=1}^{n-1}a_i^2$ is eventually less than the power of 2 in $n$, and that this somehow involves looking at $a_{2^n}$. I'll keep playing with it for a bit.

(inspired by this kind of argument that the harmonic numbers are not integers)

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