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[math-fun] Draft of August 2018 blog post
by James Propp 11 Aug '18

11 Aug '18

The draft of the essay (two short essays, actually) that I plan to post on August 17 is now available at http://mathenchant.org/038a.html and I would welcome comments. Regarding the first of the two essays (an intro and accompaniment to Thurston's video "Knots to Narnia"), I'd be grateful for references related to the video. I'd be especially curious to know whether, for every knot K, the space you get by inserting a doubly-branched K-shaped portal is a finite-to-one branched cover of 3-space. (Thurston shows that when K is an unknot you get a 2-to-1 branched cover and that when K is a trefoil you get a 6-to-1 branched cover. In my essay I show that if instead of a knot you use a two component "unlink" you get an infinite-to-one branched cover.) Also, am I right in thinking that when you use a two-component link (as shown in the article), you get a 4-to-1 branched cover? I want to make sure I've done it right before I insert my "proof" into the endnotes! As I see it, if we label the three windows A, B, and C, then we get the relations C=BA and A=CB, yielding A=BAB, so that ABAB is the identity. Thanks, Jim Propp P.S. It's no longer possible for me to get feedback via WordPress. So just send me your feedback using ordinary email.

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Re: [math-fun] Finding equidistant points among the vertices of a cube
by Dan Asimov 09 Aug '18

09 Aug '18

Another math-fun member has kindly pointed out to me that an Hadamard matrix in dimension 4K implies the maximal number (n+1) of equidistant points among the vertices of the (4K-1)-dimensional cube... ...and that since the Hadamard conjecture is that an Hadamard matrix exists in every dimension of form 4K, it would if true mean that n+1 equidistant points exist in each dimension of form 4K-1. —Dan ----- The relationship is that if there is an n x n Hadamard matrix, then there exist n orthogonal vectors among the 2^n vertices {-1, 1}^n of the n-cube. These n vectors are equal length, so their endpoints are equidistant. I ask about a related but not identical situation: when there exist n+1 equidistant points among the 2^n points of {-1, 1}^n. And in general, *how large* is the largest subset of {-1, 1}^n consisting of equidistant points? —Dan Fred Lunnon wrote: ----- See https://en.wikipedia.org/wiki/Hadamard_matrix in particular, Sylvester construction. ----- On 8/9/18, Dan Asimov <dasimov(a)earthlink.net> wrote: Suppose the metric space Q_n is the vertices of the n-dimensional cube Q_n = {-1, 1}^n with the induced metric from Euclidean space, and the questions is: Question: --------- For which n do the points of Q_n include n+1 equidistant points, the maximum possible? (And in general, what is the size S(n) of the largest equidistant subset of Q_n?) If this is in OEIS, I didn't succeed in finding it. ----- ----- -----

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Re: [math-fun] Finding equidistant points among the vertices of a cube
by Dan Asimov 09 Aug '18

09 Aug '18

The relationship is that if there is an n x n Hadamard matrix, then there exist n orthogonal vectors among the 2^n vertices {-1, 1}^n of the n-cube. These n vectors are equal length, so their endpoints are equidistant. I ask about a related but not identical situation: when there exist n+1 equidistant points among the 2^n points of {-1, 1}^n. And in general, *how large* is the largest subset of {-1, 1}^n consisting of equidistant points? —Dan Fred Lunnon wrote: ----- See https://en.wikipedia.org/wiki/Hadamard_matrix in particular, Sylvester construction. ----- On 8/9/18, Dan Asimov <dasimov(a)earthlink.net> wrote: Suppose the metric space Q_n is the vertices of the n-dimensional cube Q_n = {-1, 1}^n with the induced metric from Euclidean space, and the questions is: Question: --------- For which n do the points of Q_n include n+1 equidistant points, the maximum possible? (And in general, what is the size S(n) of the largest equidistant subset of Q_n?) If this is in OEIS, I didn't succeed in finding it. ----- -----

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[math-fun] Finding equidistant points among the vertices of a cube
by Dan Asimov 09 Aug '18

09 Aug '18

Suppose the metric space Q_n is the vertices of the n-dimensional cube Q_n = {-1, 1}^n with the induced metric from Euclidean space, and the questions is: Question: --------- For which n do the points of Q_n include n+1 equidistant points, the maximum possible? (And in general, what is the size S(n) of the largest equidistant subset of Q_n?) If this is in OEIS, I didn't succeed in finding it. —Dan

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[math-fun] Fwd: Possible relation between Catalan?s, Ramanujan-Soldner?s and Brun?s constants
by rcs＠xmission.com 06 Aug '18

06 Aug '18

Gentlemen, start your engines. --Rich ----- Forwarded message from lesniakowie(a)GMAIL.COM ----- Date: Mon, 30 Jul 2018 14:16:40 -0500 From: Marcin Le?niak <lesniakowie(a)GMAIL.COM> Reply-To: Number Theory List <NMBRTHRY(a)LISTSERV.NODAK.EDU>, Marcin Le?niak <lesniakowie(a)GMAIL.COM> Subject: Possible relation between Catalan?s, Ramanujan-Soldner?s and Brun?s constants To: NMBRTHRY(a)LISTSERV.NODAK.EDU Hello everybody, I'm new to this list and was advised (by one of the members) to share here my conjecture of a possible relation between the three constants related generally with the distribution of primes. The relation is completely surprising and unexpected (neither I nor two experts in number theory couldn't find the reason for its existence) but amazingly simple, and this is why I believe could be real... Conjecture concerns the relation between Catalanâs, Ramanujan-Soldnerâs and Brunâs constants resulting in receiving the closed formula for Brun's constant, which direct computation is not possible so far and is based on extrapolation of achievable computational results made with the help of the twin primes conjecture. Let G be the Catalanâs constant, equal to Î²(2) (Dirichlet beta function for s=2) G = Î²(2) = 0.91596559417721901505460351493238411077414937â¦ Let Âµ be the Ramanujan-Soldner constant Âµ = 1.45136923488338105028396848589202744949303228â¦ Let B2 be the Brunâs constant Conjecture: B2= (8+40(Âµ-G))/(16-(Âµ-G)) = 1.9021605831029730799822614917574361â¦ As I said this relation is so surprising, that Iâd like to encourage you all to help find its explanation - maybe it could shed the new light on the problem of distribution of the twin (and other constellations) primes. I'm very curious about your opinion about it :-) Kind regards, Marcin LeÅniak ----- End forwarded message -----

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Re: [math-fun] I just saw CA plate XKCD793.
by Bill Gosper 05 Aug '18

05 Aug '18

On 2018-08-04 14:38, Dan Asimov wrote: > Bill Gosper wrote: > ----- > ..... > ..... > (http://www.tweedledum.com/rwg/idents.htm) > ----- > > Ramanujan lives on! > > —Dan I think the only results in there worthy of Ramanujan are (d247) and (d246). (http://gosper.org/fst.pdf (J. McCarthy Festschrift), p140). What would I give to watch Ramanujan and Hardy puzzle over those! --rwg

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Re: [math-fun] math-fun Digest, Vol 186, Issue 9
by Guy Haworth 05 Aug '18

05 Aug '18

I unsubscribed some time ago but am still getting input. Are the key people on holidy?! Thanks - Guy

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[math-fun] I just saw CA plate XKCD793.
by Bill Gosper 05 Aug '18

05 Aug '18

Puzzle: It was orange on black. Seven letter vanities ("environmentals") were not introduced until decades after the switch to orange on blue. I distinctly remember an episode of Columbo where the perp had orange on blue CARSINI, a safe bet for the writers, before seven letters were legal. --rwg If I see XKCD1047, I'll know it's a fake. (Note the rollover warning about πᵗʰ root.) WRI support just told me the secret way to get Mma to display πᵗʰ root of 4. (http://www.tweedledum.com/rwg/idents.htm)

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Re: [math-fun] I just saw CA plate XKCD793.
by Dan Asimov 04 Aug '18

04 Aug '18

Bill Gosper wrote: ----- ..... ..... (http://www.tweedledum.com/rwg/idents.htm) ----- Ramanujan lives on! —Dan

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[math-fun] Simplest planar n-gon polyhedra
by Ed Pegg Jr 03 Aug '18

03 Aug '18

At https://math.stackexchange.com/questions/2869725/ I've started a sequence. What is the fewest number of planar but possibly irregular n-gons needed to make a polyhedron or toroid? 4 triangles 6 squares 12 pentagons 7 hexagons 12 heptagons Correct so far? How many octagons are needed? Ed Pegg Jr

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