In other words:

1. For given (disjoint) sets is it accepted to call a given relation R "symmetric", when: for all if then

2. For given (disjoint) sets is it accepted to call a given relation R "transitive", when: for all if and then

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 is a relation between and , the consequent of item 1 does not make much sense. The statement "for all if then " by itself normally already implies that is homogeneous.

2. This is more complicated. Here relation is apparently between and 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 has the property that 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 , and the next one, not shown, would be the equator at those visible are at Then one might guess (but it remains a guess) that  ​‑‑Lambiam 23:13, 7 May 2025 (UTC)[reply]

We know about parity. However, has anyone ever proposed a term like tri-arity that means the classification of a number n by whether n, n-1, or n+1 is a multiple of 3?? (Every integer belongs in exactly one of these 3 categories; they have properties that parallel being even and odd; the only difference is that they relate to 3 the way even and odd numbers relate to 2.) Georgia guy (talk) 17:28, 9 May 2025 (UTC)[reply]

The parity of a number is (in one-one correspondence with) its residue modulo 2, so the question can be rephrased as, is there a snappy term for "residue modulo 3"? I have never seen one. For the ternary analogon of a parity check, we find the terms "modulo-3 residue check" and "residue modulo-3 check" in the literature.  ​‑‑Lambiam 20:04, 9 May 2025 (UTC)[reply]

This question reminds me a bit of a sort of inquiry that might have been popular in the 17th–19th centuries, among figures like Henry More and Charles Saunders Peirce, who were fond of making up names for things that it wasn't clear needed names, or were even well-specified things. Of course the residue of an integer mod 3 is a well-specified thing, but it isn't clear to me that it needs a name. --Trovatore (talk) 20:33, 9 May 2025 (UTC) [reply]

It may be worth mentioning that not all residue systems are created equal; residue modulo 2 is used much more frequently and in a wider variety of ways than other residue systems. For example permutations can be assigned "even" or "odd" parity, with the rules of parity being preserved under composition, in other words even*even=even, even*odd=odd, odd*even=odd, odd*odd=even. There is no mod 3 way of doing this, nor is there for any higher order modulus. So "parity" gets its own special name due simply to the frequency of situations in which it appears. If triarity appeared as frequently then it might get official status as well. --RDBury (talk) 07:50, 10 May 2025 (UTC)[reply]

• RDBury, assuming the operation is addition; yes there is. Let's call the 3 kinds of numbers black (divisible by 3,) red (one more than a multiple of 3,) and green (one less than a multiple of 3.) So the answer can be:

• Black+black=black
• Black+red=red
• Black+green=green
• Red+black=red
• Red+red=green
• Red+green=black
• Green+black=green
• Green+red=black
• Green+green=red

(Please note that I'm just using these names for convenience; any response you have must hold regardless of what names I'm using for the 3 kinds of numbers.) Georgia guy (talk) 12:36, 10 May 2025 (UTC)[reply]

I was confused at first by RDBury's response as well, but look closer; he's talking specifically about permutations. There's a theorem that every (finite) permutation can be decomposed into a composition of permutations where you swap two elements at a time. The decomposition is not unique, but the number of pair-swaps in it always has the same parity. --Trovatore (talk) 17:57, 10 May 2025 (UTC)[reply]

More information: Permutation#Parity of a permutation. --Trovatore (talk) 17:59, 10 May 2025 (UTC) Oh actually we have a whole article: Parity of a permutation. --Trovatore (talk) 18:10, 10 May 2025 (UTC)[reply]

Here is a further exposition. Let stand for the set of permutations on a finite set of size and consider a function – that is, a way of assigning one of these three colours to each of these permutations – such that the identity is satisfied (using the addition table above). We can then prove that it assigns the colour to all permutations.

(Proof sketch: The identity permutation gives rise to the equation which is only possible if Then each involution gives rise to the equation which is only possible if The involutions generate the whole group )

On the other hand, if we use:

an addition-respecting assignment is possible, using both colours.  ​‑‑Lambiam 18:50, 10 May 2025 (UTC)[reply]