In other words:

1. For given (disjoint) sets ${\displaystyle X,Y,}$is it accepted to call a given relation R "symmetric", when: for all ${\displaystyle x\in X,y\in Y,}$if ${\displaystyle xRy}$then ${\displaystyle yRx?}$

2. For given (disjoint) sets ${\displaystyle X,Y,Z,}$is it accepted to call a given relation R "transitive", when: for all ${\displaystyle x\in X,y\in Y,z\in Z,}$if ${\displaystyle xRy}$and ${\displaystyle yRz}$then ${\displaystyle xRz?}$

However, if the term "symmetric/transitive" is not accepted for these heterogeneous relations, then do you have in mind a better name to describe them? HOTmag (talk) 10:06, 28 April 2025 (UTC)[reply]

1. Unless ${\displaystyle R}$is a relation between ${\displaystyle X\sqcup Y}$and ${\displaystyle X\sqcup Y}$, the consequent ${\displaystyle yRx}$of item 1 does not make much sense. The statement "for all ${\displaystyle x\in X,y\in Y,}$if ${\displaystyle xRy}$then ${\displaystyle yRx}$" by itself normally already implies that ${\displaystyle R}$is homogeneous.

2. This is more complicated. Here relation ${\displaystyle R}$is apparently between ${\displaystyle X\sqcup Y}$and ${\displaystyle Y\sqcup Z,}$so it is not necessarily homogeneous. There is no higher mathematical authority ruling which abuses of language are condoned and which are proscribed. Personally, I would have no qualms declaring my non-homogeneous relation satisfying this condition to be transitive, but I can give no guarantee that this might not offend some lesser god. However, it may be wise to make the reader aware of the fact that the situation is not quite normal. A transitive homogeneous relation ${\displaystyle R}$has the property that ${\displaystyle R^{2}\subseteq R,}$which one can even use as the definition of transitivity, but for a heterogeneous relation this makes no sense.  ​‑‑Lambiam 14:54, 28 April 2025 (UTC)[reply]

Thank you. HOTmag (talk) 17:50, 28 April 2025 (UTC)[reply]

Ulam numbers empirically seem to have a density of about 0.07. However, this paper in ArXiv says that they have zero density. I can't find that it has been published anywhere. What is the status of the density of Ulam numbers? Bubba73 ^{You talkin' to me?} 02:13, 29 April 2025 (UTC)[reply]

I guess it is open. Presumably, the paper was submitted to a journal. When the referees find holes in a purported proof, this is generally not made public, so we may simply not hear more about this. The paper on arXiv was originally submitted in 2020, but by now it has reached version 11, from 27 August 2023. The author identifies himself as "an ardent theory builder with very outlandish mathematical ideas drawn from intuition".^[1] Five years ago, a co-author of his on several papers^[2] published a paper "An Elementary Proof of the Twin Prime Conjecture",^[3] yet the consensus among number theorists appears to be this problem is also still open. The proof was published in a rather unknown journal. One would think the author submitted such an important result first to prestigious journals in number theory, so this strongly suggests it was rejected by these.  ​‑‑Lambiam 07:58, 29 April 2025 (UTC)[reply]

The Ulam number sequence in OEIS (OEIS:A002858) also doesn't seem to mention anything about an acceptance of the density 0 proof; rather, it just indicates that Stanisław Ulam himself believed the density to be 0, while empirical evidence suggests a density around 0.074. GalacticShoe (talk) 13:11, 29 April 2025 (UTC)[reply]

Based on the numbers up to 10¹², they appear to have a positive density. I got this data from Exploring the Beauty of Fascinating Numbers, by Shyam S. Gupta. The y-axis is density and the x-axis is the log₁₀ of the upper value.

Out to 10¹² it behaves as if the density is converging to about 0.074.... Bubba73 ^{You talkin' to me?} 01:34, 1 May 2025 (UTC)[reply]

The author is some flavor of crank, see [4] and Talk:Prime-counting function#Prime index function 100.36.106.199 (talk) 12:11, 3 May 2025 (UTC)[reply]

I don't see which one you are talking about. Is it the one who wrote that ArXiv article about Ulam numbers having zero density? Bubba73 ^{You talkin' to me?} 05:04, 5 May 2025 (UTC) -- oh, Theophilus Agama. Bubba73 ^{You talkin' to me?} 05:05, 5 May 2025 (UTC)[reply]

Yes. 100.36.106.199 (talk) 10:10, 6 May 2025 (UTC)[reply]

File:IIHF World Junior Championship.png Sagittarian Milky Way (talk) 16:44, 7 May 2025 (UTC)[reply]

All conventional projections of a sphere to the plane have a circular outline. The outline in the logo does not have a constant curvature; it is some artistic fantasy projection. One can therefore only guess at the latitudes of the parallels. If the angular distance between successive parallels is a constant ${\displaystyle \delta }$, and the next one, not shown, would be the equator at ${\displaystyle 0^{\circ },}$those visible are at ${\displaystyle \delta ,2\delta ,3\delta .}$Then one might guess (but it remains a guess) that ${\displaystyle 4\delta =90^{\circ }.}$ ​‑‑Lambiam 23:13, 7 May 2025 (UTC)[reply]