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# January 23

## Unlucky sum of three primes

Is there a number that can be written as the sum of three odd primes in exactly 13 different ways?  --Lambiam 10:46, 23 January 2023 (UTC)

If you don't find an answer here, it sounds like your question may make a good video topic for either a Numberphile, Mathologer, or Matt Parker video. Maybe you could write to them and see if they put it in the queue. --Jayron32 13:11, 23 January 2023 (UTC)
I'm pretty sure the answer is no. I ran a quick script and couldn't find any, and based on the following argument, I think that might generally suffice.
If we take some odd number ${\displaystyle n}$ and some odd prime ${\displaystyle p_{1}\leq n-6}$, then ${\displaystyle n-p_{1}}$ is an even number greater than 4 and almost certainly (cheekily assuming Goldbach's conjecture) the sum of two odd primes ${\displaystyle p_{2},p_{3}}$. Of course, we can't say anything about uniqueness of solutions if we don't establish some condition on ${\displaystyle p_{1}}$ (e.g. ${\displaystyle n=15}$ and ${\displaystyle p_{1}=3}$, ${\displaystyle p_{2}=5,p_{3}=7}$ is the same solution as ${\displaystyle p_{1}=7}$, ${\displaystyle p_{2}=3,p_{3}=5}$.) Luckily though, if we consider only ${\displaystyle p_{1}}$ between ${\displaystyle {\frac {n}{2}}}$ and ${\displaystyle n-6}$ inclusive, each ${\displaystyle p_{1}}$ gives us at least one unique solution, since it would be impossible for two of the three primes summing to ${\displaystyle n}$ to be greater than ${\displaystyle {\frac {n}{2}}}$. So the number of unique ways to have ${\displaystyle n}$ be the sum of three primes is at least the number of primes between ${\displaystyle {\frac {n}{2}}}$ and ${\displaystyle n-6}$ inclusive.
I'd have to look deeper to find any heuristics on what this might be, but I'm willing to bet that it has a growth rate that leads to the nonexistence of a number that is the sum of three odd primes in 13 different ways. GalacticShoe (talk) 21:54, 23 January 2023 (UTC)
Okay, so according to Bertrand's postulate#Erdős's theorems, "Erdős proved in 1934 that for any positive integer k, there is a natural number N such that for all n > N, there are at least k primes between n and 2n." If we consider the odd number ${\displaystyle n}$ from earlier, then since ${\displaystyle n+1}$ is even, we know that after some point, there are always at least 16 primes between ${\displaystyle {\frac {n+1}{2}}}$ and ${\displaystyle n+1}$, and removing ${\displaystyle 2}$ possible primes from ${\displaystyle n,n-2,n-4}$ (at least one of them has to be divisible by ${\displaystyle 3}$) yields ${\displaystyle 14}$ primes from which we can construct solutions. In other words, we only have to search up to a finite point to find unlucky sums of three primes before they become impossible. But as to what that finite point is? No idea, I'd have to look further. GalacticShoe (talk) 21:59, 23 January 2023 (UTC)
Effective bounds were given by Ramanujan. It appears that (assuming Goldbach) there are at least ${\displaystyle k}$ prime triplets summing up to odd ${\displaystyle n}$ when ${\displaystyle n>{\tfrac {1}{2}}R_{k}}$, in which ${\displaystyle R_{k}}$ denotes the ${\displaystyle k}$-th Ramanujan prime. Since ${\displaystyle 75>{\tfrac {1}{2}}R_{14}={\tfrac {149}{2}},}$ the search can stop quite early.  --Lambiam 23:34, 23 January 2023 (UTC)

Solved.

13 is the smallest count which doesn't occur. In PARI/GP for sums below 10000:
m=10000; v=vector(m);
forprime(p=3,m,forprime(q=p,m-p,forprime(r=q,m-p-q,v[p+q+r]++)))
v=vecsort(v); for(n=0,100,if(!setsearch(v,n),print1(n", ")))

Output: 13, 15, 17, 19, 22, 23, 26, 31, 39, 40, 41, 43, 44, 49, 51, 52, 57, 63, 65, 66, 67, 70, 71, 78, 79, 82, 83, 84, 87, 92, 94, 96, 97, 98, 99,
PrimeHunter (talk) 19:47, 28 January 2023 (UTC)

## Rule of inference vs. logical consequence

I think I understand the distinction between material conditional and logical consequence: A logically implies B iff AB is a tautology, right? But I must be confused, because this seems the same as the relationship between material conditional and rule of inference. (The article on rule of inference even seems to equate rules of inference with entailment in this section, and our article on logical consequence seems to take entailment as a synonym for logical consequence.) Yet Hilbert systems for propositional logic make a clear distinction between an axiom, which seems in this context to be a statement that some propositional formula is a tautology, and an inference rule. What am I missing? -Amcbride (talk) 16:03, 23 January 2023 (UTC) (EDIT: Just realized that middle section of rule of inference says it's using the sequent notation, which I'm used to seeing as entailment, specifically to emphasize the distinction between axioms and rules of inference... but in my confusion, this notation just seems to blur the distinction further.)

Inference rules and tautologies are very closely related. Let's consider modus ponens: ${\displaystyle P,P\to Q\vdash Q}$. There's a corresponding tautology: ${\displaystyle (P\wedge (P\to Q))\to Q}$. In general, any rule of inference gives rise to a tautology by writing that the conjunction of the hypotheses imply the conclusion -- this is called soundness of the rules of inference. Conversely, for tautologies of the form ${\displaystyle A\to B}$, you can show ${\displaystyle A\vdash B}$ by a sequence of rules of inference -- this is completeness of the rules of inference.
An important thing to understand is that tautologies are statements in the logical language, but rules of inference are not. Rules of inference are methods of combining statements to generate new statements.
Axioms are statements, but are generally not tautologies. They are additional statements which we are asserting to be true.--2600:4040:7B33:6E00:5D8:EC09:6C4:6CA0 (talk) 17:14, 23 January 2023 (UTC)
Thanks. I think part of my confusion is that, although a tautology itself is a statement in an object language, the statement that a given tautology is indeed a tautology is a metalanguage statement. Right? It's that latter animal, the assertion that a given statement is a tautology, that I'm having trouble distinguishing from a rule of inference [EDIT: when the tautology is an implication, I mean]. (And you say axioms are not generally tautologies, but for propositional logic, they are, right?) -Amcbride (talk) 17:53, 23 January 2023 (UTC)
(edit conflict) I may not be the best person to answer this, since I, like you, have been confused by the way this is defined and discussed in the literature. Here is my try:
Given a logic (a formal system), one should distinguish between the theory ${\displaystyle \Theta }$ of the logic, formed by the sentences that can be proved as theorems using its rules of inference, and its "necessity" ${\displaystyle \Delta }$, being the set of sentences that are true in all models of the logic. Normally, one should only allow ground terms here or variables that are bound by a quantifier; otherwise the notion of a sentence being true is unclear. If the logic is sound, ${\displaystyle \Theta \subseteq \Delta ,}$ that is, all provable sentences are true in all models. If the logic is complete, the converse inclusion ${\displaystyle \Delta \subseteq \Theta }$ holds. The easiest way (IMO) to think of the material conditional (aka "material implication") is as just a formula in the formal language of the logic, assuming it has an implication connective such as ${\displaystyle \to }$, in which case it is a sentence ${\displaystyle \sigma }$ of the form ${\displaystyle P\to Q.}$ Depending on the logic and on the antecedent and consequent of ${\displaystyle \sigma }$, it may or may not be an element of ${\displaystyle \Theta ,}$ and it may or may not be an element of ${\displaystyle \Delta .}$ All four cases are possible. But if it is a member of ${\displaystyle \Delta ,}$ it gets awarded a special status, variously known as "strict implication", or "entailment", or "logical consequence". The latter term is confusing; this notion of "consequence" is in terms of the models of the logic, and is not related to its deductive system with its rules of inference. It is a semantic notion; using double turnstile notation, we can express it as ${\displaystyle \vDash P\to Q.}$  --Lambiam 17:18, 23 January 2023 (UTC)
Thank you. This gives me a lot to read. I think part of my problem may be that I don't know enough about model theory or proof theory to disentangle ideas that belong with one or the other. -Amcbride (talk) 17:59, 23 January 2023 (UTC)
You don't need to know much about model theory or proof theory, as long as you keep in mind that soundness and completeness of a formal system are different properties that both cannot be taken for granted. Being true (technically known as being a tautology, sometimes denoted "${\displaystyle \vDash P}$ ") and being provable (denoted with a single turnstyle as "${\displaystyle \vdash P}$ ") are different properties of a sentence. Consider a formal system whose only sentence is ${\displaystyle \top ,}$ with the understanding that we only consider standard interpretations in which this stands for the truth value “true". Then the system has only one model, and in this model ${\displaystyle \top }$ is true, so it is a tautology. If we now choose not to add any axioms or inference rules to the system, it has no theorems, so it is not only very sound but also very incomplete. Specifically, we have ${\displaystyle \,\vDash \top }$ but ${\displaystyle \;\not \vdash \top }$.  --Lambiam 22:08, 23 January 2023 (UTC)
Thank you; I think this really does help me see it. To test my understanding, in the context of a typical propositional calculus, we could write:
• "${\displaystyle A\land (A\to B)\to B}$"
...a sentence in the object language
• "${\displaystyle A\land (A\to B)\to B\equiv \mathbf {T} }$"
...a sentence in metalanguage, equivalent to "${\displaystyle A\land (A\to B)\vDash B}$"
• "Modus ponens is a valid inference rule."
...a different sentence, also in metalanguage, equivalent to "${\displaystyle A,A\to B\vdash B}$"
Do I have that right? -Amcbride (talk) 04:44, 24 January 2023 (UTC)
Just as a distinction is made between material and logical implication, logicians make a similar distinction between material and logical equivalence. Unfortunately, there is no standard notation, and the symbol "${\displaystyle \equiv }$" is used in either of the two senses. The meaning of the second bullet is therefore not clear without context. The same applies to the use of "valid" in the third bullet. Does it mean to say that modus ponens is one of the rules of the calculus (if it isn't, a derivation using this non-existent rule does not follow the rules of the game and may be considered invalid), or is the sense that given in the article Validity (logic), in which the argument "${\displaystyle P}$ and ${\displaystyle Q}$, therefore ${\displaystyle R}$" is only considered valid when ${\displaystyle P,Q\vDash R}$? The judgement "${\displaystyle A,A\to B\vdash B}$" can be called a symbolic rendering of the modus ponens rule itself (see the infobox in our Modus ponens article) – although some formal systems allow one to derive the conclusion without the use of modus ponens, so this symbolic rendering is not entirely equivalent.  --Lambiam 08:00, 24 January 2023 (UTC)
I appreciate your continued help, Lambiam. In my second bullet I intended ${\displaystyle \equiv }$ to mean logical equivalence, not material. In my third bullet, I was confused, but I think I understand the distinction now. Stating "${\displaystyle A,A\to B\vdash B}$" is equivalent to simply saying that modus ponens is a rule of the calculus, without saying whether it is valid, right? -Amcbride (talk) 17:09, 24 January 2023 (UTC)
That is correct, with the proviso that the calculus might offer some other path for deriving ${\displaystyle B}$ from ${\displaystyle A}$ and ${\displaystyle A\to B,}$ so to be very precise this statement is equivalent to saying that the theorems of the calculus, if it does not already have modus ponens as a rule, will not be affected by adding it as a rule.  --Lambiam 18:54, 24 January 2023 (UTC)
That makes sense. Thanks again. -Amcbride (talk) 19:54, 24 January 2023 (UTC)
For an alternative rule to modus ponens, see modus tollens. See also the article on sequents, which are used in sequent calculus as formal formulas, not as metalanguage statements.  --Lambiam 20:17, 24 January 2023 (UTC)

# January 24

## Proving the limit of xⁿ by definition‏‏

This question bugs me for quite a while now, and I would like to hear an honest opinion:
Is this a valid proof, or is it somewhat "circular" by assuming the nth root function is defined and continuous?

{\displaystyle {\begin{aligned}|x^{n}-a^{n}|&={\Big |}(x-a+a)^{n}-a^{n}{\Big |}\\&=\left|\sum _{k=0}^{n-1}{\tbinom {n}{k}}a^{k}(x-a)^{n-k}\right|\\&\leq \sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}|x-a|^{n-k}\\&<\sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}\delta ^{n-k}=\varepsilon \\\delta &={\sqrt[{n}]{\varepsilon +|a|^{n}}}-|a|\end{aligned}}}

יהודה שמחה ולדמן (talk) 11:57, 24 January 2023 (UTC)

Of which statement is this meant to be a proof? ${\displaystyle \lim _{x\to a}x^{n}=a^{n}}$? If so, some conditions on ${\displaystyle a}$ and ${\displaystyle n}$ may be needed. The appeal to the binomial theorem already implies that ${\displaystyle n}$ is a non-negative integer. If ${\displaystyle a}$ and ${\displaystyle n}$ can both be equal to ${\displaystyle 0}$, you run into ${\displaystyle 0^{0}}$ being (depending on who you talk to) iffy. Apart from that, the statement follows from the continuity of multiplication. If a proof is required using the ${\displaystyle \forall \varepsilon \,\exists \,\delta }$ definition of limit, the last steps need more care and finesse. Even assuming definedness and continuity of the ${\displaystyle n}$-th root function, I don't see where the expression for ${\displaystyle \delta }$ comes from and how it guarantees the validity of the preceding equality. Using that ${\displaystyle \delta ^{n{-}k}\leq \delta }$ for ${\displaystyle 0<\delta \leq 1}$, the last inequality can be extended to
${\displaystyle \sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}\delta ^{n-k}\leq \left(\sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}\right)\delta .}$
Assuming ${\displaystyle a\neq 0}$ and ${\displaystyle n\geq 1,}$ this expression allows you to find a solution for ${\displaystyle \delta }$ that pushes this upper bound down to ${\displaystyle \varepsilon }$.  --Lambiam 12:43, 24 January 2023 (UTC)
It seems I am not helping anyone by skipping steps.
{\displaystyle {\begin{aligned}&\sum _{k=0}^{n-1}{\tbinom {n}{k}}|a|^{k}\delta ^{n-k}+|a|^{n}=\varepsilon +|a|^{n}\\&(\delta +|a|)^{n}=\varepsilon +|a|^{n}\\&\delta ={\sqrt[{n}]{\varepsilon +|a|^{n}}}-|a|\end{aligned}}}
So again: is this a valid proof or not?
יהודה שמחה ולדמן (talk) 15:45, 24 January 2023 (UTC)
The last step, in this form, obviously requires that the ${\displaystyle n}$th root of a positive number is defined. If this is not given and may not be assumed, the proof is lacking something. The proof of the lacking bit will normally be based on the continuity of not the ${\displaystyle n}$th root, but the ${\displaystyle n}$th power. Since this appears to be what the theorem is about, appealing to the conclusion of the unfinished proof for its lacking bit will indeed introduce a circularity. Strictly speaking we can do with the formally weaker existence of some ${\displaystyle y}$ such that for ${\displaystyle x>0}$ and ${\displaystyle \varepsilon >0}$ we have ${\displaystyle x^{n} This is not true in the domain of integers, so it requires something like continuity. An awkward proof can perhaps be built on the fact that the real numbers can be approximated arbitrarily closely by rational numbers.  --Lambiam 18:43, 24 January 2023 (UTC)
I saw this in a video. Could this be filling the gap we were missing?
{\displaystyle {\begin{aligned}{\bigl |}{\sqrt[{n}]{x}}-{\sqrt[{n}]{a}}{\bigr |}&={\frac {|x-a|}{\displaystyle \left|\sum _{k=0}^{n-1}\left({\sqrt[{n}]{x}}\right)^{n\,-\,1\,-\,k}\left({\sqrt[{n}]{a}}\right)^{k}\right|}}\\&\leq {\frac {|x-a|}{\left({\sqrt[{n}]{a}}\right)^{n\,-\,1}}}<{\frac {\delta }{\left({\sqrt[{n}]{a}}\right)^{n\,-\,1}}}=\varepsilon \end{aligned}}}
יהודה שמחה ולדמן (talk) 20:54, 26 January 2023 (UTC)
I don't personally feel like I'm missing a gap. This too obviously require the ${\displaystyle n}$th root of a positive number to be defined. If that is not given, the gap is this very lack of the ${\displaystyle n}$th root being defined, so this derivation won't help to fill the gap. The question is ultimately what one may and may not assume at this stage, which depends on an unknown context.  --Lambiam 22:46, 26 January 2023 (UTC)
I agree with Lambian. It is a complicated way of going around in a loop and not a proof of anything. The big trick one does with epsilon delta to get it to work easily is not to find an accurate bound but a simple bound which can be much larger. In this case if |a-x| < δ and 0 < δ < 1 then as Lambian pointed out |a-x|n < δ so you simply needs δ < ε/(|a|+1)n in your argument, or a bit less than 1 if that is greater than 1. NadVolum (talk) 00:06, 27 January 2023 (UTC)

# January 27

## Diameter of the circumscribed circle of a 11-sided Reuleaux polygon

For a Reuleaux triangle of width 1, the diameter of its circumscribed circle is approximately 1.15. (I don't have an analytical solution for this. I just drew it out and measured it.)

For an 11-sided Reuleaux polygon of width 1, what would the diameter of its circumscribed circle be?

(This is related to a question I asked on the Humanities reference desk[1]. Basically I'm trying approximate the size of the circumscribed circle of a Canadian Loonie, which happens to be an 11-sided Reuleaux polygon.) Helian James (talk) 04:56, 27 January 2023 (UTC)

What you want is the ratio of the diameter of a circle circumscribed about a regular 11-gon to its long diagonal. Or if you like, the ratio of the two longest diagonals in a regular 22-gon. This should be ${\textstyle 2{\big /}|e^{\pi i/11}+1|=\sec {\tfrac {\pi }{22}}\approx 1.0103.}$ jacobolus (t) 05:21, 27 January 2023 (UTC)
Thank you!!!Helian James (talk) 05:26, 27 January 2023 (UTC)
So if the width of the loonie is indeed 26.5 mm, then the circumcircle would have diameter a bit under 26.8 mm. –jacobolus (t) 05:28, 27 January 2023 (UTC)
Resolved

## P-recursive equation definition

I don't really understand the definition of P-recursive equation. It states that it is a linear equation but the coefficients are polynomials (or specially, sequences that are representable by polynomials)? How would that work?  AltoStev (talk) 17:27, 27 January 2023 (UTC)

A standard linear recurrence relation would look like ${\displaystyle ay_{n}+by_{n+1}+cy_{n+2}=d}$, where ${\displaystyle a,b,c,d}$ are constants. In a p-recursive recurrence relation, ${\displaystyle a,b,c,d}$ are instead polynomials in ${\displaystyle n}$. That's it.--2600:4040:7B33:6E00:5DB1:E81C:2118:36DD (talk) 18:18, 27 January 2023 (UTC)
Do you have any idea why the article does not use subscripts for the ${\displaystyle y}$ sequence? And are there uses in which ${\displaystyle f}$ is not a polynomial? The presentation
${\displaystyle \sum _{k=0}^{r}P_{k}(n)\,y_{n+k}=Q(n)}$
seems easier to grasp. One or two examples to illustrate the definition wouldn't be misplaced.  --Lambiam 00:04, 28 January 2023 (UTC)
I'd never seen P-recursive equations before clicking on the link in the OP, so I can't answer these questions.--2600:4040:7B33:6E00:B123:9399:E930:26DA (talk) 00:30, 28 January 2023 (UTC)
After looking at definitions in the literature, instead of defining "P-recursive equations", most define the notion of a "P-recursive sequence" as one obeying a homogeneous P-recursive relation, using subscripts for the sequence elements (e.g. here, where it is also stated that the nonhomogeneous case can be transformed into a homogeneous one, which I think assumes a polynomial rhs). This source, also using subscript notation, first defines the notion of a "P-recursive" recurrence relation, allowing a non-zero rhs of the equation, but requires it to be a polynomial.  --Lambiam 09:29, 28 January 2023 (UTC)

# January 28

## Interesting trigonometric identities...but why???

So this morning I "rediscovered" something I had stumbled on a few years back:

First, define ${\displaystyle A=\tan ^{-1}(1)}$, ${\displaystyle B=\tan ^{-1}(2)}$, and ${\displaystyle C=\tan ^{-1}(3)}$. Taken separately, they seem to be nothing more than a trio of nondescript irrational numbers (0.785398163397..., 1.107148717794..., and 1.249045772398..., respectively). In certain combinations however, the result is rather surprising. Notably, ${\displaystyle A+B+C=\pi }$ and ${\displaystyle {\frac {B+C}{A}}=3}$.

Very nice, but what is really going on here? I can't for the life of me remember how exactly I derived these identities in the first place. (Ah the joys of old age!) I do remember that ${\displaystyle \pi =4\tan ^{-1}(1)}$, but that doesn't really seem to help much here. Earl of Arundel (talk) 15:14, 28 January 2023 (UTC)

At the first glance, A is 45 degree (or 1/4 pi rad). At this point both equalities appear equivalent: B+C must be 3A to make the sum of three equal 4A. --CiaPan (talk) 15:31, 28 January 2023 (UTC)
Ah yes, simple algebra. Well thanks for helping to clear up the old fog. :) (Although I do wish I could remember how I found B and C to begin with!) Earl of Arundel (talk) 16:24, 28 January 2023 (UTC)
Explanation how arcus-tangents of 1, 2, and 3 add up to pi (a straight angle at O)
Make a drawing. Get a plane with Cartesian coordinates. Mark points O(0,0), P(1,0), Q(1,1), R(-1,3), and S(-1,0). Consider triangles OPQ, OQR, and ORS. Identify right angles in those triangles and then identify tangents of the three angles at O. Good luck! CiaPan (talk) 16:38, 28 January 2023 (UTC)
I see. So really this could be generalized to produce similar identities. That's really interesting. And such a simple geometric construction too. Well thank you for the enlightening perspective. Cheers! Earl of Arundel (talk) 17:17, 28 January 2023 (UTC)
Or, somewhat less exciting, by using the identity for the tangent of a sum. Let ${\displaystyle a,b,c\,}$ be a triple satisfying ${\displaystyle a+b+c=abc,}$ and define ${\displaystyle A=\tan ^{-1}a,~B=\tan ^{-1}b,~C=\tan ^{-1}c.}$ Then ${\displaystyle A+B+C=\pi .}$
For example, ${\displaystyle {\frac {7}{6}}+{\frac {4}{3}}+{\frac {9}{2}}={\frac {7}{6}}\cdot {\frac {4}{3}}\cdot {\frac {9}{2}}=7,}$ and
${\displaystyle {\begin{array}{ccl}\tan ^{-1}{\dfrac {7}{6}}&{=}&0.862170...\\\tan ^{-1}{\dfrac {4}{3}}&{=}&0.927295....\\\tan ^{-1}{\dfrac {9}{2}}&{=}&1.352127...\\&&{-}\!\!{-}\!{-}\!{-}\!{-}\!{-}\!{-}\!\!{-}~+\\&&3.141592...\end{array}}}$
--Lambiam 18:09, 28 January 2023 (UTC)
Nice! So {1, 2, 3} are the only possible integer solutions. It isn't immediately obvious how you came up with those particular values (should be an interesting after-lunch exercise anyhow). Earl of Arundel (talk) 18:48, 28 January 2023 (UTC)
Pick ${\displaystyle a}$ and ${\displaystyle b}$ such that ${\displaystyle ab\neq 1}$ and set ${\displaystyle c={\frac {a+b}{ab-1}}.}$ With some trial and error you'll find dozens of relatively simple rational triples, some much simpler than this example. For example, ${\displaystyle a=1,b=5}$ gives you ${\displaystyle c={\tfrac {3}{2}}.}$  --Lambiam 19:29, 28 January 2023 (UTC)
Neat! So in fact there are many, many such solutions. Well that's even better than the original question. What an uncanny talent for maths you have, Lambiam. I honestly envy your abilities! Earl of Arundel (talk) 23:28, 28 January 2023 (UTC)
You need not even mention any transcendental function. (6+7i)·(3+4i)·(2+9i) = -425, a negative real number. —Tamfang (talk) 00:14, 29 January 2023 (UTC)
Please see the image I added (I made it in https://www.geogebra.org/classic). --CiaPan (talk) 19:25, 28 January 2023 (UTC)
Thank you! The rough sketch I put together in MS-Paint wasn't nearly as helpful. Earl of Arundel (talk) 23:34, 28 January 2023 (UTC)
Another fun one is ${\displaystyle {\tfrac {1}{2}}\varpi =2\operatorname {arcsl} {\tfrac {1}{2}}+\operatorname {arcsl} {\tfrac {7}{23}}}$ where ${\displaystyle \varpi }$ is the lemniscate constant and ${\displaystyle \operatorname {arcsl} }$ is the inverse lemniscate sine.
But for more about the circular ones, see Machin-like_formula. –jacobolus (t) 22:31, 28 January 2023 (UTC)
Wow, I've never even heard of a "lemniscate". So this arcsl function shares very similar constructions as the atan function. Fascinating. And these Machin-like formulas are very interesting too. Especially how they (more or less) relate to the complex plane. Well thank you for the wonderful links. Great stuff! Earl of Arundel (talk) 00:02, 29 January 2023 (UTC)
The lemniscate of Bernoulli ${\displaystyle r^{2}=\cos 2\theta }$ is one kind of quartic analog of the unit-diameter circle ${\displaystyle r=\cos \theta .}$ These are both "clovers", named after the curve ${\displaystyle r^{3/2}=\cos {\tfrac {3}{2}}\theta .}$ (See Cox & Shurman, 2005.) A different quartic analog of the circle is the quartic Fermat curve ${\displaystyle x^{4}+y^{4}=1,}$ analogous to the unit circle. Both the lemniscate and the Fermat quartic are closely related to the lemniscate constant ${\displaystyle \varpi ,}$ the way the circle is related to ${\displaystyle \pi .}$ In the complex plane, these are related to the lattice of Gaussian integers the way the circle and ${\displaystyle \pi }$ are related to the ordinary integers. There is also a cubic analog ${\displaystyle x^{3}+y^{3}=1}$ which is related to the Dixon elliptic functions, and involves yet another ${\displaystyle \pi }$-analogous constant, and in the complex plane, the Eisenstein integers. –jacobolus (t) 02:46, 30 January 2023 (UTC)

This is an old puzzle just made more obscure. See [2] for an example of it on the web together with a solution. NadVolum (talk) 20:04, 28 January 2023 (UTC)

Awesome, thanks for the link! Earl of Arundel (talk) 23:37, 28 January 2023 (UTC)
While the OP enquired about ${\displaystyle \tan ^{{-}1}1+\tan ^{{-}1}2+\tan ^{{-}1}3=\pi ,}$ this puzzle establishes that ${\displaystyle \cot ^{{-}1}1+\cot ^{{-}1}2+\cot ^{{-}1}3={\frac {\pi }{2}}.}$ Of course, using ${\displaystyle \tan ^{{-}1}a+\cot ^{{-}1}a={\frac {\pi }{2}},}$ these two problems are interconvertible, but they do not seem entirely equivalent when viewed as puzzles.  --Lambiam 09:52, 29 January 2023 (UTC)

## inside-out stars

Consider a red p/q star inscribed in a blue regular p-gon. If you rearrange the vertices so that the red figure is convex, the blue edges become a p/r star. Can you easily express r as a function of p,q? —Tamfang (talk) 23:44, 28 January 2023 (UTC)

It is the modular multiplicative inverse of q with respect to p. If r comes out larger than p / 2, the traversal along the edges proceeds in the opposite sense of that of the {p / q } star; replace r by p − r to get the standard Schläfli symbol for this regular star polygon. Interestingly, some are self-dual, like {8/3} → {8/3} and {12/5} → {12/5}. This happens when p｜(q 2 − 1). Some others are self-dual after edge reversal, like {5/2} → {5/−2} and {13/5} → {13/−5}. This happens when p｜(q 2 + 1).  --Lambiam 00:17, 29 January 2023 (UTC)