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Re: [math-fun] "MALAYALAM" isn't just the longest palindrome
by Steve Witham 22 Nov '20

22 Nov '20

> From: Dan Asimov <dasimov(a)earthlink.net> > Date: 11/22/20, 1:21 AM > > Speaking of word oddities, Ararat (or tarara, as in "tarara boom-de-ay") has one T, two R's, and three A's. > > Is there any other word having its letter counts form an arithmetic sequence of at least three positive terms? > > Â—Dan In my dictionary, there are 204 words with counts (1, 2, 3), and 14 with (1, 2, 3, 4): 4 (1, 2, 3, 4) beerbibber Other than a potion in "Kingdom Come: Deliverance" and a multifunction bottle opener, see ** below. 4 (1, 2, 3, 4) chachalaca American tropical bird. 4 (1, 2, 3, 4) deadheaded 4 (1, 2, 3, 4) isoosmosis 4 (1, 2, 3, 4) kinnikinic Native American tobacco substitute. 4 (1, 2, 3, 4) kotukutuku The world's largest fuchsia. 4 (1, 2, 3, 4) peppertree 4 (1, 2, 3, 4) rememberer 4 (1, 2, 3, 4) sereneness 4 (1, 2, 3, 4) sleeveless 4 (1, 2, 3, 4) evennesses My dictionary tends to overdo suffixes. 4 (1, 2, 3, 4) keennesses 4 (1, 2, 3, 4) sanenesses 4 (1, 2, 3, 4) susurruses With two of each letter you get the arithmetic sequence 2, 2, 2, 2... (Computers can be so literal.) 5 (2, 2, 2, 2, 2) arraigning 5 (2, 2, 2, 2, 2) chimachima Yellow-headed caracara (bird) 6 (2, 2, 2, 2, 2, 2) happenchance 5 (2, 2, 2, 2, 2) horseshoer 5 (2, 2, 2, 2, 2) intestines 5 (2, 2, 2, 2, 2) nonordered 5 (2, 2, 2, 2, 2) quinaquina a.k.a. Cinchona plant yielding quinine. 7 (2, 2, 2, 2, 2, 2, 2) scintillescent scint i ll e scent 5 (2, 2, 2, 2, 2) strumstrum crude stringed instrument 5 (2, 2, 2, 2, 2) tromometer extra-sensitive seismometer Each of these words requires more than half the alphabet to spell: 14 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) ambidextrously 14 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) benzhydroxamic 14 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) dermatoglyphic 15 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) dermatoglyphics 14 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) hydromagnetics 14 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) hydropneumatic * 14 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) pseudomythical 14 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) subformatively 14 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) sulphogermanic 14 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) undiscoverably Nothing with at least three distinct letters and linearly increasing counts and more than one of the least common, nor more than four distinct letters. --Steve P.S. * Plans for a gun with hydropneumatic brakes figured in the Dreyfus Affair. ** beerbibber: The beginning of a 1927 Time Magazine article "Beer Diplomacy": To Riga, Latvia, came, some three months ago, Herr Doktor Adolph Koester, German Ambassador, lusty beerbibber. . . . He found official Latvia strongly disposed to favor the Franco-British project of a Baltic state cordon dividing Russia and Germany. Poland, close ally of France, had bestowed numerous decorations on Latvian officials. At Riga British diplomats were arbitri elegantiarum How should the German Ambassador win Latvian sympathies? Soon Herr Doktor Koester gave a true, orthodox German bier abend: a "beer evening" at which he and a group of high Latvian officials sat up with mossy... Many google hits like this: Spriest Driveadream beerbibber spriest.driveadream.com Â› beerbibber Yeah tie your fabric print. Telephony is dead. Rebel convoy is just taught his people where size does this door upon its hilltop. Somebody over here. Bases not ...

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Re: [math-fun] "MALAYALAM" isn't just the longest palindrome
by Keith F. Lynch 22 Nov '20

22 Nov '20

Tom Karzes <karzes(a)sonic.net> wrote: > I found two 10-letter (4 disctinct) words having this property: > sereneness > sleeveless Also, checking all words in /usr/share/dict/words on Panix, Actaeaceae, beerbibber, chachalaca, isoosmosis, kotukutuku, and rememberer. Next I'll check my saved email for more.

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Re: [math-fun] "MALAYALAM" isn't just the longest palindrome
by Stuart Anderson 22 Nov '20

22 Nov '20

> > That makes me wonder if there are long palindromic words in other > languages. > > If we allow words in a sentence to form palindromes, we get some long ones; Oozy rat in a sanitary zoo. Murder for a jar of red rum. Mr. Owl Ate My Metal Worm Napoleon Bonaparte: Able was I ere I saw Elba Are we not pure? “No, sir!” Panama’s moody Noriega brags. “It is garbage!” Irony dooms a man—a prisoner up to new era. The not so long but topical; Rise to vote, sir The maths problem; Never odd or even Then there's palindromic poetry; Doppelganger by James A. Lindon Entering the lonely house with my wife I saw him for the first time Peering furtively from behind a bush -- Blackness that moved, A shape amid the shadows, A momentary glimpse of gleaming eyes Revealed in the ragged moon. A closer look (he seemed to turn) might have Put him to flight forever -- I dared not (For reasons that I failed to understand), Though I knew I should act at once. I puzzled over it, hiding alone, Watching the woman as she neared the gate. He came, and I saw him crouching Night after night. Night after night He came, and I saw him crouching, Watching the woman as she neared the gate. I puzzled over it, hiding alone -- Though I knew I should act at once, For reasons that I failed to understand I dared not Put him to flight forever. A closer look (he seemed to turn) might have Revealed in the ragged moon. A momentary glimpse of gleaming eyes A shape amid the shadows, Blackness that moved. Peering furtively from behind a bush, I saw him for the first time, Entering the lonely house with my wife.

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[math-fun] An identity involving the floor function
by Neil Sloane 21 Nov '20

21 Nov '20

Dear Math Fun, The OEIS is full of assertions whose status is unclear - are they theorems or conjectures? This one is stated unconditionally but without a proof. If we had a proof it would help with quite a lot of other sequences. It is surely true, and maybe not difficult to prove. Can someone help? T(k) = k*(k+1)/2, k >= 0, is a triangular number. Claim: For n >= 1, we have Sum_{ k >= 1, stop when T(k-1) >= n } (-1)^(k+1) * ( floor ( (n - T(k-1))/k ) )^2 = Sum_{k=1..n} sigma(k). The RHS is the sum of the divisors of all numbers from 1 to n, A024916(n), which can also be written as Sum_{d=1..n} d*floor(n/d) = n^2 - Sum_{d=1..n} n%d. The summands on the LHS look like [ (n-T(k-1))/k ]^2, and the sum stops when the quantity n - T(k-1) becomes zero or goes negative. For example, when n=8, the assertion is that 8^2 - [ (8-1)/2 ]^2 + [ (8-3)/3 ]^2 = 8^2 - 3^2 + 1^2 = 56 = 1+3+4+7+6+12+8+15 . Neil

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[math-fun] dvipdf
by James Propp 21 Nov '20

21 Nov '20

Has anyone else been having problems lately with dvipdf? My install of dvipdf now converts a multi-page dvi file with actual content into a one-page file containing nothing but white space. I tried it on a file that compiled properly back in May, and it's not working now. I reinstalled TexShop and restarted my laptop, but that didn't fix the issue. (I am trying to do the final upload for an article I've been working on for seven years. So of course something like this would happen...) Jim Propp

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[math-fun] The Walrus Binary Entropy Coder
by Steve Witham 20 Nov '20

20 Nov '20

"ABSTRACT: The Walrus coder does the same job as a binary arithmetic coder-- "binary" meaning having an input alphabet of just two symbols. The Walrus has the same API as arithmetic coders, and similar coding efficiency and latency, but the Walrus maps symbols to bits differently: at each step it designates one of the two symbols "walrus" and the other "eggman." The walrus is assigned an available bit string, which is then either output (if walrus is sent) or made unavailable (if eggman is sent). ... This file is "literate executable pseudocode" in Python... An appendix contains test code... A separate Jupyter notebook [N] does some analysis [and graphs] of the Walrus's efficiency both empirically and theoretically." http://www.mac-guyver.com/switham/2020/11/Walrus Â --Steve

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Re: [math-fun] dvipdf
by Henry Baker 20 Nov '20

20 Nov '20

MacOS Catalina & Big Sur break a lot of stuff, due to the new permission mechanisms, which work in tandem with Linux file permissions, but go beyond them. (If you know anything about Android, they have a similar problem with their 'SELinux' permission structure; many people give up and turn off SELinux completely.) Sometimes going to the security & privacy section and giving the failing app permission to read/write files may work. (The app still can't go around the normal Linux file permissions.) For example, Big Sur broke Emacs, so I had to Google around until I found a fix which involved giving Emacs more permissions. At 08:03 AM 11/20/2020, James Propp wrote: >The problem affects dvipdf on my new MacBook running Catalina 10.15.7 but >not my old MacBook running High Sierra 10.13.6. > >Jim Propp > >On Fri, Nov 20, 2020 at 10:49 AM James Propp <jamespropp(a)gmail.com> wrote: >> Has anyone else been having problems lately with dvipdf? >> >> My install of dvipdf now converts a multi-page dvi file with actual >> content into a one-page file containing nothing but white space. >> >> I tried it on a file that compiled properly back in May, and it's not >> working now. >> >> I reinstalled TexShop and restarted my laptop, but that didn't fix the >> issue. >> >> (I am trying to do the final upload for an article I've been working on >> for seven years. So of course something like this would happen...) >> >> Jim Propp

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[math-fun] That InverseFourier hack seems also to solve cubics and septics.
by Bill Gosper 17 Nov '20

17 Nov '20

(But the septic solver requires a sextic solver.) Here, with crucial help from Julian (despite his disdain for preoccupation with radicals), is the cubic (with rational coefficients) solver: In[107]:= scub[x^3-8x-10] Out[107]= {(45+Sqrt[489])^(1/3)/(3^(2/3) I^(4/3))+((45-Sqrt[489])^(1/3) I^(4/3))/3^(2/3), (45-Sqrt[489])^(1/3)/(3^(2/3) I^(4/3))+((45+Sqrt[489])^(1/3) I^(4/3))/3^(2/3), (45-Sqrt[489])^(1/3)/3^(2/3)+(45+Sqrt[489])^(1/3)/3^(2/3)} In[109]:= ContinuedFraction[%%[[3]],69] Out[109]= {3,3,7,4,2,30,1,8,3,1,1,1,9,2,2,1,3,22986, 2,1,32,8,2,1,8,55,1, 5,2,28,1,5,1,1501790, 1,2,1,7,6,1,1,5,2,1,6,2,2,1,2,1,1,3,1,3,1,2,4,3,1,35657, 1,17,2,15,1,1,2,1,1} scub[q_]:= ToRadicals[Block[{rts = (#1[[1,2]] & ) /@ NSolve[q == 0,WorkingPrecision->33],foos}, foos = (rts.{{1,1,1},{1,1/I^(4/3),I^(4/3)},{1,I^(4/3),1/I^(4/3)}})^3/27; foos = (Together[#1[[1,2]]]^(1/3) & ) /@ Solve[0 == Rationalize[Expand[Times @@ (#1 - Rest@foos)]]]; Rationalize@Mean@ rts + (foos . #1 & ) /@ {{I^(4/3),1/I^(4/3)},{1/I^(4/3),I^(4/3)},{1,1}}]] That's it! For triples, the InverseFourier open codes to this little matrix multiplication foos = (rts.{{1,1,1},{1,1/I^(4/3),I^(4/3)},{1,I^(4/3),1/I^(4/3)}})^3/27; and the searches over root permutations and roots of unity degenerate. Whoa, wait a minute, it calls Solve! But only on quadratics. Which we could open-code. In[111]:= scub[Echo@MinimalPolynomial[Cos[2\[Pi]/7],x]] -1-4 x+4 x^2+8 x^3 Out[111]= {-(1/6)+(7/2 (1+3 I Sqrt[3]))^(1/3)/(6 I^(4/3))+1/6 (7/2 (1-3 I Sqrt[3]))^(1/3) I^(4/3), -(1/6)+(7/2 (1-3 I Sqrt[3]))^(1/3)/(6 I^(4/3))+1/6 (7/2 (1+3 I Sqrt[3]))^(1/3) I^(4/3), -(1/6)+1/6 (7/2 (1-3 I Sqrt[3]))^(1/3)+1/6 (7/2 (1+3 I Sqrt[3]))^(1/3)} In[114]:= FullSimplify@ArcCos@%%% Out[114]= {ArcCos[-(1/6)+(7/2 (1+3 I Sqrt[3]))^(1/3)/(6 I^(4/3))+1/6 I^(4/3) 2.37\[Ellipsis]-1.18\[Ellipsis] I], ArcCos[-(1/6)+1/6 (7/2 (1+3 I Sqrt[3]))^(1/3) I^(4/3)+2.37\[Ellipsis]-1.18\[Ellipsis] I/(6 I^(4/3))], 2π/7}. Disclaimer: For sufficiently exotic coefficients, WorkingPrecision->33 would need to increase, But I'm surprised that such a concise formula isn't more popular. —rwg

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Re: [math-fun] Zonohedral surfaces
by Dan Asimov 16 Nov '20

16 Nov '20

Thanks, George! I look forward to watching the video, which unfortunately I don't have time to do this morning. But I'm wondering a couple of things from this post: 1. How do you define the product of two polygons as a polyhedron in 3-space? Abstractly, it makes perfect sense: The product of two vertices is a vertex; of a vertex and an edge (in either order) is an edge; and of two edges is a quadrilateral. So far this is just a combinatorial object, but it could even be an abstract geometrical object, as if each polygon were in a distinct plane and the product becomes a specific subset of 2+2 = 4-space. But that seems to allow a lot of possibilities for it to end up in 3-space. How do you see that any way of doing this results in a polyhedron that's self-intersecting? 2. What does "homogeneous" mean in this situation? It could refer just to the result being combinatorially homogeneous, in that the result has maps taking vertices to vertices, edges to edges, and faces to faces while preserving incidence, that will carry any vertex to any other (transitivity on vertices). Is that it? Thanks, Dan ----- The talk Jim had referred to is now online here in a more completed form. Polyhedra fans might enjoy it: https://www.youtube.com/watch?v=_PSdVX02Vbs The question which I thought might have been intended (during the CoM discussion, or initially by Jim) was whether one could make a *homogeneous* zonohedral surface in 3D which is topologically toroidal without self-intersection. By homogeneous, I mean following the algorithm described in the video from a given initial state. (One with all vertices of degree four would be homogeneous, but that isn't a necessary condition; homogeneous examples might have vertices of any order, e.g., polar zonohedra.) One can take a product of two polygons to create a toroidal surface composed of parallelograms and having all vertices of degree four, but it will be self-intersecting. And one can approximate any continuous surface by joining homogeneous patches, e.g., gluing together small cubes. (The video shows how to join larger homogeneous patches seamlessly.) But I would conjecture that a homogeneous zonohedral surface can only be toroidal if it is self-intersecting. -----

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Re: [math-fun] Zonohedral surfaces
by Dan Asimov 16 Nov '20

16 Nov '20

Yes (I presume this means in R^3). To get on with all faces rectangles, drill a hole straight through a rectangular solid and look at its boundary. Now apply a random linear mapping and all the rectangles will probably turn into parallelograms. —Dan Jim Propp wrote: ----- George Hart’s Celebration of Mind talk (just concluded) makes me wonder: is there a non-self-intersecting torus surface made of parallelograms? -----

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