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May 23

What is the function $p$ satisfying $p(6)=11$?

Problem 8A from an exam:

How many abelian groups are there with 1000000 elements?

Solution: $1000000=2^{6}\times 5^{6}$ so the number of abelian groups is p(6) × p(6) =11 × 11 = 121.

What is the function $p$?

~2026-30088-06 (talk) 12:04, 23 May 2026 (UTC)

It's the partition function. See also Abelian group#Classification. --RDBury (talk) 13:04, 23 May 2026 (UTC)

PS. I'm pretty sure the phrase "up to isomorphism" should be included; it's more or less implied by context, but not actually stated. --RDBury (talk) 13:08, 23 May 2026 (UTC)

May 30

What about the application of the Solomonoff induction to the person to person information exchange?

It seems it adds some principal bottleneck to the human to human communications. You have to guess what the other person means.

More details here in Russian: https://transitional-writes.dreamwidth.org/41031.html It seems it is proof of the famous Mises calculation argument theorem. The only mean to circumvent this bottleneck is to use physically preserve money like gold, uranium, platinum.

Can we at least discuss this topic? ~2026-32197-56 (talk) 10:08, 30 May 2026 (UTC)

You appear to be asking about applying Solomonoff's theory of inductive inference to support Von Mises's Economic calculation problem. Please translate your comments from Russian into English, if you want our help. JRSpriggs (talk) 12:35, 30 May 2026 (UTC)

Thank you for your reply.

Yes, that's exactly what I'm asking for.

Could we move step by step?

1. First question is the inductive bottleneck in the human to human communications caused by the

properties of the inductive inference and perception per se.

See https://en.wikipedia.org/wiki/Solomonoff%27s_theory_of_inductive_inference

If we apply this theory to the "bitstream" that is seen by the peers in the human communccation

we can infer that there is fundamental uncertainty that we get all information right from the peer.

2. The second question is the application of the first step to the problem of the economic calculation.

If we can not trust peers we have to use something that we can trust.

Some primitives of the decentralized control that could be trust like gold.

But before the second question let's discuss the first one.

Maybe we will need some additional steps and discussions between first two questions. ~2026-32197-56 (talk) 13:52, 30 May 2026 (UTC)

The same query has been posted on the Science Reference desk. Please keep the discussion to one location to prevent duplication, forking, and excessive use of valuable volunteer editor's' time. {The poster formerly known as 87.81.230.195} ~2026-27434-43 (talk) 14:40, 30 May 2026 (UTC)

See discussion at Wikipedia:Reference desk/Science#What about the application of the Solomonoff induction to the person to person information exchange?. --Guy Macon (talk) 15:10, 30 May 2026 (UTC)

Maximal V=L without indiscernibles

Let Silver_α stand for the α-th Silver indiscernible (see Zero sharp). Assuming ZFC and Hugh Woodin's Ultimate-L conjecture. What can we deduce about the theory of L_Silver0? JRSpriggs (talk) 15:05, 30 May 2026 (UTC)

I can't remember (assuming I actually once knew) how to prove it, but I'm fairly sure that if $\alpha$ is a Silver indiscernible for $L$, then $L_{\alpha }$ is an elementary submodel of $L$, and in particular the first-order theory (without parameters) of $L_{\alpha }$ is identical to that of $L$. I don't think Ultimate L has anything to do with this.

Guessing a little at what you might be getting at, given the subject heading: You have to realize that $L$ doesn't know what its indiscernibles are. That is, there's no formula $\varphi$ such that $L\models \varphi (\alpha )$ if and only if $\alpha$ is a Silver indiscernible for $L$. If there were, then $L$ could define its least Silver indiscernible, and then it wouldn't be indiscernible from the second-least one. --Trovatore (talk) 18:49, 30 May 2026 (UTC)

Yes, I also remember reading that L_Silverα is supposed to be an elementary submodel of L. But I am not sure that the attributes of L are independent of the choice of V. That is why I tried to specify a particular version of V by specifying Hugh Woodin's conjecture. The name he gave to the conjecture maybe confusing in this context. As I understand it, he was trying to specify a model which is "close" to HOD, the hereditarily ordinal definable sets, and has a very regular structure, potentially including all large cardinals consistent with the axiom of choice.

Of course, if one chops off L at an ordinal which is not a Silver indiscernible, then it will have a different theory which is less comprehensive.

I think that below Silver₀, there will probably be approximately-elementary submodels. That is, some subset of the formulas will be reflected in the submodel, perhaps all formulas with fewer than a certain number of symbols. I specified Silver₀ because I wanted to keep the issue simple. JRSpriggs (talk) 22:46, 30 May 2026 (UTC)

$L$ is basically unconcerned with what's going on in $V$. It's constructed rigidly step-by-step along the ordinal numbers, looking only at what you've constructed so far and not caring at all about what other stuff might be going on in the universe. So I don't think "Ultimate L" is really going to affect anything. --Trovatore (talk) 23:14, 30 May 2026 (UTC)

I remember reading something about L[U] where U is an ultrafilter associated with one measurable cardinal and then using a second measurable cardinal to create a club which avoids most regular ordinals. The indiscernibles in such a case might not be the same as the usual Silver indiscernibles. If you cut your model off at the wrong kind of cardinal, it might give you an L with the wrong attributes. Another situation is L[P(ω)] which satisfies the AD rather than AC. JRSpriggs (talk) 01:45, 31 May 2026 (UTC)

Hmm. This one did stump me for a bit. It's certainly true that you can cut the universe off at a cardinal that is inaccessible in $L$ but not Mahlo in $L$, and then you have a model in which the class of regular cardinals is not stationary. But it's also true that all Silver indiscernibles are Mahlo in $L$.

I think the resolution must be that you can't say "ORD is Mahlo" as a first-order statement of set theory, because it requires quantifying over club proper classes. Therefore the non-Mahloness of ORD in your truncated model doesn't have to reflect to the indiscernibles.

It's still possible I've missed something but that's what I'm going with for now. --Trovatore (talk) 05:04, 31 May 2026 (UTC)

If Ord is not a fixed point of α -> Silver_α, then does that not mean that we would have a fake zero sharp and fake indiscernibles? So then the theory of L would be wrong. JRSpriggs (talk) 12:05, 31 May 2026 (UTC)

There is in fact something I'm missing here and I haven't been able to figure out overnight. I really don't think that 0# depends on the upper reaches of the universe. It's definable at a low level of the analytic hierarchy and, once you've got it, is extremely stable. But I'm running into an issue trying to figure out what happens if you work inside a model in which, say, there's a greatest inaccessible in $L$. That would be an element of $L$ that's definable in $L$, so it has to occur before the first Silver indiscernible. On the other hand the Silver indiscernibles themselves are all inaccessible in $L$. Clearly I'm rusty. I'll think about it some more. --Trovatore (talk) 17:54, 31 May 2026 (UTC)

I think the answer must be that if $\alpha$ is an inaccessible of $L$ such that $L_{\alpha }\models {\text{ZFC + there is a greatest inaccessible}}$, then $L_{\alpha }[0\#]\not \models {\textrm {ZFC}}.$. Presumably 0# gives you a way to witness that $L_{\alpha }[0\#]$ does not satisfy Replacement. But I'm still missing some details. --Trovatore (talk) 18:14, 31 May 2026 (UTC)

If L_α has a greatest inaccessible in it, then α<Silver₀, that is, zero sharp does not "exist" in that model. Even equality to Silver₀ would imply that there is an unbounded class of inaccessibles in the model.

Of course, L does not know which ordinals are Silver indiscernibles, and it does not know which formulas they satisfy (i.e. which subset of ω constitutes zero sharp). But in V outside of L, we could know some small part of that information. That is what I was asking for originally. JRSpriggs (talk) 23:05, 31 May 2026 (UTC)

I finally found the link to a description of Woodin's conjecture and its opposite: "Cardinals beyond Choice". Please see here. JRSpriggs (talk) 23:29, 31 May 2026 (UTC)

Regarding "you can cut the universe off at a cardinal that is inaccessible in L but not Mahlo in L". If you are cutting it off at an ordinal, β, which is say greater than Silver_ω, then β would not even be an initial ordinal in V (assuming V is a full model of ZFC with any necessary embellishments), so V_β would not be a model of ZF. JRSpriggs (talk) 13:09, 1 June 2026 (UTC)

To see this, let β be the second L-inaccessible after Silver_ω. It will not be L-Mahlo. Its V-cardinality will be the same as that of Silver_ω. So V_β will not satisfy the axiom of replacement, although it does satisfy all the other axioms of ZFC. The L of V_β is L_β, but L_Silver0 will not be an elementary submodel of L_β. So any attempt to build a club of "indiscernibles" in L_β would have to go wrong. JRSpriggs (talk) 00:46, 2 June 2026 (UTC)

I am not sure, but I am beginning to suspect that what I called Silver_α maybe the same as the least α-Erdos cardinal. JRSpriggs (talk) 19:18, 4 June 2026 (UTC)

After re-reading our article on Erdos cardinals, I now think that I was wrong. If zero sharp "exists", then every uncountable initial ordinal is a Silver indiscernible and there are probably many others between them. If zero sharp does not "exist", then there are no Silver indiscernibles. Either way, the notion becomes trivial, unworthy of inclusion in our list of large cardinal properties which I was hoping to do. JRSpriggs (talk) 12:59, 5 June 2026 (UTC)

Except Silver₀, as the least indiscernible it deserves to be included in the list of large cardinals. It could be thought of as the representative of zero sharp (which is described as not a cardinal). It has the bizarre property that if it exists then it and all smaller large cardinals are merely countable, but if it does not exist, then most of those smaller large cardinals are uncountable. It might be the same as the least ω₁-Erdos cardinal, the least iterable cardinal, or both. JRSpriggs (talk) 17:53, 5 June 2026 (UTC)

June 2

Provable vs true but not provable.

After reading popular literature on provability and about godel and tarski and forgetting most of it, I'm left with wondering: Just as the set of rational numbers is much smaller than the irrationals, is it true that the set of provable mathematical statements is much smaller than the set of true but unprovable statements? And can that be proved?Rich (talk) 06:06, 2 June 2026 (UTC)

Since "statements" usually consist of strings in some formal language, they'll form a countably infinite set. So cardinality will be countably infinite for both kinds of statements. Sesquilinear (talk) 07:59, 2 June 2026 (UTC)

It would depend on which field of mathematics you are discussing (number theory, analysis, geometry, category theory, set theory, etc.) and what metric you are using to measure the quantity of statements. Does the negation of a formula count the same as the formula? How much does the weight of a formula decrease as the number of quantifiers in it increases? In more concrete areas like number theory, the proportion of provable statements might be higher. In more abstract areas like set theory, there would be many statements which arguably do not even have a truth-value. JRSpriggs (talk) 14:27, 2 June 2026 (UTC)

Let S(n) stand for the set of all true statements that can be expressed using no more than n symbols, and let p(n) be the fraction of statements in S(n) that are provable. We may wonder what happens to p(n) as n → ∞. One issue is that provability is not an absolute notion, but is relative to a given proof system. No true statement X is unprovable in the sense that for any given sound proof system Σ we can make a system Σ_X that is also sound but in which X is provable, simply by adding it to the axioms. Since the number of (originally) unprovable statements in S(n) is finite, only a finite number of such extensions is needed to make all true statements provable, so there exists a sound proof system for which p(n) = 1. (This existence proof is seriously nonconstructive.)  ​‑‑Lambiam 22:21, 2 June 2026 (UTC)

Here is a proof by hand-waving that p(n) → 0 as n → ∞ if the proof system (assumed to be consistent and sufficiently powerful) is not allowed to depend on n. "Sufficiently powerful" means that Gödel's first incompleteness theorem applies, so there is at least one unprovable true statement.

Consider now an oversimplified proof system that has two zero-argument logical constants, say ${\mathsf {T}}$ and ${\mathsf {?}},$ and one binary logical operator, $\land .$ Let us use the number of zero-argument constants as a measure of the size, so ${\mathsf {T}}\land ({\mathsf {?}}\land {\mathsf {T}})$ has size $n=3.$ Using the notation of natural deduction, our little system has two rules:

${\frac {~}{\mathsf {T}}}~~~~~~~~~~~~~~~{\frac {P,Q}{P\land Q}}.$

So ${\mathsf {T}}$ is an axiom; it needs no premises. Clearly, a statement is provable in this system iff it contains no ${\mathsf {?}}.$ The semantic interpretation of the logical constants is that ${\mathsf {T}}$ and ${\mathsf {?}}$ are different representations of the truth value "true", while $\land$ represents conjunction. The role of ${\mathsf {?}}$ in this oversimplified system is to represent some unprovable true statement.

In this system, all statements that can be formulated are true, but only a fraction is provable. To be precise, $S(n)$ contains $2^{n}C_{n{-}1}$ statements, in which $C_{k}$ stands for the $k$-th Catalan number. But only $C_{n{-}1}$ of those are provable in the system, so $p(n)=2^{{-}n}.$

Clearly — here the professor paused a second while sternly gazing across the audience and using his hand to underline the weight of the declaration — this example illustrates a phenomenon that we will see in any proof system.  ​‑‑Lambiam 15:19, 3 June 2026 (UTC)

June 4

Calculating diminishing returns formula for game stat caps

Hi everyone, I'm trying to design a custom RPG mod and need some help figuring out the exact math for a stat curve.

Right now, I have a "Haste" stat that increases a character's attack speed. If a player has 0 Haste, their attack cooldown is 1.0 second. I want to implement a system of diminishing returns so that no matter how much Haste a player stacks, the attack cooldown can never drop below 0.2 seconds (effectively a hard asymptote at 0.2).

I know I can use a hyperbola for this, but I'm struggling to parameterize it. Ideally, I want 100 points of Haste to bring the cooldown down to 0.6 seconds, while 500 points should bring it down to around 0.3 seconds, scaling smoothly in between.

What is the standard algebraic formula to express this kind of curve where $y$ approaches a minimum limit as $x$ approaches infinity, given these specific data points?

Thanks! ~2026-33106-46 (talk) 08:09, 4 June 2026 (UTC)

I'm betting this is for an RPG; in my experience devs use all sorts of complex expressions for curves like this, including piecewise defined functions that mere mortals will struggle to understand. So I don't think there is a "standard" formula, but if it were me then I'd use a Linear fractional transformation, in other words a function of the form (ax+b)/(cx+d). Their graphs, aside from some degenerate cases, are hyperbolas with a vertical and horizontal asymptote. It turns out these are uniquely defined by three values, and you've specified three values L(0) = 1, L(100) = .6 and L(∞) = .2. I get L(x) = (x+500)/(5x+500); this has L(500) = 1/3, which I hope is close enough to your desired value of .3.

To determine values of a, b, c, d from the three given points. First, note that by dividing the top and bottom by a constant, we can assume any value for a, so take a=1. That gives (x+b)/(cx+d). From L(0) = 1 we get b/d = 1, d=b, and we have (x+b)/(cx+b). From L(∞) = .2 we get 1/c = .2, c=5, and L(x) = (x+b)/(5x+b). From L(100) = .6 you get 100+b = .6(500+b), 100+b = 300+.6b, .4b=200, b=500. You can use a similar calculation to work out an expression with L(0), L(∞), and L(some other value) all given. It should be noted that fitting a curve to a set of values is more of an art than a science, and there are many possible tools and approaches you could use, but as I said, this is the approach I would use. --RDBury (talk) 09:24, 4 June 2026 (UTC)