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, explicit mathematical 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.

  • 1
    $\begingroup$ Before asking your question -- did you check Wikipedia? -- en.wikipedia.org/wiki/Partition_function_(number_theory) $\endgroup$
    – Stefan Kohl Mod
    Dec 16, 2022 at 8:54
  • $\begingroup$ @StefanKohl I've read through it many times. It only contains generating functions, recurrence relations and approximation functions, as well as some congruences and some stuff about strict partitions. There is no exact formula in there, even though there exists one, courtesy of Rademacher. However, I think Rademacher's formula is not closed-form, but I am pretty sure mine is. But that then seems WAY too unrealistic to be true, so I'm wondering if there something about my formula that makes it less impressive or something. $\endgroup$ Dec 16, 2022 at 14:45
  • $\begingroup$ I suggest having a look at www2.math.upenn.edu/~wilf/PIMS/PIMSLectures.pdf especially page 14 $\endgroup$ Dec 19, 2022 at 6:13
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    $\begingroup$ My previous remark was meant to dispute the claim that Rademacher's formula is not closed form. I suppose the link I gave is not relevant to the current form of the question. In any event, if no one is giving you the sort of answer you want, and several users are telling you that they need to know what your method is before they can answer your question, it seems to me to be quite wrong-headed to insist that these users are wrong. $\endgroup$ Dec 20, 2022 at 23:33
  • $\begingroup$ @GerryMyerson How is it not closed form? It is an infinite sum. $\endgroup$ Dec 22, 2022 at 15:11
  • $\begingroup$ Please, go to the link that I gave you a few days ago, it should clear up your confusion. $\endgroup$ Dec 22, 2022 at 16:02


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