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November 8

dS = AdS

A de Sitter space and an anti-de Sitter space (in their more general forms: see EOM: De Sitter space and EOM: Anti-de Sitter space) may each be defined as any pseudo-Riemannian manifold that is isometric to a quasi-sphere embedded in a pseudo-Euclidean space with the induced metric tensor. The sign of this metric tensor is immaterial and is thus only a matter of convention. Yet neither WP nor EOM acknowledges this strong equivalence: they both describe these spaces as having positive and negative curvature respectively, as though this distinguishes them. If one attaches the labels "space-like" and "time-like", it is evident to me that it is the curvature from the "space-like perspective" being referred to, and that if one swaps the labels, the sign of this curvature changes. Yet from a geometric perspective, these labels are superfluous. The typical physicist might be surprised by this dS = AdS equivalence. What am I missing here? Can we find references that deal with this? —Quondum 14:31, 8 November 2022 (UTC)[reply]

If both are isometric to a quasi-sphere, they should be isometric to each other. Do they have the same light cones? --Lambiam 15:53, 8 November 2022 (UTC)[reply]

Inherently (given they have the same metric up to a sign: the light cone is the set of null geodesics through a point), yes. What might be confusing is that if one applies "space"/"time" labels, the re-labelling would change say three "space" and two "time" dimensions to three "time" and two "space" dimensions (and the sign of the "curvature", as I see it). —Quondum 18:36, 8 November 2022 (UTC)[reply]

To assign a physical meaning, the tangent spaces should be Minkowski spaces, having three space dimensions and one time dimension, distinguished by having another sign in the signature of the Minkowski metric than the other three dimensions. It does make a difference whether the curvature of a spacelike (three-dimensional) section, an ordinary Riemannian manifold, is positive or negative; this is an intrinsic property of the space that does not depend on how the space is embedded in a higher-dimensional flat space. --Lambiam 23:19, 8 November 2022 (UTC)[reply]

This is a mathematical question, not a physics question, so the restriction does not apply (though in physics, lower- and higher-dimensional spaces without such restriction are used and also referred to as dS and AdS). I explicitly opened with "in their more general forms", as at EOM. The concepts of "space-like" and "time-like" do not apply in a purely geometric context. When we have a manifold with a definite metric, we have only one "sort" of dimension to reference, and sign of curvature can be given a canonical definition. This is not true with indefinite spaces (in a dS or AdS space we simultaneously have "totally geodesic submanifolds" that have positive and negative curvatures – equal but for the sign – with spherical and hyperbolic geometries respectively).

For concreteness (and to show how blindingly obvious my point is), we can define two manifolds to match the definitions at EOM, each with the induced metric:

dS: The manifold $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{4}^{2}-x_{5}^{2}=1/c>0$ in $\mathbf {R} ^{5}$ with metric $dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}-dx_{4}^{2}-dx_{5}^{2}$ ( $n=2,p=2$ ).
AdS: The manifold $x_{5}^{2}+x_{4}^{2}-x_{3}^{2}-x_{2}^{2}-x_{1}^{2}=-1/c<0$ in $\mathbf {R} ^{5}$ with metric $dx_{5}^{2}+dx_{4}^{2}-dx_{3}^{2}-dx_{2}^{2}-dx_{1}^{2}$ ( $n=2,p=2$ ).

The variables have been re-indexed so that they are algebraically obviously equivalent (except that the metric differs by a sign, but we know that this is immaterial). Any choices of

n

and

p

work; their values swap between the two manifolds. —Quondum 23:59, 8 November 2022 (UTC)[reply]

The four-dimensional de Sitter spacetime and anti-de Sitter spacetime are different, being distinguishable by the sign of their curvature. Both are specific instances of a more general geometric object, namely a space that is isometrically embeddable as a pseudo-hypersphere in a pseudo-Euclidean space of one dimension higher. So it looks like the EOM describes this generalization twice, in different entries. --Lambiam 20:38, 9 November 2022 (UTC)[reply]

Yes, that is what I'm saying: the two-parameter dS and AdS families are actually the same family, and I have read nothing that recognizes this.

Agreed, the 3+1 dS and AdS cases of physics are distinct members of the family. The dS case has three dimensions from which the curvature seems positive and one from which it is negative, and the AdS case has this reversed. The bigger family makes it clear that describing the curvature as categorically "positive" and "negative" is not correct. Reducing these to 1+1 as is reasonable in physics, we have the same space whether we call it dS or AdS, and the distinction is now only which direction you label as "space-like", with the sign of the curvature conventionally referring to only the space-like curvature.

Try to convince a physicist that the curvature is both negative and positive in this way (and that selecting which to describe it is only a convention), though ... Thanks – I feel affirmed in my analysis. —Quondum 23:02, 9 November 2022 (UTC)[reply]

I am not sure the difference between the four-dimensional dS and AdS cases is one of systematically reversing the signs. They appear to be embedded in different pseudo-Euclidean spaces. Quoting from: Bang-Yen Chen, “Marginally trapped surfaces and Kaluza-Klein theory”, April 2009, International Electronic Journal of Geometry 2(1):1–16^[1]:

De Sitter space-time can be defined as a hypersurface of Minkowski space. Take Minkowski space-time

\mathbb {E} _{1}^{5}

with the standard metric

g=-dt^{2}+dx^{2}+dy^{2}+dz^{2},

the de Sitter space-time is the submanifold described by the hyperboloid

-t^{2}+x^{2}+y^{2}+z^{2}=c^{2},

where

c

is some positive constant. The metric on de Sitter space-time is the metric induced from the ambient Minkowski metric.

Similarly, an anti de Sitter space-time can be realized as a hypersurface of the pseudo-Euclidean space

\mathbb {E} _{2}^{5}

with index 2 described by

-t^{2}-x^{2}+y^{2}+z^{2}=-c^{2}

where

c

is some positive constant.

--Lambiam 03:13, 10 November 2022 (UTC)[reply]

The source has an ignorable error of one too few variables in these expressions. dS₄ and AdS₄ (resp. S⁴
₁ and H⁴
₁ in the notation of the source) are different spaces: S⁴
₁ ≠ H⁴
₁. I have not suggested trying to draw any equivalence between these two cases: I am only interested in the interpretation of the sign of the "curvature", which I am saying has no natural interpretation (possibly outside of the cases S^s
₀ and H^h
₀).

I am saying that S^s+h
_h = H^s+h
_s for all s and h (full dS–AdS equivalence), where s denotes "spherical" dimensions and h denotes "hyperbolic" dimensions, and that sources call the curvature of S^s+h
_h positive and of H^s+h
_s negative, despite them being identical geometrically. —Quondum 14:37, 10 November 2022 (UTC)[reply]

November 9

between chessboard distance and taxicab distance in higher dimensions

Looking at Chebyshev distance, which also links to Manhattan distance, I thought about intermediates. In three dimensions there is what I privately call the garnet distance: max(|∆x|+|∆y|,|∆x+∆z|,|∆y+∆z|) – it defines a ball in the shape of a garnet crystal, a rhombic dodecahedron. Generalizing to n dimensions, the k-garnet distance is the maximum of the sum of any k coordinate distances, 1≤k≤n (1: Chebyshev; n: Manhattan).

Has anyone made use of such intermediate measures? —Tamfang (talk) 18:01, 9 November 2022 (UTC)[reply]

I think you meant to write

\max(|\Delta x|+|\Delta y|,|\Delta x|+|\Delta z|,|\Delta y|+|\Delta z|).

--Lambiam 20:02, 9 November 2022 (UTC)[reply]

Er, yes. —Tamfang (talk) 08:46, 10 November 2022 (UTC)[reply]

November 10

Locality-preserving functions

Having a problem with locality-preserving functions. In particular, a locality-preserving hash function $f:{\mathcal {M}}\rightarrow \mathbb {R}$ (where ${\mathcal {M}}=(M,d)$ is a metric space) is a hash function satisfying $\forall p,q,r\in {\mathcal {M}},d(p,q)<d(q,r)\Rightarrow |f(p)-f(q)|<|f(q)-f(r)|$ , and more generally, we can say that a locality-preserving function $f:{\mathcal {M}}\rightarrow {\mathcal {K}}$ for metric spaces ${\mathcal {M}}=(M,d_{M}),{\mathcal {K}}=(K,d_{K})$ is a function satisfying $\forall p,q,r\in {\mathcal {M}},d_{M}(p,q)<d_{M}(q,r)\Rightarrow d_{K}(f(p),f(q))<d_{K}(f(q),f(r))$ .

It can be seen that any such function must be injective. If $f(p)=f(q)$ but $p\neq q$ , then for any $r\in {\mathcal {M}}$ with $d_{M}(p,r)\neq d_{M}(q,r)$ (for example, $p$ or $q$ itself), we must have either $d_{M}(p,r)<d_{M}(q,r)$ or $d_{M}(p,r)>d_{M}(q,r)$ , but $d_{K}(f(p),f(r))=d_{K}(f(q),f(r))$ , which is contradictory.

The area I'm having trouble with is that I have no idea if locality-preserving functions have to be continuous. I'm pretty sure I have it narrowed down to one problem which I am completely stuck on: if $f:{\mathcal {M}}\rightarrow {\mathcal {K}}$ is a locality-preserving function, is it possible for there to be some point $x\in {\mathcal {M}}$ such that $x$ is an accumulation point but $f(x)$ is an isolated point? Similar problem in establishing whether $f$ is an open mapping, where the problem becomes whether it is possible for $x$ to be an isolated point while $f(x)$ is an accumulation point.

As a quick aside, I'm fairly sure that if ${\mathcal {K}}$ (resp. ${\mathcal {M}}$ ) is compact, then the answer is no for continuity (resp. open mapping) GalacticShoe (talk) 04:22, 10 November 2022 (UTC)[reply]

Neither one needs to hold. For instance, let

{\mathcal {M}}=\{0\}\cup \left\{{\frac {1}{n}}\colon n\in \mathbb {Z} ^{+}\right\}

,

{\mathcal {K}}=\ell ^{\infty }(\mathbb {R} )

under the supremum norm, and

f(0)=e_{1}

,

f\left({\frac {1}{n}}\right)={\frac {n+1}{n}}e_{n+1}

(where

e_{i}

is the sequence with 1 in the

i

-th position and 0 elsewhere).

f

is a locality-preserving function and 0 is an accumulation point of

{\mathcal {M}}

, but

f(0)=e_{1}

is an isolated point of

f({\mathcal {M}})

. The inverse of

f

is also a locality-preserving function from

f({\mathcal {M}})

back to

{\mathcal {M}}

. (Wrong, see below for a correct example.) BentSm (talk) 05:58, 10 November 2022 (UTC)[reply]

Writing

f:{\mathcal {M}}\to {\mathcal {S}},

is the codomain

{\mathcal {S}}

in this proposed counterexample one of scalar values? --Lambiam 10:27, 10 November 2022 (UTC)[reply]

No. (I was basing my answer from his (extended) definition of a locality-preserving function.) BentSm (talk) 01:08, 11 November 2022 (UTC)[reply]

Hey BentSM, thanks for the response! Just wanted to clarify a point of confusion; is the distance under the supremum norm defined as

d_{sup}(x,y)=sup_{n}|x_{n}-y_{n}|

, and if so, would

|0-{\frac {1}{3}}|={\frac {1}{3}}>{\frac {1}{6}}=|{\frac {1}{3}}-{\frac {1}{2}}|,d_{sup}(f(0),f({\frac {1}{3}}))={\frac {4}{3}}<{\frac {3}{2}}=d_{sup}(f({\frac {1}{3}}),f({\frac {1}{2}}))

not render

f

non-locality preserving? Thanks again! GalacticShoe (talk) 18:07, 11 November 2022 (UTC)[reply]

You're quite right. This does work, though:

{\mathcal {M}}=\{0\}\cup \left\{{\frac {1}{n}}\colon n\in \mathbb {Z} ^{+}\right\}

,

{\mathcal {K}}=\ell ^{\infty }(\mathbb {R} )

under the

L^{1}

norm, and

f(0)=e_{2}

,

f\left({\frac {1}{n}}\right)={\frac {1}{n}}e_{1}+e_{n+2}

. If

x\neq y

,

d_{1}(f(x),f(y))=2+|x-y|

(

d_{1}(f(x),f(x))=0

, of course). BentSm (talk) 00:32, 12 November 2022 (UTC)[reply]

Suppose we want to hash (one-dimensional) real values into integer-valued buckets. So the desired hash function has signature

f:\mathbb {R} \to \mathbb {Z} .

A prime candidate is the rounding function defined by

f(x)=\lfloor x+0.5\rfloor .

If any function from the reals to the integers is a locality-preserving hash, this should be one. But (under the usual metric in which

d(x,y)=|x-y|

) we have

d(3.4,3.6)=0.2<d(3.6,3.9)=0.3,

whereas

|f(3.4)-f(3.6)|=|3-4|=1>|f(3.6)-f(3.9)|=|4-4|=0.

This suggest strongly that the given definition is off. Perhaps it should be the other way around: values mapped to the same bucket should be close. That would explain the suggested relation to space-filling curves: themselves being continuous, they are surjective and so have a (not unique) inverse. Such an inverse can be viewed as a hash, and pairs of points whose hashes are close are themselves close. Naively reversing the implication does not work, however:

|f(2.8)-f(3.4)|=|3-3|=0<|f(3.4)-f(3.6)|=|3-4|=1,

whereas

d(2.8,3.4)=0.6>d(3.4,3.6)=0.2\,.

So we may want a condition in the shape of

|f(x)-f(y)|<\varepsilon \,\Rightarrow \,d(x,y)<\delta .

--Lambiam 12:11, 10 November 2022 (UTC)[reply]

I do agree that the definition is rather unusual. It is certainly is not that of the two references listed in the corresponding section of the Wikipedia article referenced. I tried doing some searching, which turned up more references to "locality-preserving" than direct definitions; see, e.g., [2], [3], [4], [5]. For what it's worth, none of what I found matched up with the definition given in the article (except for one site that cites Wikipedia). BentSm (talk) 03:38, 11 November 2022 (UTC)[reply]

For example, the Hilbert curve satisfies this "inverse continuity" condition if we put

\delta ={\sqrt {c\varepsilon }},

where I think we can take

c={\tfrac {27}{5}}.

--Lambiam 16:11, 10 November 2022 (UTC)[reply]

Slopes and angles

If a line had a slope of S (assuming the line goes through the origin and S cannot be negative), how can I use that to find out the angle that the line makes with the X or Y-axis? Primal Groudon (talk) 14:18, 10 November 2022 (UTC)[reply]

By consulting our article Slope: it gives the relation between the slope of a line and its angle of inclination. It is not relevant whether the line goes through the origin, and the relation is valid regardless of the sign of the slope. --Lambiam 16:35, 10 November 2022 (UTC)[reply]

The angle of the line to the X axis will be the arctan of the slope. That's because t the line will always be the hypotenuse of a right triangle where the X axis is one leg and a line parallel to the Y axis is the other leg. Since tangent Θ = Y/X = slope, to find the angle Θ you take the inverse tangent (arctan) of the slope. --2600:1004:B024:84EE:F02:CC26:4A9A:E65 (talk) 21:54, 10 November 2022 (UTC)[reply]

There's two angles corresponding to a slope differing by 180°. Very often you'll want the one between -90° and +90°. I'm not sure why you specified that the slope is not negative, that doesn't matter much. If anything the slope that is a bit peculiar is the vertical slope going straight up and down where the ratio is infinity. NadVolum (talk) 01:19, 11 November 2022 (UTC)[reply]

November 11

Roman numerals

Is there a way to express numbers in Roman numerals after 3,999,999? Finnish Wikipedia says that vinculum is used below the "letter" to make numerals from 4,000,000 to 3,999,999,999. But English Wikipedia says nothing about that. Is there really an method to this? --40bus (talk) 13:39, 11 November 2022 (UTC)[reply]

According to this source, there were no standard ways of expressing numbers higher than 1,000,000 in Roman numerals. --Lambiam 18:36, 11 November 2022 (UTC)[reply]

I thought that vinculum below was a standard, and wanted to know how to make numbers from 4,000,000,000 to 3,999,999,999,999.--40bus (talk) 21:41, 11 November 2022 (UTC)[reply]

First, I believe the passage Lambiam cites is intending to refer to amounts representable as a single character with vinculum. Second, as it says at Roman numeral#Large numbers, none of the notations devised to extend the range of Roman numerals was ever standardized, so I think 40bus is asking for something that doesn't exist. I can't cite my source, but I remember seeing it stated somewhere that the largest number known to have been expressed in ancient times in Roman numerals was 2,300,000 -- represented not using a vinculum but as ↈↈↈↈↈↈↈↈↈↈↈↈↈↈↈↈↈↈↈↈↈↈↈ. --174.89.144.126 (talk) 03:25, 12 November 2022 (UTC)[reply]

November 12

Graham's number g₁

After reading about Graham's number g₁, I learnt that a 3.6 trillion-digit number is still insufficient for it. So, limiting my non-math mind to understanding just g₁, I decided to take a number with centillion (short-scale) zeros (which, if I'm not mistaken, is 10 to centillionth power). What fraction or percentage of g₁ does that number constitute? Otherwise, is it possible to represent a fraction/percentage of g₁ using large number names? Thanks. 212.180.235.46 (talk) 12:18, 12 November 2022 (UTC)[reply]

The number is incomprehensibly larger than 10 to the centillionth power. Expressing the latter number as a fraction of g₁ would require more zeros after the decimal point than the number of quarks that can fit in 10 to the centillionth observable universes. (Insert sound effect of head exploding.) --Lambiam 15:25, 12 November 2022 (UTC)[reply]

(edit conflict) Graham's number, and even the number of digits in Graham's number defy attempts to quantify it in ordinary notation. The percentage you're talking about would be equally hard to comprehend because it would be so small. The good news is that, other than being an example of a extremely large number used in a mathematical proof, it has no real use and the numbers you encounter in real life are more tractable in terms of notation. Humans really aren't evolved to comprehend large numbers. I can imagine the number of people in a stadium, say 100,000. Then the number of people on Earth is about a stadium full of people for each person in a stadium (10,000,000,000), but that's just a calculation, not something I can actually imagine. --RDBury (talk) 15:27, 12 November 2022 (UTC)[reply]

As to imagining Earth's population canned and ready for shipment, see "7.3 Billion People, One Building", --Lambiam 15:56, 12 November 2022 (UTC)[reply]

Graham's number itself is kind of mostly of historical interest. The bound for that problem has since been greatly improved, and proofs of other problems have since involved even larger numbers. And given that the lower bound for the problem is only 13, there's always the chance that the actual answer to the problem that Graham's number came from is laughably small. :) Double sharp (talk) 00:22, 14 November 2022 (UTC)[reply]

Graham's number is g₆₄, not the g₁ the OP mentions. But

g_{1}=3\uparrow \uparrow \uparrow \uparrow 3

. -- SGBailey (talk) 13:10, 14 November 2022 (UTC)[reply]

Does it make much difference for us ;-) As a person opening a dam said after saying, this dam holds ten million gallons of water, and the engineer beside him said ten thousand million, Oh what is a thousand in so many millions ;-) By the way, the number of people in the world is projected to go over 8 billion sometime thie week. NadVolum (talk) 13:28, 14 November 2022 (UTC)[reply]

"Any large number is finite, and you can start thinking about it as 3." – John Conway. Double sharp (talk) 17:16, 14 November 2022 (UTC)[reply]

I think the OP is asking "what proportion of g₆₄ is g₁?" Basically, the relationship between 1 and g₁ is the same as the relationship between g₁ and g₂; which is to say that if you start at 1, and build the set of operations to get to g₁, you then take that entire set of operations and apply them g₁ to get to g₂. Then, you take everything you did to get from 1 all the way to g₂ and apply that to g₂, and you get g₃. Repeat that process, each "g" step applying the full set of operations you took to get to g_n, and applying them to g_n to get to g_n+1, and do that ALL THE WAY to g₆₄. That's Graham's number. It should be noted that if I wanted to compare g₁ to g₂ as a fraction, I would need more atoms than exist in the known universe just to make the ink required to write the denominator of THAT fraction. These scales are stupid. --Jayron 32 18:46, 14 November 2022 (UTC)[reply]

November 14

Wall-Sun-Sun primes

Why have we conjectured the existence of infinitely many Wall-Sun-Sun primes even though none are known?? Georgia guy (talk) 17:10, 14 November 2022 (UTC)[reply]

Well, one of the oldest (and most important) theorems regarding prime numbers is Euclid's theorem, which has proven that there must be an infinite amount of prime numbers. In general, it is usually conjectured that for any well-defined subset of prime numbers, such as Mersenne primes, there would also be an infinite number of those as well, though these are often in the form of conjecture rather than proof. See, for example, the Lenstra–Pomerance–Wagstaff conjecture about Mersenne primes, or the twin prime conjecture or more generally Polignac's conjecture. It should be noted that proof of existence does not require any actual examples. On the balance, it is likely that Wall-Sun-Sun primes exist, though given that every single finite number, no matter how large, is by definition infinitesimally small compared to actual infinity. Who knows, maybe the first one exists at an order of magnitude so large it cannot be expressed even if we used all of the matter in the universe. But it may still exist. I'm not sure that Wall-Sun-Sub primes have been proven to exist as yet, but often times, if one exists, then it is likely infinitely many do as well. --Jayron 32 17:24, 14 November 2022 (UTC)[reply]

Courtesy link: Wall–Sun–Sun prime. I'm not saying it's the case here, but sometimes you can make heuristic arguments that the number of certain primes is infinite. For example if you know how fast a sequence grows, you know the "probability" of a number of a certain size being prime, and you can combine this information to produce a series whose sum is an estimate for the number of primes in the sequence. If the series diverges then you might conjecture that the number of primes is infinite. The article isn't too clear on the evidence for and against the conjecture (there is a link though), but one possibility is that there seems to be infinitely many near-Wall–Sun–Sun primes. In the absence of any evidence to the contrary, the "probability" of a near-Wall–Sun–Sun prime being an actual Wall–Sun–Sun prime is 1/3, so one might expect a third of the entries in this seemingly infinite sequence to be primes we're looking for, even though none have shown up to the current limit of calculations. Another piece of evidence on the "for" side is that Wall–Sun–Sun prime are the case k=1 of k-Wall–Sun–Sun primes, and these are known to exist for k=2 through 10, with k=11 being the next k for which the question is open. It would seem odd if there existed k-Wall–Sun–Sun primes for most values of k but not for a few "random" exceptions (through stranger things have happened).

I have a hard time getting enthused about conjectures relating to prime numbers though, there are many of them and it seems that the mathematics needed to resolve them hasn't been invented yet. In cases where weaker versions have been proved, the proofs are so complex and lengthy that it's difficult to follow or appreciate them. Analytic number theory opened up new avenues of proof, and is the key to Dirichlet's theorem on arithmetic progressions. But absent the creation of new branches of mathematics, I think progress will be incremental. --RDBury (talk) 06:47, 15 November 2022 (UTC)[reply]

Something regarding a prime p has p possible values: 0 to p-1. If it's 0 then p is a Wall-Sun-Sun prime. The values appear to be randomly distributed and there is no apparent reason to think that 0 is impossible so for a random prime p, the expected chance of a Wall-Sun-Sun prime is 1/p. The estimated number of Wall-Sun-Sun primes is then the sum of 1/p over all primes p and this sum is infinite (this is the divergence of the sum of the reciprocals of the primes). But all primes below 300×10¹⁵ have been tested without finding one, shouldn't we start to suspect there is something unknown which prevents the 0 value? Not really, the sum diverges very slowly. The sum of 1/p for all tested primes is only around 3 and the values keep looking random. If the search limit could be extended to 10³⁰, which is unlikely to ever happen if they still have to be tested one by one, then the expected number of primes only increases by around 0.5. We will probably never find a Wall-Sun-Sun prime but keep expecting there are infinitely many. If we do find one then it will probably be due to new theory and not an increase in computer speed or search time. I don't agree with Jayron32 that we should usually expect infinitely many primes of a given form. When mathematicians make such predictions it is usually because of heuristic arguments like above: Sum "the expected chance" of individual candidates and show this sum is infinite. I use quotation marks because a specific number either has a property or not, and "the expected chance" is a qualified guess based on incomplete information. Maybe it later turns out to not be as qualified as we thought. We usually only talk of "the expected chance" until we have actually tested the number. What is the "expected chance" that 4 is prime? A common general estimate for n says 1/log(n) which is 0.72 for 4. Does that mean 4 has 72% chance of being prime? Some may argue this is actually a fair estimate. Others may say it's obviously composite so the chance is 0. The sum of infinitely many positive numbers can be finite. For example, 1/2 + 1/4 + 1/8 + ... = 1. It's a finite sum similar to this which makes most mathematicians expect that the number of Fermat primes is finite. You will hear more conjectures of the form "There are infinitely many primes of this form" but that's because such forms often seem more "interesting" and we discuss them more. The two main reasons for not expecting infinitely many primes are a little "boring": The sequence grows rapidly (like doubling the number of digits each time), or we can easily prove a factor of each number after the first one or two, e.g. n²-1 = (n-1)×(n+1). Numbers of form 2^k+1 is a combination of these: If k has an odd factor above 1 then we can prove a factor of 2^k+1 so the only primality chance is for the rapidly growing Fermat numbers 2^2ⁿ+1. This and the apparent coincidence that it's prime for n = 0 to 4 makes it more interesting. It also has a history going back to the 17th century and remains unsolved. PrimeHunter (talk) 08:41, 15 November 2022 (UTC)[reply]

Your heuristic argument is one I hadn't thought of; it's simple and elegant but I think your point is that it's only an argument, not a proof, and probably not convincing enough to counter the existing evidence to the contrary. Perhaps a more relevant calculation is the probability of getting 0 hits when the probability of a hit is 1/p and trials run over all primes less than 300×10¹⁵. Re-reading the original post, it asks "Why do we conjecture ... ?" I think the OP is under the impression that there is agreement (among "we") on what should be a conjecture and what shouldn't be. A conjecture is just an opinion about what might be true, and if enough prominent people work on resolving it then it may gain some notability, but that doesn't imply everyone agrees with the opinion. On a related note, I was just looking at the article on Landau's problems, a set of 4 conjectures proposed 110 years ago, though most date from earlier. Landau stated they were "unattackable at the present state of mathematics". Math is still 0 for 4 on them, and while some have been "attacked" since, and partial results obtained, I think it's fair to say that no beachheads have been established and significant progress is still doubtful. That's just four of the 60+ unsolved problems on prime numbers in List of unsolved problems in mathematics. --RDBury (talk) 10:13, 15 November 2022 (UTC)[reply]

PS. I just did a ball-park estimate on the probability: Assuming that a hit for p is random with probability 1/p then the probability of getting no hits for any p < 300×10¹⁵ is about 1%. This is low, but so low as to strain credulity when you consider that if there had been a hit then this discussion would probably not be occurring. --RDBury (talk) 10:44, 15 November 2022 (UTC)[reply]

@RDBury: I guess you meant to say "not so low as to strain credulity". I view this as an example of the multiple comparisons problem. We look for primes in lots of sequences. Some have a little more primes so far than expected, some a little fewer. This one has 3 fewer which just happens to place it at 0 and cause increased interest. To be more specific than my first post, it is known that if p is prime then a certain number

F_{\alpha (p)}

is divisible by p. If it's also divisible by p² then p is called a Wall–Sun–Sun prime.

{\frac {F_{\alpha (p)}}{p}}

is an integer so the question is whether this integer is divisible by p. It appears random so the estimated chance is 1/p. If we try to divide it by p then there are p possible values of the remainder: 0 to p-1. These remainders look random. The wanted 0 value is the one I talked about in the first post. We haven't seen it yet but so what? PrimeHunter (talk) 19:55, 15 November 2022 (UTC)[reply]

World-class mathematicians will only publicly venture conjectures as such if they have a strong hunch, based on their experience and expertise, that its dictum is true. It is a reasonable question to ask what makes them think so, but they themselves may not have the answer, just like a chess grandmaster may "feel" a move is strong while unable to explain cogently why. --Lambiam 10:32, 15 November 2022 (UTC)[reply]

Mathematical conjectures in general can be based on different things like the intuition of a mathematician. When mathematicians state conjectures about prime numbers, it is usually based on specific heuristics of the form: "If we assume these numbers behave randomly (possibly apart from a known pattern we compensate for) then the estimate becomes ...". The primality chance 1/log(n) of a random n is central in such conjectures. PrimeHunter (talk) 19:33, 15 November 2022 (UTC)[reply]

Such statements about expected chances can be made formal. Let

r_{n}\,{\hat {=}}\,\lceil {\sqrt {n}}\rceil .

Then the probability that a number drawn at random from the discrete uniform distribution on the interval

[n-r_{n},n+r_{n}]

is prime is asymptotically equal to

{\tfrac {1}{\log n}}.

--Lambiam 10:20, 15 November 2022 (UTC)[reply]

November 15

Largest prime with a known index

While the largest known prime number is 2^82,589,933 − 1, its exact index in the sequence of all prime numbers is not known. What is the largest prime number for which its exact index in the sequence of all prime numbers is known? GeoffreyT2000 (talk) 16:20, 15 November 2022 (UTC)[reply]

It's not a simple question, it seems, but this Reddit thread from 2016 discusses it and may lead you to some interesting places. This Quora thread from the same time period also discusses the matter, and seems to have some better evidence for a prime that meets your definition. The second thread cites work by Tomás Oliveira e Silva that has set a lower bound for such a sequence of primes somewhere on the order of 4 x 10¹⁸. --Jayron 32 17:11, 15 November 2022 (UTC)[reply]

Those sources are about the largest prime for which all smaller primes have been identified. The question only asked for the index, meaning we only have to know the number of smaller primes. There are faster ways to compute this number than to compute and count each of the primes. See Prime-counting function. The current record is pi(10²⁹) = 1,520,698,109,714,272,166,094,258,063.[6] To make it an index of a prime, compute the largest prime below 10²⁹ which is 10²⁹−27. You could technically improve the record slightly by computing and counting the next few primes above 10²⁹ but that is not considered an official record. Any such "record" could keep being broken in a small fraction of a second if you assume the previous count was correct. PrimeHunter (talk) 19:22, 15 November 2022 (UTC)[reply]

Well, there you go. Thanks for the clarification. --Jayron 32 19:33, 15 November 2022 (UTC)[reply]

November 16

Vetting a new result?

This is about Yitang Zhang's recent preprint of a new result about Laundau-Siegel zeroes, that (if correct) makes some progress towards resolving the Riemann hypothesis. You can find more info with a web search if you don't know about it. I don't claim to understand the proof, but a look at the paper (and similar comments on mathoverflow) indicate that it's basically a brute force calculation. You start by bounding some quantities with funny-looking constants like 0.504 or 2022, crunch through 100 pages of formulas, and end up with a thing that shows what you wanted to prove. It's like in calculus where you pick constants at the beginning to make the thing at the end be less than epsilon. That is, it doesn't depend on fancy abstract or subtle considerations all that much.

The preprint has been up for a couple of weeks now, so I'm surprised that nobody seems to be commenting on it. This is different from some other preprints claimed to solve significant open problems, where there is immediately a bunch of attention, typically followed by someone finding a mistake; or even Zhang's earlier now-famous result on prime gaps, which immediately launched a swirl of rumors with big shots saying they had checked the paper and it looked good, modulo the possibility of some subtle error needing closer scrutiny to notice. Green and Tao's theorem on arithmetic progressions in the primes was maybe before my time, but I think it also kicked up a stir with lots of people examining the preprint.

Anyone know if the current silence about Zhang's new paper is typical, with the other cases being exceptional? Has anyone looked at or heard anything about any issues the paper might have? Thanks. 2601:648:8201:5E50:0:0:0:DD22 (talk) 02:39, 16 November 2022 (UTC)[reply]