I was looking at the notes of a calculus' student and one problem their teacher stated was to find a way to algebraically define the list of numbers $\frac{7}{3}, \frac{5}{4}, 1, \frac{8}{9}, ...$ into a sequence. Is there a "logical" answer for this? To me, one could add literally any other number into the list and find a 4-degree polynomial arbitrarily. This could be done with any finite list of 4 numbers however there are cases where 'logic' or 'common sense' is appealed (e.g. finding the next term of the sequence $1, 2, 3, 4, ...$).

I tried to analyze their differences in a spreadsheet and found nothing, as well as the ratio between the consecutive terms. Any ideas?

Edit: Here's an image of the notes (they're from a class in spanish).

enter image description here

It was part of a task where they were asked to find the $5311th$ term, this can only be doable by finding an algebraic expression.

  • $\begingroup$ Just to make sure-you're certain that's the order the numbers were given in? $\endgroup$ – A-Level Student Oct 28 '20 at 23:08
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    $\begingroup$ "To me, one could add literally any other number into the list ..." ... You're correct about that, which is why Find-the-next-term-of-the-sequence questions tend to get panned here. That said ... I notice that the terms use most of the integers from $1$ to $9$, except for $2$ and $6$. Significant? Impossible to say. ... Are you certain the student's notes fully captured the teacher's intent? Perhaps the teacher is anticipating making the same any-number-will-do point to the class. $\endgroup$ – Blue Oct 28 '20 at 23:12
  • $\begingroup$ @A-levelStudent I double checked, that is in fact the order. $\endgroup$ – NotAMathematician Oct 28 '20 at 23:13
  • $\begingroup$ I added a screenshot of the task, just the way it was stated. $\endgroup$ – NotAMathematician Oct 28 '20 at 23:18
  • $\begingroup$ (b) is pretty easy: denominator increases by 2, numerator magnitude is 4 times denominator plus 1, numerator sign alternates. Maybe (a) is something as ridiculously contrived as that. $\endgroup$ – Andy Walls Oct 28 '20 at 23:23

Hint: It might be easier to consider non-simplified fractions as terms of the sequence:


  • $\begingroup$ This is often integral to solving these types of problems. $\endgroup$ – Andrew Chin Oct 28 '20 at 23:22
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    $\begingroup$ This is clearly the 'correct' take, but I think I'm going to leave my answer around to show how frustrating this kind of problem can be absent better guidance. $\endgroup$ – Steven Stadnicki Oct 28 '20 at 23:26
  • $\begingroup$ Yeah now I feel silly. Great answer! $\endgroup$ – NotAMathematician Oct 29 '20 at 1:21

A bit of thought process: I notice that the denominators seem to be jumping around a little bit, which seems unlikely — so my guess would be that several of these are 'artificially' reduced. A bit more pondering gives that $3|15$, $4|16$, and $9|18$, so that $1$ in the sequence is probably masking $\frac{17}{17}$. Writing the terms this way gives $\frac{35}{15}$, $\frac{20}{16}$, $\frac{17}{17}$, and $\frac{16}{18}$. Unfortunately, that sequence of numerators isn't much more compelling; first differences give 15, 3, 1, which fits far too many sequences. My urge would be to say $\frac{16}{19}$ based on a read of the differences as $2^{a_n}-1$, but I'd be hard-pressed to defend that well.


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