Do all sequences of numbers created by taking a number, performing an operation on it, then repeating this process $n$ times, have a corresponding formula for finding the $n^{\operatorname{th}}$ term without doing all the math in between? Like asking a question and skipping to the answer?
2026-04-13 04:35:35.1776054935
Do all number series have an $n^{\operatorname{th}}$ term function?
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1
Well, that's quite an abstract question, but if you are asking if having a starting point $x_1=a$, say and having some relationship between the elements of the sequence $(x_n)$ will always give a formula for $x_n$ then the answer is NO in the general case. Here is the following problem:
Let $x_1=a$. Define $x_{n+1}=\frac{x_n}{2}$ if $n$ is even and $x_{n+1}=3x_n+1$ otherwise. Then if the answer to your question was YES then you could say whether the sequence could reach $1$ for any starting point $a$. The thing is, this problem is called the Collatz conjecture and it's open, i.e. no one can solve it fully. So I guess that means NO.
Another problem you might encounter is if you take repeated integrals, i.e. say you start with $x_1=1$ and take $x_{n+1}=\int_0^\infty y^{x_n-1}e^{-y}dy$ that is the Gamma function ($\Gamma(x_n)$). Since it doesn't have a closed form (different from the notation) then you want be able to have a decent expression.
However, if you have a linear relationship between finitely many of the entries in the sequence, then yes, you can use auxiliary equations to find a closed form provided you can solve a system of equations.
Hopefully that gave you an idea even though it's quite vague (I noticed you said just numbers in the beginning so I've assumed that means real\complex and not only positive integers).
Also, this might help: https://www.reddit.com/r/math/comments/11vkk8/do_all_recursive_sequences_have_an_explicit_form/