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December 19

Who is the following unknown?

When asked "WHO IS YOUR X?" (X still being unknown to me but is known to the respondents), here are the answers I get:

A answers: "A"

B answers: "C"

C answers: "C"

D answers: "F"

E answers: "F"

F answers: "F"

To sum up, the special phenomenon here is that, everybody has their own X (usually), and if any respondent points at another respondent as the first respondent's X, then the other respondent must point at themself as their X.

I wonder who the unknown X may be, when I only know that X is a natural example from everyday life. I thought about a couple of examples, but none of them are satisfactory, as follows:

X is the leader of one's political party, or X is one's mayor, and the like, but all of these examples attribute some kind of leadership or superiority to X, whereas I'm not interested in this kind of solution - involving any superiority of X.

Here is a second solution I thought about: X is the first (or last) person born in the year/month the respondent was born, and the like. But this solution involves some kind of order (in which there is a "first person" and a "last person"), whereas I'm not interested in this kind of solution - involving any order.

Btw, I've published this question also at the Miscellaneous desk, because this question is about everyday life, but now I decide to publish this question also here, because it's indirectly related to a well known topic in Math. 79.177.151.182 (talk) 13:27, 19 December 2024 (UTC)

Head of household comes to mind as a fairly natural one. The colours then correspond to different households which can be just one person. One objection is that "head of household" is a fairy traditional concept. With marriage equality now being the norm it's perhaps outdated. --2A04:4A43:909F:F9FF:397E:BBF9:E80B:CB36 (talk) 15:11, 19 December 2024 (UTC)

I have already referred to this kind of solution, in the example of "my mayor", see above why this solution is not satisfactory. 79.177.151.182 (talk) 15:31, 19 December 2024 (UTC)

The question has been resolved at the Miscellaneous reference desk.

Resolved

79.177.151.182 (talk) 15:48, 19 December 2024 (UTC)

X may well be 'the oldest living person of your ancestry'. --CiaPan (talk) 20:46, 19 December 2024 (UTC)

Resolved or not, let's try to analyze this mathematically. Given is some set $S$ and some function $f:S\to S.$ For the example, $S=\{{\mathsf {A}},{\mathsf {B}},{\mathsf {C}},{\mathsf {D}},{\mathsf {E}},{\mathsf {F}}\},$ with $f({\mathsf {A}})={\mathsf {A}},$ $f({\mathsf {B}})=f({\mathsf {C}})={\mathsf {C}},$ $f({\mathsf {D}})=f({\mathsf {E}})=f({\mathsf {F}})={\mathsf {F}}.$

Knowing that "everybody has their own X (usually)", we can normalize the unusual situation that function $f$ might not be total in two ways. The first is to restrict the set $S$ to the domain of $f,$ that is, the set of elements on which $f$ is defined. This is possible because of the condition that $f(u)=v$ implies $f(v)=v,$ so this does not introduce an undue limitation of the range of $f.$ The second approach is to postulate that $f(u)=u$ whenever $f(u)$ might otherwise be undefined. Which of these two approaches is chosen makes no essential difference.

Let $T=f(S)$ be the range of $f$, given by:

$T=\{v\in S:(\exists u\in S:f(u)=v)\}.$

Clearly, if $f(v)=v,$ we have $v\in T.$ We know, conversely, that $v\in T$ implies $f(v)=v.$

Let us also consider the inverse image of $f$, given by:

$f^{-1}(v)=\{u\in S:f(u)=v\}.$

Suppose that $f^{-1}(v)\neq \emptyset .$ This means that there exists some $u\in f^{-1}(v),$ which in turns means that $f(u)=v.$ But then we know that $f(v)=v.$ Combining this, we have,

$f(v)=v\Leftrightarrow v\in T\Leftrightarrow f^{-1}(v)\neq \emptyset .$

The inverse-image function restricted to $T,$ to which we assign the typing

$f^{-1}:T\to \mathbb {P} (S),$

now induces a partitioning of $S$ into non-empty, mutually disjoint subsets, which means they are the classes of an equivalence relation. Each class has its own unique representative, which is the single element of the class that is also a member of $T$. The equivalence relation can be expressed formally by

$u\sim v\Leftrightarrow f(u)=f(v),$

and the representatives are the fixed points of $f.$

Applying this to the original example, $T=\{{\mathsf {A}},{\mathsf {C}},{\mathsf {F}}\},$ and the equivalence classes are:

• $\{{\mathsf {A}}\},$ with representative ${\mathsf {A}},$
• $\{{\mathsf {B}},{\mathsf {C}}\},$ with representative ${\mathsf {C}},$ and
• $\{{\mathsf {D}},{\mathsf {E}},{\mathsf {F}}\},$ with representative ${\mathsf {F}}.$

Conversely, any partitioning of a set defines an equivalence relation; together with the selection of a representative for each equivalence class, this gives an instance of the situation defined in the question.  --Lambiam 20:47, 19 December 2024 (UTC)

FWIW, the number of such objects on a set of size n is given by OEIS: A000248, and that page has a number of other combinatorial interpretations. If you ignore the selection of a representative for each class, you get the Bell numbers. --RDBury (talk) 00:35, 21 December 2024 (UTC)

December 20

Give a base b and two base b digits x and z, must there be a base b digits y such that the 3-digit number xyz in base b is prime?

Give a base b and two base b digits x and z (x is not 0, z is coprime to b), must there be a base b digits y such that the 3-digit number xyz in base b is prime? 1.165.207.39 (talk) 02:10, 20 December 2024 (UTC)

In base 5, $3Y1_{5}=76+5Y$ is composite for all base-5 Y. GalacticShoe (talk) 03:39, 20 December 2024 (UTC)

$3Y1_{7}$ also offers a counterexample. While there are many counterexamples for most odd bases, I did not find any for even bases.  --Lambiam 09:58, 20 December 2024 (UTC)

December 23

Is it possible to make Twisted Edwards curve birationally equivalent to twisted weirestrass curves ?

Is there an equation fo converting a twisted Edwards curve into a tiwsted weierstrass form ? 2A01:E0A:401:A7C0:6D06:298B:1495:F479 (talk) 04:12, 23 December 2024 (UTC)

According to Montgomery curve § Equivalence with twisted Edwards curves, every twisted Edwards curve is birationally equivalent to a Montgomery curve, while Montgomery curve § Equivalence with Weierstrass curves gives a way to transform a Montgomery curve to an elliptic curve in Weierstrass form. I don't see a definition of "twisted Weierstrass", so I don't know if you can give an extra twist in the process. Perhaps this paper, "Efficient Pairing Computation on Twisted Weierstrass Curves" provides the answer; its abstract promises: "In this paper, we construct the twists of twisted Edwards curves in Weierstrass form."  --Lambiam 10:35, 23 December 2024 (UTC)

December 24

How did the Romans do engineering calculations?

The Romans did some impressive engineering. Engineers today use a lot of mathematical calculations when designing stuff. Calculations using Roman numerals strike me as being close to impossible. What did the Romans do? HiLo48 (talk) 05:50, 24 December 2024 (UTC)

The kind of engineering calculations that might have been relevant would mostly have been about statics – specifically the equilibrium of forces acting on a construction, and the ability of the design to withstand these forces, given its dimensions and the mechanical properties of the materials used in the construction, such as density, modulus of elasticity, shear modulus, Young modulus, fracture strength and ultimate tensile strength. In Roman times, only the simplest aspects of all this were understood mathematically, namely the statics of a construction in which all forces work in the same plane, without torque, and the components are perfectly rigid. The notion of assigning a numerical magnitude to these moduli and strengths did not exist, which anyway did not correspond to precisely defined, well-understood concepts. Therefore, engineering was not a science but an art, mainly based on experience in combination with testing on physical models. Any calculations would mostly have been for the amounts and dimensions of construction materials (and the cost thereof), requiring a relatively small number of additions and multiplications.  --Lambiam 19:03, 24 December 2024 (UTC)

Calculating with Roman Numerals might seem impossible, but in some ways it's simpler than our positional system; there are only so many symbols commonly used, and only so many ways to add and multiply them. Once you know all those ways you efficiently can do calculations with them, up to the limits imposed by the system.

And for a lot of things they relied on experience. Romans knew how to build circular arches, but rather than do calculations to build larger arches, or ones with more efficient shapes they used many small circular ones which they knew worked, stacked side by side and sometimes on top of each other. See e.g. any Roman aqueduct or the Colosseum. For materials they probably produce them on site or close by as they're needed.--2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C (talk) 20:12, 24 December 2024 (UTC)

To add to the above, it wasn't really the Romans that were great innovators in science, engineering, etc.. I think more innovation and discovery took place in Ancient Greece and Ancient Egypt. Certainly Greece as we have a record of that. Egypt it's more that they were building on such a monumental scale, as scale no-one came close to repeating until very recently.

Romans were military geniuses. They conquered Greece, and Egypt, and Carthage, and Gaul, and Britannia, and everywhere in between. They then built forts, towns, cities and infrastructure throughout their empire. They built so much so widely that a lot of it still stands. But individually a lot of it isn't technically impressive; instead it's using a few simple patterns over and over again.--2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C (talk) 12:09, 25 December 2024 (UTC)

See Roman abacus. catslash (talk) 22:03, 25 December 2024 (UTC)

It has to be said that in Roman times calculations for architecture were mostly graphical, geometrical, mechanical, rather than numeric. In fact, from the perspective of an ancient architect, it would make little sense translating geometrical figures into numbers, making numeric calculations, then translating them into geometrical figures again. Numerals become widely used tools only later, e.g. with the invention of Analytic Geometry (by Descartes), and with logarithms (Napier); and all these great mathematical innovations happened to be so useful also thanks to the previous invention of the printing press by Gutenberg --it's easier to transmit information by numbers than by geometric constructions. One may even argue that the invention of the printing press itself was the main reason to seek for an adequate efficient notation for real numbers (achieved by Stevinus). pma 21:22, 1 January 2025 (UTC)

This is a great answer! Tito Omburo (talk) 21:38, 1 January 2025 (UTC)

Architectural calculations, mentioned by Vitruvius, are not what I think of as engineering calculations. The former kind is about form. The latter kind should provide answers questions about structural behaviour, like, "Will these walls be able to withstand the outward force of the dome?" Can such questions be addressed with non-numerical calculations?  --Lambiam 23:40, 1 January 2025 (UTC)

Are these sequences mod any natural number n periodic?

The period of Fibonacci number mod n is the Pisano period of n, but are these sequences mod any natural number n also periodic like Fibonacci number mod n?

1. Lucas number
2. Pell number
3. Tribonacci number
4. Tetranacci number
5. Newman–Shanks–Williams number
6. Padovan sequence
7. Perrin number
8. Narayana sequence
9. Motzkin number
10. Bell number
11. Fubini number
12. Euler zigzag number
13. Partition number OEIS: A000041
14. Distinct partition number OEIS: A000009

42.76.153.22 (talk) 06:06, 24 December 2024 (UTC)

For 1. through 4., see Pisano period#Generalizations. Although 5. through 8. are not explicitly listed, I'm pretty sure the same argument applies for their periodicity as well. GalacticShoe (talk) 07:05, 24 December 2024 (UTC)

For 13, I'm not sure, but I think the partition function is not periodic modulo any nontrivial number, to the point that the few congruences that are satisfied by the function are also very notable, e.g. Ramanujan's congruences. GalacticShoe (talk) 07:23, 24 December 2024 (UTC)

It's possible that a sequences is eventually periodic but not periodic from the start, for example powers of 2 are periodic for any odd n, but for n=2 the sequence is 1, 0, 0, ..., which is only periodic starting with the second entry. In other words a sequence can be become periodic without being pure periodic. A finiteness argument shows that 1-8 are at least eventually periodic, but I don't think it works for the rest. (Pollard's rho algorithm uses this finiteness argument as well.) It says in the article that Bell numbers are periodic mod n for any prime n, but the status for composite n is unclear, at least from the article. Btw, Catalan numbers not on the list?. --RDBury (talk) 08:08, 24 December 2024 (UTC)

If only the periodicity of Bell numbers modulo prime powers were known, then periodicity for all modulos would immediately result from the Chinese remainder theorem. GalacticShoe (talk) 01:29, 29 December 2024 (UTC)

PS. 1-8 are pure periodic. In general, if the recurrence can be written in the form F(k) = (some polynomial in F(k-1), F(k-2), ... F(k-d+1) ) ± F(k-d), then F is pure periodic. The reason is that you can solve for F(k-d) and carry out the recursion backwards starting from where the sequence becomes periodic. Since the previous entries are uniquely determined they must follow the same periodic pattern as the rest of the sequence. If the coefficient of F(k-d) is not ±1 then this argument fails and the sequence can be pre-periodic but not pure periodic, at least when n is not relatively prime to the coefficient. --RDBury (talk) 18:12, 24 December 2024 (UTC)

Ah, what was tripping me up was showing pure periodicity, recursing backwards completely slipped my mind. Thanks for the writeup! GalacticShoe (talk) 18:31, 24 December 2024 (UTC)

good questions. What about TREE(n) mod k, for arbitrary fixed k?Rich (talk) 23:03, 27 December 2024 (UTC)

Link: TREE function. There are a lot of sequences like this where exact values aren't known, Ramsey numbers are another example. It helps if there is a relatively simple recursion defining the sequence. --RDBury (talk) 11:50, 28 December 2024 (UTC)

For 9. Motzkin numbers are not periodic mod 2. Motzkin numbers mod 2 are OEIS:A039963, which is OEIS:A035263 with each term repeated (i.e. $0\rightarrow 00,1\rightarrow 11$.) OEIS:A035263 in turn is the sequence that results when one starts with the string $1$ and successively maps $0\rightarrow 11,1\rightarrow 10$ (e.g. $110\rightarrow 101011$.) It is clear that if OEIS:A035263 were periodic with period $n$, then the periodic string $X$ of length $n$ would need to map to string $XX$, but this is impossible as the last character of $X$ is always the opposite of the last character of the map applied to $X$. Thus OEIS:A035263 is nonperiodic, and neither is OEIS:A039963. GalacticShoe (talk) 01:08, 29 December 2024 (UTC)

This paper (linked from OEIS) goes into more detail on Motzkin numbers. I gather the sequence might be called quasi-periodic, but I can't find an article that matches this situation exactly. A003849 is in the same vein. --RDBury (talk) 16:59, 30 December 2024 (UTC)

For 11., the article seems to suggest that Fubini numbers are eventually periodic modulo any prime power. I'm pretty sure this means that they the numbers eventually periodic mod any number $n$, since the lcm of the eventual periods modulo all prime power divisors of $n$ should correspond to the eventual period modulo $n$ itself, with the remainders being obtainable through the Chinese remainder theorem. However, the wording also seems to suggest that periodicity modulo arbitrary $n$ is still conjectural, so I'm not sure. GalacticShoe (talk) 02:44, 29 December 2024 (UTC)

You have answered all questions except 12 and 14, and 9 and 13 are the only two sequences which are not periodic mod n (except trivial n=1), 12 (Euler zigzag numbers) is (sequence A000111 in the OEIS), which seems to be periodic mod n like 10 (Bell numbers) (sequence A000110 in the OEIS) and 11 (Fubini numbers) (sequence A000670 in the OEIS), but all of these three sequences need prove, besides, 14 (Distinct partition numbers) (sequence A000009 in the OEIS) seems to be like 13 (Partition numbers) (sequence A000041 in the OEIS), i.e. not periodic mod n. 1.165.199.71 (talk) 02:27, 31 December 2024 (UTC)

December 31

Generating a point on the Y axis from regular pentagon with point on X axis

For a $K$ consisting of points in R^2, define the function B such that $B(K)$ as the Union of $K$ and all points which can be produced in the following way. For each set of points A, B, C, & D from $K$ all different so that no three of A, B, C & D are co-linear. E is the point (if it exists) where ABE are colinear and CDE are co-linear.

If $K_{0}$ = the vertices of a regular Pentagon centered at 0,0 with one vertex at (1,0), does there exist N such that $B^{N}(K_{0})$ includes any point of the form (0, y)? (extending the question to any N-gon, with N odd) Naraht (talk) 05:16, 31 December 2024 (UTC)

I think you meant to write $B^{N}(K_{0}).$  --Lambiam 07:55, 31 December 2024 (UTC)

Changed to use the Math.Naraht (talk) 14:37, 31 December 2024 (UTC)

I'm not 100% sure I understand the problem, but try this: Label the vertices of the original pentagon, starting with (1, 0), as A, B, C, D, E. You can construct a second point on the x-axis as the intersection of BD and CE; call this A'. Similarly construct B', C', D', E', to get another, smaller, regular pentagon centered at the origin and with the opposite orientation from the the original pentagon. All the lines AA', BB', CC', DD', EE' intersect at the origin, so you can construct (0, 0) as the intersection of any pair of these lines. The question didn't say y could not be 0, so the answer is yes, with N=2.

There is some theory developed on "straightedge only construction", in particular the Poncelet–Steiner theorem, which states any construction possible with a compass and straightedge can be constructed with a straightedge alone if you are given a single circle with its center. In this case you're given a finite set of points instead of a circle, and I don't know if there is much theory developed for that. --RDBury (talk) 13:12, 1 January 2025 (UTC)

Here is an easy way to describe the construction of pentagon A'B'C'D'E'. The diagonals of pentagon ABCDE form a pentagram. The smaller pentagon is obtained by removing the five pointy protrusions of this pentagram.  --Lambiam 16:53, 1 January 2025 (UTC)

January 1

What is the first number not contained in M136279841?

See (sequence A268068 in the OEIS), the first number not contained in M74207281 is 1000003, but what is What is the first number not contained in M136279841 (the currently largest known prime)? 61.224.131.231 (talk) 03:34, 1 January 2025 (UTC)

The corresponding sequence (11, 3, 8, 7, 6, 10, 4, 9, 1, 5, 25, 31, 39, ...) is not in OEIS. Finding the answer to your question requires an inordinate amount of computing power. The decimal expansion of this Mersenne prime has some 41 million digits, all of which need to be computed. If this is to be done in a reasonable amount of time, the computation will need the random access storage of at least some 22 million digits.  --Lambiam 10:10, 1 January 2025 (UTC)

I'm not seeing that this question requires an inordinate amount of computing power to answer. 41 million characters is not a very large set of data. Almost all modern computers have several gigabytes of memory, so 41 million characters will easily fit in memory. I took the digits of M136279841 from https://www.mersenne.org/primes/digits/M136279841.zip and searched them myself, which took a few minutes on a consumer grade PC. If I have not made a mistake, the first number that does not appear is 1000030. The next few numbers that do not appear are 1000073, 1000107, 1000143, 1000156, 1000219, 1000232, 1000236, 1000329, 1000393, 1000431, 1000458, 1000489, 1000511, 1000514, 1000520, 1000529, etc. CodeTalker (talk) 03:59, 2 January 2025 (UTC)