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Background: Although this site is most-often used for specific one-off questions, many of the highest scored questions (also on MathOverflow), which gather a lot of attention to the site are about informal lists. So, in the theme of, but in contrast to, past questions like:

Not especially famous, long-open problems which anyone can understand,

Open Problems list

Question: What are some conjectures of your own?

Specifically, I'm looking for conjectures / problems that

  • Have taken up your interest especially (perhaps it was made aware to you, and / or few others),
  • Not particularly in public knowledge,
  • And that you think are novel (and are comfortable sharing!).

Basically, problems that you find yourself coming back to, that a fellow bored mathematician would find interesting too. All fields welcome.

Note 06/09/23: Let me be clear to appease @user1729 who was worried that people were answering in the form of "I'm great because I came up with this amazing conjecture!", this is not for self-gratification, but to share a problem which is personal to you, for the sake of sharing not because it's yours.

Motivation: If you couldn't already tell, I enjoy recreational mathematics, but all my current problems either bore me or have hit a dead end. I enjoy looking at the big unsolved problems (well-known or not), but sometimes I wish I had a list of very small open-problems which are just as novel, but not as hard or daunting. I'm sure others would appreciate such a list.


As a show of good faith (that I'm genuinely interested in looking into your problems recreationally rather than farming points), this is a Community Wiki question (so accumulated upvotes don't inflate my reputation, if any).


Edit: After discussing the reasons for past-closure on meta here, it's come to my attention that some have interpreted this question to put emphasis on the fact that these conjectures are yours, which is not the point. My word-choice is instead to emphasize the fact that the universe of math is so broad and intricate, that I imagine many people have stumbled upon miniature problems which others have not, and that is what I would like to provide this space for. Nobody "owns" conjecture, but many have discovered conjectures that are not known beyond themselves. This is the place to share them.

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    $\begingroup$ I am not happy with the closure of this question. See What's the deal with closing "What are some conjectures of your own"? on MetaMSE. $\endgroup$
    – Graviton
    Jun 9, 2023 at 1:40
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    $\begingroup$ @Hopeful: Again, why obfuscate the search for these conjectures here, where they cannot be answered? It would be better for those who have these conjectures to post questions of their own, and possibly get answers. As fun as this collection of conjectures may be, for people searching for the answers to these conjectures, this collection will impede their search. To me, this is why this question was closed twice. Placing a bounty not only hinders community moderation of a controversial question, but also attracts more "answers" that should probably be posted as separate questions instead. $\endgroup$
    – robjohn
    Jun 12, 2023 at 9:21
  • $\begingroup$ You should add to your bullet points: "That would be significant finding, if true." -- so you can get people to take a little more risk with their secret conjectures. $\endgroup$
    – Marcos
    Jul 26, 2023 at 23:05

27 Answers 27

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While I was in university I conjectured that the numbers $2, 4, 8, 64$ and $2048$ are the only powers of $2$ whose base-$10$ representation consists of solely even digits. I wrote an IBM-$360$ Assembler program to search for more; it didn’t find any.

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    $\begingroup$ See OEIS sequence A068994. $\endgroup$ Jun 7, 2023 at 23:43
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    $\begingroup$ It is completely mindblowing to me that this is open $\endgroup$ Jun 8, 2023 at 0:40
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    $\begingroup$ I wonder if this has anything to do with an idea that the existence of an odd digit increases the probability that the next power of 2 will have an odd digit. So, for larger powers with many odds, they will stay there forever. $\endgroup$ Jun 8, 2023 at 0:48
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    $\begingroup$ @DavisYoshida Usually these base-10 specific claims are more difficult than you'd think, because the choice of base is mathematically pretty arbitrary. $\endgroup$
    – Brady Gilg
    Jun 8, 2023 at 17:17
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    $\begingroup$ An equivalent form of this conjecture is: When are there powers of 2 made up solely of digits less than 5? (From the original conjecture, factor out a 2 and see what's left.) If we pretend that the digits are distributed uniformly randomly, this is effectively asking to flip a coin and land heads every time - once for each of its $\lfloor n \log_{10} 2\rfloor +1$ digits. For $2^n$ that's a probability of $\frac{1}{2^{\lfloor n \log_{10} 2\rfloor +1}}$. Increasingly unlikely, but hey, we're flipping infinitely often so who knows? $\endgroup$
    – Merosity
    Jun 9, 2023 at 6:34
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My most recent conjecture is that $$f(n)=\sum_{j=1}^n j!^2=1!^2+2!^2+\cdots +n!^2$$ has no prime factor $p\le n$ for any positive integer $n$. This was inspired by a post about this sum where an other user conjectured that $f(n)$ is squarefree for every positive integer $n$ which also seems to be the case.

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    $\begingroup$ $1248829$ divides $f(1248829)$, see OEIS sequence A290250. $\endgroup$ Jun 8, 2023 at 1:18
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    $\begingroup$ @JamesRettie Good find. How on earth did u find that on the oeis, like what search terms did u used? $\endgroup$ Jun 18, 2023 at 18:18
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    $\begingroup$ @TitoPiezasIII I did a GOOGLE search for: sum of squares of factorials. On the first page of the search results I got a link to a Wolfram article called: Factorial Sums. That Wolfram article contains a link to the OEIS article that James Rettie mentioned. $\endgroup$ Jun 21, 2023 at 6:40
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    $\begingroup$ @WillOctagonGibson: Ah, our friend Google. I knew the search capabilities of OEIS are not that powerful. :) $\endgroup$ Jun 21, 2023 at 6:44
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I have a conjecture about Lie algebras and their faithful representations by matrices. How big do you have to choose the matrices for a given Lie algebra of dimension $n$? For example, the Heisenberg Lie algebra needs $3\times 3$ matrices, and no smaller ones will do.

More formally, let $\mu(\mathfrak{g})$ be the minimal dimension of a faithful $\mathfrak{g}$-module for $\mathfrak{g}$. By Ado's theorem his is an integer valued invariant. My conjecture is the following:

Conjecture: Let $A(n)$ denote the maximum of $\mu(\mathfrak{g})$ where $\mathfrak{g}$ ranges over all $n$-dimensional Lie algebras. Then the growth of $A(n)$ is exponential.

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  • $\begingroup$ Thanks. What I call in my book an "Optimistic hypothesis" is probably false but if it were true, then even the abc conjecture (I know Oesterlé, from Paris) would be proven. It is very difficult a proof about my conjecture (for example $9P$ is a polynomial having degree $81$ and $nP$ has degree $n^2$) $\endgroup$
    – Piquito
    Jun 7, 2023 at 17:02
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    $\begingroup$ @Balajisb It is already hard work to obtain good bounds for $\mu(L)$ for some low-dimensional examples of nilpotent Lie algebras. The problem is connected to many other interesting topics (Milnor's problem for fundamental groups of complete affinely flat manifolds, Yang-Baxter equations, crystallographic actions etc.), and many experts claim that this is a hard problem. $\endgroup$ Jun 13, 2023 at 12:47
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This was the topic of my recently completed Masters dissertation and as far as I am aware it hasn't really been studied by anyone else (probably because it is a bit weird). The conjecture is essentially that it is always possible to design a competition meeting certain requirements that will always choose a sensible winner out of a group of people.

A competition in the sense of the conjecture is essentially a sorting network (https://en.wikipedia.org/wiki/Sorting_network) except instead of running on a linear order it runs on a tournament (https://en.wikipedia.org/wiki/Tournament_(graph_theory)) and instead of needing to fully sort the elements, it only needs to output a king (http://mathonline.wikidot.com/tournament-kings).

There are two questions that are closely related, one is can you design a competition if you know what the input tournament will be, and the other is where instead you have to be able to find a king in an arbitrary tournament of size $n$.

It turns out that if either of these is always possible for all $n$ then the other is also possible (one direction is trivial and the other follows from one of the main conclusions of my dissertation).

The reason I became interested in this is because I noticed that a lot of sports have tournament structures where which matches are going to be played is known from the start but not who will be playing in each match, the simplest example is a knock-out tournament. But I also noticed that knock-out tournaments need not produce "sensible" winners if the grouping is bad and winning isn't transitive.

The reason that I use king instead of more natural choices like "the person who wins the most matches" is that it turns out that for large enough groups of people it isn't possible to create tournaments that always choose a sensible winner with that definition, additionally king is in a certain formal sense the weakest notion of "sensible winner".

I was able to prove a lot in my dissertation (tournaments which in a certain sense can't be broken up into smaller blocks always have a competition which chooses a sensible winner), but I currently don't really know how to go any further. I know this was pretty rambly but I am really interested in this and I would love to see it be worked on more.

EDIT: Due to a few people asking to read my dissertation I have uploaded a slightly modified copy to onedrive here https://1drv.ms/b/s!At9zHjetG0NvdrtyNyplvy8grz8?e=dA9APC It should be noted that this is not well written and lacks diagrams so it is of limited use, it is my plan to rewrite it at some stage to be more presentable. As far as I am aware there are no major errors (there is a minor typographical error on the first line of page 18, it should read $\mathcal{C}$ instead of $\mathcal{B}$). If you do want to understand the proofs drawing diagrams of simple cases helps a lot, however the notation is unnessecarily confusing so unless you are really interested it is not really worth it. Finally be aware that the majority of section $5$ is essentially unproven as it relies on claim $58$ which is unproven.

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    $\begingroup$ It's always been a minor bother to me that many tournaments produce a winner by assuming that all players can be ordered from worst to best. We've all played Rock-Paper-Scissors, where the three options can be compared pair-wise to have a winner, yet no option is universally best. Yet, this simple logic apparently goes over many tournament-rule-makers' heads (not to mention elections), which is frankly dehumanizing (because human-ability is absolutely not some nice linearly ordered structure).. Anyway, I'm happy that I'm not the only one who has noticed this, and has put good research into it! $\endgroup$
    – Graviton
    Jun 8, 2023 at 2:29
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    $\begingroup$ @Graviton That's an interesting point. RPS is in a certain sense intransitive, which means there isn't a dominance order on the moves you can make. For many problems I encounter I find posits that are non-total. Further, many things have a stochastic ordering rather than a deterministic one. $\endgroup$
    – Galen
    Jun 8, 2023 at 4:40
  • $\begingroup$ relevant?: de.wikipedia.org/wiki/Arrow-Theorem $\endgroup$
    – chaosflaws
    Jun 8, 2023 at 9:20
  • $\begingroup$ @Stef If you give me an email or something I can send you a copy. If you otherwise have a good way for me to distribute it I can do that. It is my plan to rewrite it so it is cleaner and easier to read and then submit it to a journal but it has been a year and I have been too busy so I don't actually know if that will ever happen. $\endgroup$
    – Fishbane
    Jun 8, 2023 at 13:23
  • $\begingroup$ @chaosflaws Arrow's impossibility theorem is related in the sense that they both ask about choosing a candidate from a set of options in a way that is "sensible" in relation to some "ordering" on the candidates, but other than that there are no real similarities and as far as I can tell neither has any impact on the other. $\endgroup$
    – Fishbane
    Jun 8, 2023 at 13:30
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I conjectured that $41!$ is the largest non-pandigital factorial number. I asked a question about this here, although I had made the conjecture a couple years before that (circa 2018).

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  • $\begingroup$ Almost certainly true since there is no counterexample upto $(10^5)!$. We can formulate similar conjectures for the Fibonacci-numbers , the powers of two etc. $\endgroup$
    – Peter
    Jun 10, 2023 at 15:08
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Let $A$ be a cube-free integer such that the elliptic curve $X^3+Y^3=AZ^3$ has rank $\gt0$ over $\mathbb Q$ (i.e. there are infinitely many integer triplets $(X,Y,Z)$ in the curve).

My conjecture is that the only values of $A$ for which there are values of $Z$ divisible by $A$ are $A=6,9$ and $12$.

If true, this conjecture would have very important consequences, in particular a proof of Fermat's Last Theorem generalized to distinct exponents.

The curves $X^3+Y^3=AZ^3$ for $A=6,9,12$ have rank $1$ over $\mathbb Q$ and their generators are $P_6=(37,17,21)$, $P_9=(2,1,1)$ and $P_{12}=(89,19,39)$. The points of least height for which we have $A|Z$ are $2P_6,9P_9$ and $4P_{12}$. Naturally, to calculate $nP$ for $X^3+Y^3=AZ^3$, it is necessary to know the formulas corresponding to these curves, which differ from the usual ones for the Weierstrass form.

In my book, "Invitación al estudio de la aritmética de curvas elípticas" (september, 1993) I posted these "exceptional" examples that I had a hard time finding and, in those days, I did not know of a conjecture by Paul Erdös that my three examples proved. Two years after, July 1995, in the Volume 27, issue 4, pages 317-318 of the Bulletin of the London Mathematical Society, a paper by A. Nitaj was published in which my three examples appeared as a proof. I'm not saying that it was plagiarism, but I'm trying to say that finding other "exceptional" values of $A$ must be very difficult because if it wasn't, then Nitaj would have exposed other examples.

enter image description here

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My conjecture is that every group of order $n!$ has a subgroup of order $n$.

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    $\begingroup$ Here is some discussion about this conjecture. $\endgroup$
    – Dan
    Jun 7, 2023 at 22:03
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Conjecture: If a regular $n$-gon contains a regular $m$-gon, where $n$ and $m$ are coprime, with no sides coinciding, the maximum number of points of contact between them is four.

I thought this would be fairly easy to prove, but apparently it isn't.

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I have two conjectures that I feel should be answered somewhere but I haven't seen anything yet. Intuitively, they feel true and I expect them to be.

Let $M$ be an Hadamard space. That is, $M$ is a complete, simply connected Riemannian manifold with non-positive curvature. Let $x_1,...,x_n \in M$. Finally, let $A$ be the closed geodesically convex hull of $\{x_k\}$.

  1. Is the boundary of $A$ the union of geodesic segments with finite length? I expect issue dealing with boundary cusps, but I'm talking about the smooth part of the boundary. I expect yes, but have no clue how to show it. There are very few papers on the boundaries of geodesically convex hulls on manifolds.

  2. Fix $x \in A$. Does there exists weights $w_1,...,w_n \geq 0$ with $w_1+...+w_n=1$ such that $x = \arg \min_{y \in M} \sum_{k=1}^n w_k d^2(y,x_k)$? In other words, is each $x \in A$ some weighted frechet mean of $\{x_k\}$? I expect yes. (Note: the converse is true)

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  • $\begingroup$ About 1 I don't have a clue either, but 2 sounds like it should be fairly straightforward to prove with an induction step ($x$ lies on a geodesic between $\hat{x}^\text{L}$ and $\hat{x}^\text{R}$ for which we know the $\hat{w}_k^\text{L}$ and $\hat{w}_k^\text{R}$, from this show that there exist also the $w_k$ for $x$) and wrapping it all up in a completeness argument. Have you run into any particular trouble with that approach? $\endgroup$ Jun 8, 2023 at 16:45
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    $\begingroup$ The issue is the competition could possibly be a subset of the convex hull. The convex hull in manifolds have very very weird behaviors. For instance, the convex hull of 3 points in Euclidean space is a triangle. The boundary of the triangle is 1 dimensional. The boundary of a convex hull of 3 points in $\mathbb{PC}^3$ is 4 dimensional, if I recall correctly. $\endgroup$ Jun 9, 2023 at 18:26
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Here's a recent one of mine:

For every integer $n > 1$, there exists an integer $k$ such that the rightmost $7$ in the decimal expansion of $3^k$ is in the $(10^n)$'s position.

Thus for $n = 4$ we can take $k = 50$ because $3^{50} = 717897987691852588770249$ has its rightmost $7$ in the $(10^4)$'s position.

See OEIS sequence 363196

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    $\begingroup$ This is very interesting. Anything special about 7 in particular, or could we ask the same question about other digits? $\endgroup$
    – Graviton
    Jun 8, 2023 at 6:34
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    $\begingroup$ You could ask about other digits. It's very easy to find conjectures like this that are extremely likely to be true, but very unlikely to be provable in the current state of the art. $\endgroup$ Jun 8, 2023 at 15:42
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    $\begingroup$ This is true because you can force $3^k$ to end with any finite string of digits, as long as that string of digits is equal to 1, 3, 9, or 27 mod 80. $\endgroup$
    – Nitrodon
    Jun 9, 2023 at 14:45
  • $\begingroup$ @Nitrodon I'm struggling a little on how to prove this. I recognise that $80 = 3^4-1$, and the fact that it divides $10000$ seems significant. Any hints on how to proceed? $\endgroup$ Aug 28, 2023 at 3:15
  • $\begingroup$ Show by induction that $3^{10^{i-2}} \equiv 2 \cdot 10^{i-1} + 1 \mod 10^i$ for $i \ge 4$ and $3^{25 \times 10^{i-3}} \equiv 5 \cdot 10^{i-1} + 1 \mod 10^i$ for $i \ge 5$. This implies that the multiplicative order of $3 \mod 10^i$ is $5 \cdot 10^{i-2} = 10^i/20$. So $1/20$ of the numbers $0$ to $10^i - 1$ are powers of $3 \mod 2^i$. Which $1/20$? Well, all powers of $3$ are $1, 3, 9$ or $27$ mod $80$, and those constitute $1/20$ of the numbers $9$ ti $10^i - 1$. $\endgroup$ Aug 28, 2023 at 5:41
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Let $n \in \mathbb{N}$ and $S \subseteq \mathbb{Z}/n\mathbb{Z}$ such that for all $x, y \in S, x + y \neq 0$. Then $|(S + S) \cup S| \geq 2 |S|$.

I found this when thinking about Seymour's Second Neighborhood Conjecture. It's a special case of that one but seems like it should be much easier. I've hit my head against it for a while with no success.

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    $\begingroup$ At least when $n$ is prime, I think you can apply the Cauchy-Davenport theorem to $$T = S \cup \{0\}$$ $\endgroup$
    – abacaba
    Jun 12, 2023 at 19:24
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This is from some recreational math that I had done. A while ago I had looked into number bases, and found a way to do imaginary number bases. I even wrote a paper on it, but couldn’t afford to publish it. I also started looking into complex bases, but didn’t figure it out (as far as I know, there isn’t a general method). I haven’t poked at it since.

But when I was looking into it, it appeared to me that the infinite repeating nines thing ($0.\bar{9} = 1$) that happens in regular bases does NOT happen in complex bases – so my conjecture was that such a repeating infinite nine phenomena breaks/isn't true when in a complex base. I’m not a mathematician, so what follows won’t be a rigorous write up.

For instance, take the value of $17.\overline{\textrm{P}}$ $(17.\textrm{PPPPP}...)$ in base twenty-five, using the digits 0123456789ABCDEFGHIJKLMNP (skipping O because it looks too similar to 0). This is the same as $32.\overline{9}$ in base ten. The infinite repeating nines thing means that this value is the exact same as 18 in base twenty-five (or 33 in base ten).

But in a complex base, such a rule breaks (the conjecture). Here's an example I found via brute force. I'll use subscripts to say which number is in which base: $18_{25} = 33_{10}$. The number $(1 + i)_{10}$ in base $-3 + 4i$ (which also uses 25 digits*) is $17.\overline{\textrm{P}}$ That is, it’s $17.\textrm{PPPPP}...$ But it can’t be $18_{-3+4i}$, that’s actually $(5+4i)_{10}$. It can’t be $17_{-3+4i}$, that’s $(4+4i)_{10}$. It can’t be $16_{-3+4i}$ either, that's $(3+4i)_{10}$. It’s also not $17.\overline{0\textrm{P}}_{-3+4i} ([3.7 + 4.9i]_{10})$ nor is it $17.\overline{\textrm{P}0}_{-3+4i} ([1.3 + 0.1i]_{10})$.

On the other hand, I could be completely wrong about this as I might be looking at it superficially, but... that's what conjectures are: guesses!


*The number of digits that a base $b$ uses is dependent on kind of number $b$ is. For a purely real $b$, it is $\lceil |b| \rceil$. For imaginary and complex bases, it is $\lceil |b|^2\rceil$. So base twenty-five will use $25$ digits, and base $-3+4i$ will use $\lceil | -3+4i | ^ 2 \rceil = 25$ digits as well. And they are the same. Both base $25$ and base $-3+4i$ use the same twenty-five digits. The value of the digits never changes, it is only the base that changes. And this is only true for bases $|b| > 1$. For bases $0 < |b| < 1$, their inverse is used to determine how many digits they use instead.

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  • $\begingroup$ Can you make the 25 digits you use for base -3+4i more explicit? $\endgroup$
    – Bob Dobbs
    Jun 9, 2023 at 7:25
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    $\begingroup$ @BobDobbs Done. $\endgroup$
    – Status
    Jun 10, 2023 at 14:50
  • $\begingroup$ $-3-3i$ is carried and it has no representation with positive powers. Right? $\endgroup$
    – Bob Dobbs
    Jun 10, 2023 at 15:44
  • $\begingroup$ @BobDobbs Yes, $-3-3i$ is $0.\overline{\textrm{P}}$. Looks like the same is true for $-2-2i \;(0.\overline{\textrm{G}})$ and $-1-i \;(0.\overline{8})$. $-4-4i$ on the otherhand, is $15\textrm{I}$. I don't have a general method to convert to this base, however. The above was done working backwards. $\endgroup$
    – Status
    Jun 11, 2023 at 14:17
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Conjecture #1: That there are infinitely-many, regular polytopes for any dimension n. "Regular" here is re-defined (for dimensions greater than 1):

  • consisting of faces (polytopes of dimension n-1) that are all equal (this attempts to satisfy the conventional (2d) requirement of equilateral).
  • at the same point on each face, has a line perpendicular to the face that goes through the center (should be considered equal to the traditional requirement of equiangular).

(Note: This is an attempt at making a generalization for the word regular to span any dimension. Also, to account for the very nice solids like the 30-faced, diamond-faced polyhedron or the 60-faced, deltoid hexacontrahedron.)

Conjecture #2: Define an interesting prime as a prime who's reciprocal has a period roughly equal to itself. (Ex. 1/7 has a period of 6, or 1/113 has a period of 112. All other primes, notably, seem to have a period of, at most, half of this, hence the interesting part.)

The conjecture is that the sum of these interesting prime reciprocals converges to either 1 or $\phi$ -- two numbers uniquely related to the concept of reciprocity.

(Full disclosure, I'm hiding the example of primes {2, 3, 5) which are non-repeating or have a period which isn't near enough to the prime number itself to qualify unless adding it to the series magically makes it add to $\phi$ or 1 -- which would, then, be exceedingly INTERESTING.)

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  • $\begingroup$ I'm having trouble visualizing one of these for 3 dimensions which isn't a platonic solid. Is the difference that they can be non-convex? $\endgroup$ Jun 8, 2023 at 0:43
  • $\begingroup$ @DavisYoshida I'm not OP but I believe so. Check out Kepler–Poinsot polyhedron, Regular compounds, and Regular skew apeirohedron. Surely at least one of these (perhaps all) fulfill OP's definition, to aid visualization. $\endgroup$
    – Graviton
    Jun 8, 2023 at 2:36
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    $\begingroup$ @DavisYoshida I don't think the faces are required to be regular polytopes themselves. So an octahedron build from isoscele triangles should be an example (essentially pull the Platonic octahedron a little longer at two opposing corners). $\endgroup$
    – quarague
    Jun 8, 2023 at 7:32
  • $\begingroup$ @quarague: Thank you for pointing out the isosceles-faced octahedron. I'll have to update my second requirement, for I don't believe this solid would count as "regular" in my mind's view of the argument -- for the reason that it would allow infinitely-many 8 sided, triangular solids that would all be considered regular. $\endgroup$
    – Marcos
    Jun 8, 2023 at 17:05
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    $\begingroup$ @DavisYoshida: They must all be convex to satisfy the second requirement. An example of a non-platonic solid is the deltoid (kite)-faced hexacontahedron with 60-sides. There's also a 30-sided diamond(rhombus?)-faced solid which is very nice. $\endgroup$
    – Marcos
    Jun 8, 2023 at 17:07
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In 2014 Benjamin and Snyder[1] developed a neusis construction for the regular hendecagon ($11$-gon). The conjecture is that $11$ is the only non-Pierpont prime number of sides that allows such a neusis construction.

Basis for the conjecture

Let $p$ be a prime for which the maximum prime factor of $p-1$ is $5$ (with a lower maximum prime factor neusis or Euclidean constructibility is well known; with a higher maximum neusis construction is known to fail). Then the neusis construction of a regular $p$-gon will require solution of a quintic equation which, in Benjamin and Snyder's methodology, ultimately rests on solving a resolvent equation of seventh degree.

In modulo-$p$ variables, these equations take the form

$(x-r)^5\equiv0\bmod p$ Eq. 1

$(3r-1)^4(25r^3-40r^2+21r-2)\equiv 0\bmod p$ Eq. 2

With $p=11$, the corresponding real-variable equations lead to Eq. 2 being reducible, specifically having a squared cubic-polynomial factor; therefore the construction of the regular hendecagon can proceed. To get this result in the $\bmod 11$ domain the factor $25r^3-40r^2+21r-2$ must have a double root $\bmod 11$ to go with the four $\bmod11$-congruent roots of $(3u-1)\equiv0$. Indeed it does:

$25r^3-40r^2+21r-2\equiv3(r+1)^2(r+3)\bmod 11$

But the cubic factor $25r^3-40r^2+21r-2$ fails to have a double root in any prime modulus greater than $11$, so no more double cubic root in the real domain. When I tried the Benjamin-Snyder method with $p=31,41,61$ I found the resolvent irreducible in the real domain, killing off any attemped construction (the $\bmod p$ congruent roots of $(3u-1)^4\equiv0$ do not correspond to identical real roots). With no evident alternative way forward the regular hendecagon seems to be uniquely neusis-constructible.

Reference

  1. Benjamin, Elliot; Snyder, C. Mathematical Proceedings of the Cambridge Philosophical Society 156.3 (May 2014): 409-424.; https://dx.doi.org/10.1017/S0305004113000753
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After finding in this paper that an ODE can hold solutions with finite duration in time if and only if it has a Non-Lipschitz component in time (were uniqueness could be broken), so no linear ODE could have them (their solutions are never-ending in time), neither any ODE with solutions through power series since matching a constant value violates the Identity theorem, so this dismissed 99% of the systems I saw in engineering about accurately describing something that stop moving, as the things I experience in everyday phenomena.

So after learning in MSE in the tag finite-duration about these kind of ODEs with finite extinction times, and after realizing that traditional approximations like "small angle"/"linearizations" or second order expansions destroy the Non-Lipschitz component, so making impossible to them to experience a finite stopping time, now I conjecture a correction for the 2nd order approximation of every classic system as: $$y''+ay'+by+c\ \text{sgn}(y')=0$$ for some constants $a,\ b,\ c$, which have these interesting properties:

  1. It has a finite extinction time $T<\infty$ determined by the initial conditions at time $t_0$, were uniqueness is hold on $t\in[t_0,\ T)$
  2. It is able of having a final value $y(T)$ different from zero, which is not possible for traditional ODEs that can only rest at equilibrium (in the case they vanish at infinity)
  3. And last one, it has closed-form solution!, so I really don't know why it is isn't contained in every physics book.

The nonlinear term was motivated from the Couloumb's Friction (check this answer for details), but due their properties I believe it should be considered in ODEs for describing phenomena that stop moving.

Right now I am trying to understand if it could have a meaningfull use in physics (maybe it really don't add too much value), also what happen in PDEs, how it will behave under forcing and how this could be extended into control theory.

I know maybe this is probably too simple for the brilliant minds here, but is something have got my mind the last year, and I don't know is everybody is aware of it.

By the way, for me this research is recreational, so every bit of help is appreciated.

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    $\begingroup$ Are you referring to steady state conditions? $\endgroup$ Jun 12, 2023 at 9:33
  • $\begingroup$ @HopefulWhitepiller Don't really know if only involves steady-state conditions. I think is more fundamental, as example think in the classic example of the "small-angle approximation" of the nonlinear pendulum: $$y''+ay'+by=0$$ Its solution is never-ending in time, so in phase-space the trajectory is curling toward the origin without never achiving it... that don't look right to me at least $\endgroup$
    – Joako
    Jun 12, 2023 at 15:58
  • $\begingroup$ @HopefulWhitepiller I believe is interesting since it sepparates the description of phenomena in two kinds: those which are neverending (don't know exactly which ones are, but I imagine like an electron which born with the universe will have a good representation), and those phenomena that stop moving (like a classic pendulum): they are different in their description from the very beginning in the math: if stop moving its ODE must be Non-Lipschitz, and since its speed profile is compact supported its spectra must be wideband, among others restrictions. Current models don't model this. $\endgroup$
    – Joako
    Jun 12, 2023 at 16:05
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I suspect the following conjecture I asked as a question is true: The Connell Sequence and Redheffer Matrices.. By no means is this a deep problem, or in the research domain. I suspect the square numbers play a roll, but I am uncertain.

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Let the monster :

$$f(x)=\left(\left(x!\cdot\left(2x!\right)^{\frac{x!!}{2}}!\cdot\left(3\left(2x!\right)^{\frac{x!!}{2}}!\right)^{\frac{x!!!}{3}}!\cdot\left(4\left(3\left(2x!\right)^{\frac{x!!}{2}}!\right)^{\frac{x!!!}{3}}!\right)^{\frac{x!!!!}{4}}!\cdot\left(5\left(4\left(3\left(2x!\right)^{\frac{x!!}{2}}!\right)^{\frac{x!!!}{3}}!\right)^{\frac{x!!!!}{4}}!\right)^{\frac{x!!!!!}{5}}!\cdot\left(\left(6\left(5\left(4\left(3\left(2x!\right)^{\frac{x!!}{2}}!\right)^{\frac{x!!!}{3}}!\right)^{\frac{x!!!!}{4}}!\right)^{\frac{x!!!!!}{5}}!\right)^{\frac{x!!!!!!}{6}}!\right)!\cdot\left(\left(7\left(6\left(5\left(4\left(3\left(2x!\right)^{\frac{x!!}{2}}!\right)^{\frac{x!!!}{3}}!\right)^{\frac{x!!!!}{4}}!\right)^{\frac{x!!!!!}{5}}!\right)^{\frac{x!!!!!!}{6}}!\right)^{\frac{x!!!!!!!}{7}}!\right)!\right)...\right)$$

Then denotes by $B$ the lemniscate constant we have as conjecture :

$$\lim{x\to 0}\left(\frac{f\left(x\right)}{f\left(0\right)}\right)^{\frac{1}{x}}=^?B-3/10$$

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An integer distance graph is a graph whose vertices are the integers $\mathbb{Z}$, and for which an edge connects any two integers whose distance is in some chosen finite set of positive integers $D$. We denote this graph $G(\mathbb{Z},D)$. In "Colouring prime distance graphs", Erdös, Eggleton and Skilton show that $G(\mathbb{Z},D)$ can be colored periodically with $\chi(G(\mathbb{Z},D))$ colors (i.e. there exists some coloring with $\chi(G(\mathbb{Z},D))$ colors and some $p\in \mathbb{Z}^+$ such that $\forall n, n$ and $n+p$ are colored the same)

My conjecture is the higher-dimensional version of this result. Given a finite set of positive real numbers $D$, we can define the distance graph $G(\mathbb{Z}^n,D)$ to be the graph whose vertices are $\mathbb{Z}^n$ and for which edges connect vertices whose Euclidean distance lies in $D$. I conjecture that for any distance set $D$, $G(\mathbb{Z}^n, D)$ can be colored periodically with $\chi(G(\mathbb{Z}^n,D))$ colors.

The simple pigeonhole-method based proof for $n=1$ seems to break down for higher dimensions, and beyond that I have no idea how one might proceed.

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Some years ago I asked a question about the following conjecture:

Let $\,n = (2k+1)^2 \,\, $with $k\in \mathbb{N}$ and so $n>1$, and let $$\,\,A = \sum_{d \in \mathbb{N}; \ d|n} d.$$ Then $n^m$ is never divisible by $A$ for every $m \in \mathbb{N}$

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In the past I had played with the binomial expansion for my curiosity.

Let $p$ be an odd prime. From the relation $(1 + \mu)^{p − 1} = 1$, after binomial expansion and multiplication with $\mu^k$, we get $p\mu^k=-\sum_{i=2}^p{p\choose i}\mu^{k+i}$ for each $k\in\Bbb N$. Then, by iteration, $p\mu$ can be written in the form $p\mu= \sum_{n=0}^\infty a_n\mu^{p+n}$ where $|a_n|\leq\frac{p-1}{2}$. Note that $a_0=-1, a_1=\frac{p-1}{2}$. I called this complete reduction.

For $p = 3$, the complete reduction is periodic with period $2$ and repeating coefficients $−1, 1$. For $p = 5$, the complete reduction is periodic with period $6$ and repeating coefficients $−1, 2, −2, 1, 0, 0.$ For $p = 7$, I computed the first $28$ coefficients of the complete reduction as below and I am not sure if it is fully correct: $$−1,3,3,2,2,3,1,−2,0,1,1,−2,−2,0,3,−1,2,1,−3,−1,3,0,2,0,2,−1,− 2,1$$ I couldn’t observe a periodicity of the sequence $(a_n)$. Probably it is not periodic. And maybe it is easy to prove that it is not. On the other hand, I had conjectured that $(a_n)$ is periodic for all odd prime numbers $p$.

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Let $F_n$ be the the $n$-th fibonacci number.

Conjecture : For every integer $b>1$ there is a positive integer $n$ (in fact I conjecture infinite many positive integers $n$) such that $F_n+b^n$ is a prime number.

With search limit $n=6\ 000$ , we get the following table of solutions ("?" means that there is no prime for $n\le 6\ 000$) :

2  1
3  3
4  1
5  3
6  1
7  15
8  4
9  1389
10  1
11  2496
12  1
13 ?
14  2
15  18
16  1
17  24
18  1
19 ?
20  2
21  1242
22  1
23  756
24  2
25  18
26  2
27  9
28  1
29  3
30  1
31  2664
32  10
33  69
34  4
35  891
36  1
37  96
38  10
39  15
40  1
41  21
42  1
43 ?
44  4
45  3
46  1
47  45
48 ?
49  1794
50  8
51 ?
52  1
53  2715
54  2
55  48
56  2
57 ?
58  1
59 ?
60  1
61 ?
62  4
63  3
64  8
65  3
66  1
67  15
68  4
69  3
70  1
71  3
72  1
73  9
74  2
75  18
76  4
77  165
78  1
79  63
80  8
81  147
82  1
83  3
84  2
85 ?
86  67
87 ?
88  1
89  1029
90  2
91  921
92  4
93 ?
94  2
95  84
96  1
97 ?
98  5
99 ?
100  1

Who can find a solution for the remaining cases ?

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My interest in statistical association led me to define an independence gap, $\phi,$ which appears to have tighter bounds than I expected. I think these bounds relax to what I originally had expected as the number of variables increases.

Precisely:

$$\limsup \limits_{n \rightarrow \infty} \phi_{X_1, \ldots, X_n}(x_1, \ldots, x_n) = 1$$

$$\liminf \limits_{n \rightarrow \infty} \phi_{X_1, \ldots, X_n}(x_1, \ldots, x_n) = -1$$

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I posit that there is a 3D equivalent of ghost diagrams which gives IFS representations, and I haven't found that 3D equivalent yet due to lack of time.

I found a conjecture that the fastest way to traverse a 3d and higher dimension voxel space in all directions, known as a flood fill algorithm, is to have three copies of the voxel space which are updated every time 1/ a new voxel is found, 2/ a voxel has its neighbors explored and found 3 / a copy of all currently found voxels.

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    $\begingroup$ What is IFS an acronym for in this context? $\endgroup$
    – Galen
    Jun 9, 2023 at 14:42
  • $\begingroup$ @Galen probably Iteraded function system, which is a way to construct fractals such as the sierpinski gasket, julia set, barnsley fern, and most fractal covers you see on books about quantum mumbo-jumbo. $\endgroup$ Nov 29, 2023 at 21:00
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I saw a claim in an article on Science News (2020) [1] concerning a jet from an AGN being pushed backward into a second jet, and my first thought was precession of the AGN, instead of an especially dense, fast, and powerful local intergalactic medium.

My hypothesis is based on a pic of galaxy PKS 2014-55 [1][2], and is that some SMBHs have a visible rotation of some sort, be it a wobble (unlikely in a massive rotating object), a precession (assumes a gyroscopic effect), or other mechanism.

The pic in question shows a 2.5 giga light year jet at one angle, a 1 giga light year jet at another angle, and a turbulent cloud that might be a 3rd jet or a transition. 2 of the jets had an somewhat distinct terminus, suggesting that they were still continuing outward. Based on the displayed force and trajectories, I'm strongly discounting "forced backward stream".

In other images ((Centaurus A[3]), (Hercules A[4]), (FR-I source 3C31, 3C353, 3C288, WAT source 3C465[5]), we have seen a plume effect. These plumes appear to show where the forces on the ionized gases of the jets have dissipated. The surrounding gas appears to be holding the dissipating jets in place. The dispersion pattern from the end of each linear jet shows how the origin galaxy and jet have moved on. Further, those plume origins point out the strength of the relevant original magnetic influence on the respective jets. That influence is sufficient to maintain the jet's alignment with its source galaxy despite extending 10+ times the galaxy's diameter. Additional examples are seen in Boccardi, et. al.'s paper (2017) [9][10] on an AGN as seen in radio frequencies and S.Kohler's paper (2016) [11].

While "wobble" might fit better for the angle of separation, precession has been seen in massive gyroscopes near peak motion.) With that in mind, I further suggest that the bubble jets of the Milky Way [12] look like a fast gyroscopic precession. This creates a void whose boundaries are the edge effects of the jet.

Subsequent research, to nail down the exact references, revealed that my conjecture supports V.Singh (2016)[13].

(I own none of the images or websites that are linked to in the comments.)

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I asked fairly recently (here is the link to the post in question) about the prime factorisation of integers of the form $\operatorname{lcm}(1, \dots, k) \pm 1$:
I mainly conjectured that these numbers were all squarefree except for $k = 8$, and that their number of prime divisors counted with multiplicity was bounded from above by $\bigg\lfloor \frac{\ln(k)}{\ln(2)} \bigg\rfloor = \max\{\alpha \in \mathbb{N} \,|\, 2^\alpha \,\leq\, k\}$.
The second conjecture was shown to be false for $k = 359$ due to the $-1$ variant having $10$ divisors but it's still quite intriguing how "low" the number of prime divisors is for all the tested numbers considering their sheer size, although it does make sense considering $\operatorname{lcm}(1, \dots, k)$ has so many divisors relatively to them.

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  • $\begingroup$ A conjecture that first fails for k=359 belongs in the "first fail" collection: conjectures that seem to be true, but aren't. $\endgroup$ Jun 14, 2023 at 4:08
  • $\begingroup$ @richard1941 It is perfectly reasonable to have doubts with a conjecture which fails that "early" (I have my own doubts), but my focus here was moreso on quantifying the intuition that these numbers have a low amount of prime divisors rather than the exact bounding function, though I'll admit I probably failed to convey that in this post (I just wanted to keep things relatively short). $10$ divisors for a number of the same size as $\operatorname{lcm}(1, \dots, 359) = 2^8 \cdot 3^5 \cdot 5^3 \cdot 7^3 \cdot 11^2 \cdot \dots$ is still rather impressive (in my opinion). $\endgroup$
    – Bruno B
    Jun 14, 2023 at 6:18
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Small thought/ conjecture : The number with largest number of distinct positive integer factors in any interval of the form $[n,2n]$ is of the form $\prod_{i \in S} m_i$ where $m_i$ is a primordial number and $S$ is a subset of positive integers.

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Let $n\ge 4$ and two vectors $x$ and $y$ in $\mathbb{R}^n$ that satisfy

  1. $\sum_{i=1}^{n}{x_{i}^2}=\sum_{i=1}^{n}{y_i}^2=1$
  2. $\sum_{i=1}^{n}{x_{i} y_i}=0$
  3. $\sum_{i=1}^{n}{x_{i}}=\sum_{i=1}^{n}{y_i}=0$

With these conditions, prove or disprove that $$\sum_{i=1}^n\sum_{j=1}^n{(|x_{i}-x_{j}|-|y_{i}-y_{j}|)^2}\geq 4$$

This one is linked to the following question in math stack exchange and also I have suggested it as a puzzle here IBM Ponder This, you can look at some of the nice and interesting solutions that were proposed.

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