Is there any discernible pattern for the number list $\frac{7}{3}, \frac{5}{4}, 1, \frac{8}{9}, ...$

Question

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).

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.

Answer 1

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

$\frac{7}{3},\frac{10}{8},\frac{13}{13},\frac{16}{18}$...

Answer 2

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.