Here's my proposed question (edit: now posted) (edit 2: posted version now edited):

I have found an algorithm for counting partitions, and I've been able to express it into a mathematical formula, $P(n) = \sum_{i=1}^{x} p_i$. The formula has less than $P(n)$ computation steps (that is, it counts partitions in batches), but it is still huge, and I have yet to simplify it. Therefore, I would like to know whether or not other such formulas have been found, in case one of them is the same as mine. If so, I could perhaps spare myself the hassle of pursuing a dead-end/already-explored end/etc. In summary, I just want to know if my discovery is a rediscovery (and thereupon, find out more about the value of the discovery). So, are the formulas that count all the partitions of $n$ (in batches)?

Is this too vague/broad? It would be a better question if I could share more, but I do not want to disclose too much information about my research outside of a proper publication.

This is how my question looks now:

Does there exist a formula of this form:

$$P(n) = \sum_{i=1}^{x\le P(n)} p_i$$

Where $P(n)$ is the partition function, and the $p_i$ are batches of partitions of quantity $\ge 1$. The sum is just a

closed-form, explicitmathematical expression of an algorithm that counts through all the partitions, but given the algorithm's nature, that counting can be somewhat compressed.If no such algorithm has been found, my question defers to whether or not an algorithm for counting partitions one-by-one has been found. I'd think so, but I haven't seen one in my research.

It has been closed due to lacking clarity. All I want is an explanation for what is missing to give an answer. What is confusing? No-one has explained what is confusing/missing, except for one user saying the answer was trivially yes based on what I find to be a non-sensical reading of the question (that is e.g. $P(n) = \sum_{k=1}^n P_k(n)$, thus the answer is yes). I don't see why I have to publish my research just to give a concrete example of a form that seems simple enough.