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

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.

• Just to make sure-you're certain that's the order the numbers were given in? – A-Level Student Oct 28 '20 at 23:08
• "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. – Blue Oct 28 '20 at 23:12
• @A-levelStudent I double checked, that is in fact the order. – NotAMathematician Oct 28 '20 at 23:13
• I added a screenshot of the task, just the way it was stated. – NotAMathematician Oct 28 '20 at 23:18
• (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. – Andy Walls Oct 28 '20 at 23:23

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