October 2019 - mathfun@mailman.xmission.com

[math-fun] Unitary Transforms of Elliptic Curves?
by bradklee＠gmail.com 11 Oct '19

11 Oct '19

The proof that canonical or symplectic transformations leave invariant the real period of an elliptic curve is not too difficult using Jacobian matrices of determinant 1. Generalization of the proof seems to imply the corollary that complex determinant transformations from unitary act as rotation of the period parallelogram through the complex plane. It would be nice to have an intelligible reference on the full theorem, if it is correct. A few years ago I skimmed through every “Yellow Peril” textbook I could find on the subject (including Silverman), and didn’t find the answers I was looking for. Did I miss something, or am I just worried about another fact that is “obviously” right (or wrong)? — Brad

1 0

[math-fun] heptagonal stopsign nerdsnare
by Bill Gosper 11 Oct '19

11 Oct '19

Remember that XKCD where they blindside a passing nerd with a paralyzing math puzzle? add what for the next one? <http://gosper.org/hepstop7.bmp> What is the scale ratio of consecutive stopsigns? And consecutive "trumpets"? The guy behind you is honking his! —rwg

2 3

[math-fun] a(n), a(n+1) and a(n+2) don't share any digit
by Éric Angelini 10 Oct '19

10 Oct '19

Hello Math-Fun, How many terms in the hereunder finite seq? « a(n), a(n+1) and a(n+2) don't share any digit, with a(1) = 0. » Best, É.

5 5

[math-fun] Dragon functional equations
by Bill Gosper 10 Oct '19

10 Oct '19

A couple of weeks ago, Julian found Drag[(m + 1/2 + 1/4*(-1)^m)/2^n] == 1/2 (Drag[m/2^n] + Drag[(m + 1)/2^n]), for positive integers m, n, where Drag(t) is the usual Dragon-filling map of [0,1] into the bounding box [-1/3 - i/3, 7/6 + 2i/3]. This describes where along the Dragon is the *m*th vertex of the (self-avoiding) *n*th order "median curve". Using Julian's remarkable piecewiserecursivefractal interpolator-inverter applied to the Dragon's skin, as defined by http://larryriddle.agnesscott.org/ifs/heighway/heighwayBoundary.htm, I just noticed Out[161]= ConditionalExpression[dragskin[1/3 - t] ==3/2 + I/2 - dragskin[1/3 + t], Abs[t] <= 1/6] In[125]:= ConditionalExpression[I dragskin[1 - 3 t] == dragskin@t, 0 <= t <= 1/12] In[126]:= ConditionalExpression[I ( dragskin[1/2 - t] - 1) -> dragskin[1/2 + t] - 1, Abs@t <= 1/4] Plus an intriguing symmetry I haven't yet quantified, smooth scaling? <http://gosper.org/skin 012o3.png> The whole boundary is dragskin([0,1]). Unfortunately, Riddle's (really Mandelbrot's?) clever construction draws the skin at an unboundedly nonuniform rate. I.e., it has dimension D (an unlovely cubic surd between 1 (length) and 2 (area)), but its fill rate, measure(dragskin(t),t1<t<t2)/(t2-t1)^D, gets arbitrarily small. (And arbitrarily large?) But these functional relations raise my hope of finding some continuous, monotone u(t) to give dragskin(u) a uniform fill rate. —rwg Mandelbrot once challenged me to find the skin dimension. I disbelieved the irreducible cubic from a 5⨉5 determinant. Mandelbrot wrote only terse expressions. So I found a 6⨉6. Same damn cubic. I called him up. "Yes, that's the cubic." "How the heck did you get that? it's not your style!" "I just looked at how the wiggles fit into each other." It just occurs to me that I must have had two different, uniformly filling skin descriptions to write those determinants! Oh, to be young again.

3 3

Re: [math-fun] A Permutation Cipher Text
by Keith F. Lynch 10 Oct '19

10 Oct '19

Brad Klee <bradklee(a)gmail.com> wrote: > https://0x0.st/zwRU.txt > Easy to brute force if you get the gist of the permutation > scheme but looks difficult to reverse engineer otherwise. That looks very familiar. I notice that common letters remain common, that common digrams either remain common or are reversed, and that words and reversed words sometimes appear. I've seen that before, when I played around with taking a block of text and using each successive possible ASCII character in turn as a delimeter and reversing the text between every instance of that character, treating the end as being followed by the beginning (a "necklace"). For instance: reverse [e] revesre [r] rsevere [s] esrerev [v] erersev But I tried that, and it didn't work. Nor did running through the ASCII sequence in reverse order. Maybe another one of the 128! possible orders for seven-bit ASCII. It doesn't help that there are two non-ASCII characters in the text. Some years ago it occurred to me that exclusive-oring a text with a subset of the square roots, in binary, of the first N primes, followed by the above scrambling algorithm (in ASCII order), followed by the same exclusive-oring again, would be a pretty good encryption algorithm. Does anyone disagree?

2 1

[math-fun] The Triangle
by ed pegg 09 Oct '19

09 Oct '19

Some guy has been busy. https://arxiv.org/abs/1910.03379 --Ed Pegg Jr

2 1

Re: [math-fun] How many 5 digit #s are (US) zip codes?
by Leo Broukhis 08 Oct '19

08 Oct '19

Andre, By "upside-down" I mean the rotation of the (vertical) plane by 180 degrees around a (horizontal) line perpendicular to the plane. :) Leo On Sun, Sep 22, 2019 at 4:45 PM Andres Valloud < avalloud(a)smalltalk.comcastbiz.net> wrote: > Hi Leo, could you please clarify for me what you mean when you say > 'upside-down'? > > On 9/22/19 0:17 , Leo Broukhis wrote: > > Out of two strings matching [0125689]{5}, where the second string > > corresponds to the upside-down reading of the first, no more than one > > string will be a valid zip code. > > > > Leo > > > > On Fri, Sep 13, 2019 at 11:51 AM Andres Valloud > > <avalloud(a)smalltalk.comcastbiz.net > > <mailto:avalloud@smalltalk.comcastbiz.net>> wrote: > > > > Regarding the subject, it appears US (basic) zip codes are > effectively > > character digit strings of length 5 (that is, [0-9]{5}). Some state > > there are 42k active zip codes, presumably of the five digit variety. > > > > On 9/13/19 6:28 , Bill Gosper wrote: > > > G4G once had an entertainer named "Mr. Zipcode"(?). > > > You tell him yours and he tells you where, and then goes on to > > describe > > > places to eat, etc. (He struggled with Ray Solomonoff's obscure > > town in > > > upstate New York. Or maybe some offstage confederate struggled > with > > > https://m.usps.com/m/ZipLookupAction?search=zip .) > > > > > > Impressive, but maybe not 10^-5 impressive. > > > I just mistakenly tried 44007, which seems to be Nantes, > > > the site of a French Revolution atrocity. Does Mr. Zipcode mention > > > atrocities? —rwg > > > _______________________________________________ > > > math-fun mailing list > > > math-fun(a)mailman.xmission.com <mailto: > math-fun(a)mailman.xmission.com> > > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com <mailto:math-fun@mailman.xmission.com> > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > >

2 2

[math-fun] Space-Fillers : 2X2 Combinatorics.
by Brad Klee 06 Oct '19

06 Oct '19

1. A path goes through a unit square, starting and ending on a vertex. 2. The path is self-similar, with deflation of 1 square to 2X2. 3. The path goes through all four sub-squares, each exactly once. 4. Consecutive sub-squares are adjacent at a vertex on the path. 5. The curve cannot cross itself. Q. Restricting to 1/4^n Z-function, how many paths are possible? When inflating from 1 square to 4, there are 5 valid paths between adjacent corners, and 2 valid paths between opposite corners: Approximating Polygons: https://0x0.st/zwup.png . Considering that paths between adjacent corners have L&R variants, an upper limit for the total count is: 5^8*5^(12*2)*2^(8*2) ~ 10^27 . I wanted to reduce the count to the number of distinct curves, but the space appears too large. Maybe a fun idea for a computer App: Allow user to choose templates until the set closes, then print. It is possible to compute all (1/9^n)-sampled approximating polygons for 3x3 via backtracking algorithm, but unfortunately I am already late for a game of Go with one of my friends around town. Cheers --Brad

1 1

[math-fun] a(n)+a(n+1) is pandigital
by Éric Angelini 06 Oct '19

06 Oct '19

Hello Math-Fun, a(1)=1 no duplicate term lexicographically first sequence such that a(n) + a(n+1) is pandigital: How many (positive) terms in S? S=1, 1023456788, 10, 1023456779, 19, 1023456770, 28, 102345661, 37,... (the -9 pattern between even indexed terms is obvious and so is the +9 pattern between the odd indexed ones) Best, É.

1 0

[math-fun] ries near miss
by Bill Gosper 04 Oct '19

04 Oct '19

Out[10]= {-(Log[7]/(2 ProductLog[-1, -(((5/(1 + E^3 Sqrt[\[Pi]]))^(1/4) Log[7])/( 2 \[Pi]))])), (8 Sqrt[3/23] \[Pi] Sinh[(Sqrt[23] \[Pi])/6])/(-1 + 2 Cosh[(Sqrt[23] \[Pi])/3])} In[11]:= N@% Out[11]= {0.368412535931434, 0.368412535931434} In[12]:= Equal @@ %% Out[12]= False The 2nd expression is, amazingly, Integrate[QPochhammer[q,q],{q,0,1}]. https://en.wikipedia.org/wiki/Euler_function I can't quite coax it out of ries. Does anyone have a closed form for Integrate[DedekindEta[Log[q]/2/I/π],{q,0,1}] ~ .3416968346182687559823? It will probably involve Gamma, which is not yet in ries. —rwg

1 2