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We have sequences $a_n$ such that:

  • $a_1=1, a_2=2,$
  • $a_{n+k}=a_n$ - for some unknown $k \in \mathbb{N}$. $n=1,2,3,...$

And for the sequences $b_n=a_{n+2}-a_{n+1}+a_n$ we know that:

$$ b_{n+1}=\frac{1}{2}(b_n^2+1). $$

Find $a_n$ - ?

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Do you mean the limit? If so, note that both ${a_n}$ and ${b_n}$ approach the same limit, say $L$, because ${a_{n + 2}}$ and ${a_{n + 1}}$ also approach $L$. –  glebovg Nov 14 '12 at 22:24
    
I mean general formula for $a_n$. $a_3 -?, a_4 -?$ and so on... –  Gordon Nov 14 '12 at 22:30
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Is this homework, in which case you should add the homework tag? And what have you tried? –  Mark Bennet Nov 14 '12 at 22:30
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1 Answer

up vote 2 down vote accepted

One answer is for $a_n$ to be the sequence $1,2,2,1,0,0$ repeating.

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Thank you. It's very helpfull. –  Gordon Nov 14 '12 at 22:35
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@Gordon Aren't you curious how such an answer might be found, or whether there are other answers? –  Mark Bennet Nov 15 '12 at 7:39
    
No, thanks. :) I know all answers and the way how to found it. –  Gordon Nov 15 '12 at 7:58
    
Gordon, if you know all the answers, and how to find them, can I encourage you to post your answer to the question, as a service to those who will come after us? –  Gerry Myerson Nov 15 '12 at 10:19
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Ok. Here my thoughts. If $b_n > b_{n+1}$, then $b_n>(1+b_n^2)/2$, and $(b(n)-1)^2 <0$. So, we have that $b_n$ is periodic, nondecreasing, therefore $b_n$ is a sequence of constants. It's possible only if all $b_n =1$. So we have that $b_1=1$. And the only answer is $a_n= (1,2,2,1,0,0,...)$ repeating. –  Gordon Nov 15 '12 at 15:49
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