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1/2, 1/2, 5/8, 5/7, 17/22, 13/16,...

I notice the top numbers are all primes but could not find how that helps. At first I thought maybe it is similar to a Fibonnaci type sequence because of the first two fractions, but I couldnt find a way to extend that to the 3rd term and beyond. I tried also writing as linear combination of the previous two and looking for pattern but did not find anything. Do you have any ideas?

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    $\begingroup$ Where do these come from? $\endgroup$
    – gmvh
    Commented Apr 9, 2021 at 13:47
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    $\begingroup$ there is not enough information in this question for a sensible answer. $\endgroup$
    – Carlo Beenakker
    Commented Apr 9, 2021 at 14:26
  • $\begingroup$ @CarloBeenakker I am trying to find the next term of the sequence. $\endgroup$
    – kiwani
    Commented Apr 9, 2021 at 14:48
  • $\begingroup$ the next term could be anything, the problem is not well defined. $\endgroup$
    – Carlo Beenakker
    Commented Apr 9, 2021 at 15:04
  • $\begingroup$ @CarloBeenakker what am I missing? This is a contest problem. There is an answer but I just can’t find it. $\endgroup$
    – kiwani
    Commented Apr 9, 2021 at 15:05

1 Answer 1

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Fiddling with OEIS I get $$ \frac{n^2+1}{n^2+n+2},\qquad n=0,1,2,3,4,5 $$ so I guess the next term is $$ \frac{6^2+1}{6^2+6+2} = \frac{37}{44} $$

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  • $\begingroup$ Oh thank you so much. I went to OEIS and I didn’t see how it would detect fractional sequences. How did you do that? $\endgroup$
    – kiwani
    Commented Apr 9, 2021 at 15:21
  • $\begingroup$ I did the numerators in OEIS, it said: largest odd divisor of $n^2+1$. So I tried numerator $n^2+1$. Then what would the denominator be to match the given sequence? Then I looked that up in OEIS (linked in the question). $\endgroup$
    – Gerald Edgar
    Commented Apr 9, 2021 at 15:23
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    $\begingroup$ The question as stated is unsolvable. Maybe a better version would be: how can I use OEIS to conjecture a formula for this sequence? $\endgroup$
    – Gerald Edgar
    Commented Apr 9, 2021 at 15:26
  • $\begingroup$ Okay thanks Gerald $\endgroup$
    – kiwani
    Commented Apr 9, 2021 at 15:27
  • $\begingroup$ That works but I don’t see how someone would find this with pen and paper. I have read about the method of finite differences. Would this be the correct approach? $\endgroup$
    – kiwani
    Commented Apr 9, 2021 at 15:36

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