January 2017 - mathfun@mailman.xmission.com

[math-fun] Newman = -1/Oldman
by Bill Gosper 30 Mar '17

30 Mar '17

This must have been found by Newman himself but I only just noticed. Enumerate the nonnegative rationals by iterating "Newman's function": In[96]:= NestList[1/(2*Floor[#] + 1 - #) &, 0, 9] Out[96]= {0, 1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3} "Oldman's function" runs the sequence back to 0, counting the rationals: In[97]:= NestList[2*Ceiling[1/#] - 1 - 1/# &, 4/3, 9] Out[97]= {4/3, 1/4, 3, 2/3, 3/2, 1/3, 2, 1/2, 1, 0} E.g., 0 is the zeroth rational, 4/3 is the ninth. Oldman is completely unnecessary! Just negate and reciprocate the last value, and Newman runs backwards: In[100]:= NestList[1/(2*Floor[#] + 1 - #) &, -3/4, 9] Out[100]= {-3/4, -4, -1/3, -3/2, -2/3, -3, -1/2, -2, -1, ComplexInfinity} (with a much louder completion announcement). Or if you prefer, Oldman runs the negative rationals forwards: In[99]:= NestList[2*Ceiling[1/#] - 1 - 1/# &, -1, 9] Out[99]= {-1, -2, -1/2, -3, -2/3, -3/2, -1/3, -4, -3/4, -5/3} obviating Newman. --rwg

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[math-fun] Panel puzzle
by James Propp 18 Mar '17

18 Mar '17

You can join four isosceles right triangles along hinges (I use tape) to form a square, and then join four such square panels (again with tape) to form a long rectangle, and lastly join the first and last panels together with tape to form a closed band. If you color the two sides of the band with different colors, it's a fun puzzle to turn the band inside out, exchanging the colors. I learned about this in the 80s, and amused myself by learning how to do the moves behind my back. Does anyone know the name and/or provenance of this puzzle? Jim

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Re: [math-fun] Gap sequences for Thue-Morse
by Fred Lunnon 25 Feb '17

25 Feb '17

For case w = 1 , the resulting gap sequence is apparently the standard square-free ternary sequence S_n = 1 + T_n - T_{n-1} , with T_n denoting Thue-Morse sequence. This is indeed an "automatic sequence" in the sense of Allouche & Shallit's book of that title (2003), and generated by the width-4 iterated morphism 0 -> 03, 1 -> 02, 2 -> 21, 3 -> 20 followed by terminal symbol-to-symbol encoding 3 -> 1 . If I have understood JS correctly, it follows that both other cases w = 01, 010 are also automatic sequences, and so generated in a similar fashion. How wide are their morphisms? An ominous feature of the recursions I found earlier is the division by 3 rather than 2 involved in computing n" , suggesting that the width must be at least 6 ; if the same morphism also serves case w = 0 (and 1), the width goes up to at least 12 . WFL >> >> On 1/21/17, Jeffrey Shallit <shallit(a)uwaterloo.ca> wrote: >>> I just want to point out that there is a technology that can answer >>> questions like this purely mechanically, by a machine computation. >>> Given any particular w, or even for all w of a given length, the gaps >>> between occurrences are specified by a finite automaton, which can be >>> computed "automatically". We can also produce a so-called "linear >>> representation" that specifies gap positions and/or their first >>> differences. See the papers on my home page, particularly >>> E. Charlier, N. Rampersad, J. Shallit, Enumeration and decidable >>> properties of automatic sequences, Int. J. Found. Comput. Sci., 23 >>> (2012), 1035-1066. >>> >>> Jeffrey Shallit >>> https://cs.uwaterloo.ca/~shallit/papers.html >>> >>> >>> On 1/21/17 9:26 AM, Fred Lunnon wrote: >>>> As an indirect result of contemplating Jim Propp's Mathematical >>>> Enchantments >>>> >>>> https://mathenchant.wordpress.com/2017/01/16/avoiding-chazakah-with-the-pro… >>>> I became distracted (the way one does) into examining the sequence of >>>> gaps >>>> between repetitions of a given word w in the Thue-Morse sequence. >>>> >>>> Obviously the gap sequence will vary according to w . But it turns >>>> out that every such sequence is related by linear transformation and >>>> index shift to one of just three particular cases: w = 0, 01, 010 >>>> --- see for example sect. 4 of >>>> Lubomira Balkova, Edita Pelantov a, Wolfgang Steiner >>>> "Return words in the Thue-Morse and other sequences" >>>> >>>> https://hal.archives-ouvertes.fr/file/index/docid/90970/filename/BPS.pdf >>>> >>>> Initial segments, compressed into infinite words --- >>>> Thue-Morse: >>>> 01101001 10010110 10010110 01101001 10010110 01101001 01101001 >>>> 10010110 ... >>>> Gaps w = 0 : >>>> 21020121 01202102 01202101 21020121 01202101 21020120 21020121 >>>> 01202102 ... >>>> Gaps w = 01 : >>>> 11201102 11202110 11201102 11011202 11201102 11202110 11202112 >>>> 01102110 ... >>>> Gaps w = 010 (halved): >>>> 21032301 21030123 21032301 23210301 21032301 21030123 21030121 >>>> 03230123 ... >>>> >>>> From case w = 0 , the gap sequence for w = 1 follows by >>>> transposing >>>> 0 with 2 : which may be considered linear S_n -> 2 - S_n , or else >>>> the >>>> limit of shifts n -> n + 2^k for k = 0,1,2,... Either way, the >>>> result >>>> commences >>>> 0 1 2 0 2 1 0 1 2 1 0 2 0 1 2 1 ... >>>> 'Ang abaht --- >>>> < = > < > = < = > = < > < = > = ... >>>> --- haven't we seen something very like this somewhere else recently? >>>> >>>> So it would be reasonable to expect the other two cases to have some >>>> simple relation to Thue-Morse as well. Yeah, right --- the most >>>> concise >>>> (conjectural) recurrences I have been able to come up with for S_n so >>>> far are the following horrors. >>>> >>>> Given index n > 0 , >>>> for 0 < n <= 8 , set explicitly S_1,...,S_8 = >>>> 1, 1, 2, 0, 1, 1, 0, 2 for w = 01 , >>>> 2, 1, 0, 3, 2, 3, 0, 1 for w = 010 ; >>>> define k, n', n" via >>>> 4 n' < n <= 8 n' = 2^k ; >>>> n" = (n' - (-1)^k)/3 for w = 01 , >>>> (n' + 2(-1)^k)/3 for w = 010 ; >>>> then for n > 8 , S_n = >>>> S_(n - 4*n') for 4 n' < n <= 6 n' , >>>> S_(n - 3*n' + n") for 6 n' < n <= 7 n' , >>>> S_(n - 5*n' + n") for 7 n' < n <= 8 n' . >>>> >>>> Fred Lunnon >>>> >>>> _______________________________________________ >>>> math-fun mailing list >>>> math-fun(a)mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>> >>> >> >

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[math-fun] Algebra question
by David Wilson 01 Feb '17

01 Feb '17

Consider the set S of sequences where the nth element is a residue of the nth prime. Natural elementwise addition and multiplication operations can be defined on S. With respect to these operations, S has unique additive and multiplicative identities (0 and 1), as well as additive and multiplicative inverses. S also has a subset isomorphic to the integer, where integer k corresponds to the sequence whose nth element is k modulo the nth prime. Not being a number theorist, I was wondering about the algebraic structure of this set.

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[math-fun] Should ET know number theory?
by Henry Baker 31 Jan '17

31 Jan '17

In trying to figure out how mankind might talk with an extra-terrestrial, it seems to be a given that the extra-terrestrial would have to know about prime numbers and divisibility. Is there a simple short argument for why that should be the case? After all, humans seem to have gotten along quite a while w/o knowing anything about prime numbers, and primes only seem to have forced themselves into engineering relatively recently. There's also a potentially serious problem about the representation of numbers. We love our positional notation, but where's the proof that there aren't more interesting representations? After all, the Egyptions used (or at least some small fraction (!) of them did) used "Egyptian fractions", which is similar to partial fraction decomposition. There may be a civilization that used some sort of floating point notation (with the exponent & mantissa represented in some unspecified form) from the very start, and may never have cared at all about "integers". Of course, if ET lives in the same universe, then ET would have to know about quantum theory, so how much of quantum theory is understandable w/o any serious knowledge of primes & divisibility?

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Re: [math-fun] no politics on math-fun, please
by Henry Baker 31 Jan '17

31 Jan '17

I wasn't involved (this time)... At 01:19 PM 1/31/2017, you wrote: >I strongly discourage political discussion on Math-Fun. >People (Henry?) are free to contact Fred to help organize a new group >dedicated to the forbidden topics, and may announce it here once.

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[math-fun] no politics on math-fun, please
by rcs＠xmission.com 31 Jan '17

31 Jan '17

I strongly discourage political discussion on Math-Fun. People (Henry?) are free to contact Fred to help organize a new group dedicated to the forbidden topics, and may announce it here once. I will be mulling about the possibility of a tetrahedron having a 27-point sphere. Rich

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[math-fun] Torus tiling problem
by Dan Asimov 31 Jan '17

31 Jan '17

----- QUESTION: Given the square torus T^2, what is the smallest number of triangles in a tiling of the torus by triangles such that they are all acute? I need to define a few terms for this particular question: * The "square torus" T^2 is the result of identifying opposite edges of a square. (Equivalently, it is the metric space obtained from the quotient of the plane R^2 by its integer subgroup Z^2.) * A "tiling of the torus by triangles", for the purposes of this question, is any union of triangles that equals the torus, such that a) Any two triangles have disjoint interiors, and AAAA b) No triangle's vertex lies on the interior of any triangle's edge. ----- It's well known that the square can be tiled in this sense with 8 acute triangles and no fewer. The standard example (https://www.ics.uci.edu/~eppstein/junkyard/acute-square/8-square.gif) gives, by identifying opposite edges of the square, a tiling of the square torus using 8 acute triangles. Note: The most general flat torus is obtained by identifying opposite edges of a parallelogram. Clearly, any non-rectangular flat torus can be tiled in the sense of this question using only 2 acute triangles. Can the upper bound of 8 for the square torus be improved on? And what about a non-square rectangular torus? —Dan

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[math-fun] First they came ...
by Fred Lunnon 31 Jan '17

31 Jan '17

https://terrytao.wordpress.com/2017/01/31/open-thread-for-mathematicians-on… << Of the 774 posts on this blog, I estimate that about 99% of the posts indeed relate to mathematics, mathematicians, or the administration of this mathematical blog, and only about 1% are not related to mathematics or the community of mathematicians in any significant fashion. This is not one of the 1%. >> https://en.wikipedia.org/wiki/First_they_came_... WFL

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Re: [math-fun] Torus tiling problem
by William R Somsky 31 Jan '17

31 Jan '17

Or simpler yet for the rectangle. Rotate so it's taller than wide. Draw the top edge, bottom edge, horizontal bisector and both diagonals. On Jan 31, 2017 1:43 AM, "William R Somsky" <wrsomsky(a)gmail.com> wrote: Rectangular non-square can be done w/ four. Eg on rectangle ([0,2],[0,4]) draw (0,0)-(2,0) (0,2)-(2,2) (0,4)-(2,4) (0,2)-(1,0)-(2,2) (0,2)-(1,4)-(2,2) On Jan 31, 2017 1:19 AM, "William R Somsky" <wrsomsky(a)gmail.com> wrote: > Thinking more on this, I don't think the "clearly all non-rectangular > Tori" is so clear. What about for very skew parallelograms with large > aspect ratios? Neither diagonal splits them into acute triangles. (Think of > a parallelogram with sides of 1 and 4 and angles of 45 and 135 degrees) > > On Jan 31, 2017 1:02 AM, "William R Somsky" <wrsomsky(a)gmail.com> wrote: > >> What I meant was: take your torus, cut and unroll it to be your square >> ABCD. Now, as a square, clip off BCE and glue it to the other side with BC >> on AD to make a parallelogram. Divide it on its diagonal as roll it back >> into a torus. But now, spelling it out, I see my flaw: identifying the >> edges of the new parallelogram, CD wouldn't get pasted back onto BA, but >> into the shifted EABE. Yeah, I think it's the same as you spelled out or >> maybe a "dual" where I don't violate the triangle rule but end up creating >> a different torus. >> >> On Jan 30, 2017 21:37, "Dan Asimov" <dasimov(a)earthlink.net> wrote: >> >> WRS, I don't know if I'm understanding this from your description. >> >> I'm thinking of the square being say ABCD (ccw starting from upper right). >> >> B————E————A >> | / \ | >> | / \ | >> | / \ | >> |/ \| >> C—————————D >> >> Then say let E be the midpoint of AB and F be the midpoint of CD. Finally >> one triangle is CED and the other is what remains: the triangles EBC >> and AED joined along their common edge BC = AD by the identification of >> the square to make a torus. >> >> IF that is the same idea as you are suggesting, it has the flaw that the >> vertex E (of triangle CED) is identified with an interior point of the >> edge CD. >> >> Which violates condition b) below. >> >> Or am I misunderstanding? >> >> —Dan >> >> >> > On Jan 30, 2017, at 8:13 PM, William R Somsky <wrsomsky(a)gmail.com> >> wrote: >> > >> > Two, I believe. Unless I'm mistaken, you just trim a triangle off the >> left >> > edge of the square, move it to the right side matching the identified >> > edges, and you have a parallelogram, which is trivially >> > two-acute-triangularizable. >> > >> > On Jan 30, 2017 16:44, "Dan Asimov" <dasimov(a)earthlink.net> wrote: >> > >> >> ----- >> >> QUESTION: Given the square torus T^2, what is the smallest number of >> >> triangles in a tiling of the torus by triangles such that they are all >> >> acute? >> >> >> >> I need to define a few terms for this particular question: >> >> >> >> * The "square torus" T^2 is the result of identifying opposite edges >> of >> >> a square. (Equivalently, it is the metric space obtained from the >> quotient >> >> of the plane R^2 by its integer subgroup Z^2.) >> >> >> >> * A "tiling of the torus by triangles", for the purposes of this >> >> question, is any union of triangles that equals the torus, such that >> >> >> >> a) Any two triangles have disjoint interiors, >> >> >> >> and >> >> >> >> b) No triangle's vertex lies on the interior of any triangle's edge. >> >> ----- >> >> >> >> It's well known that the square can be tiled in this sense with 8 acute >> >> triangles and no fewer. The standard example ( >> https://www.ics.uci.edu/~ >> >> eppstein/junkyard/acute-square/8-square.gif) gives, by identifying >> >> opposite edges of the square, a tiling of the square torus using 8 >> acute >> >> triangles. >> >> >> >> Note: The most general flat torus is obtained by identifying opposite >> >> edges of a parallelogram. Clearly, any non-rectangular flat torus can >> be >> >> tiled in the sense of this question using only 2 acute triangles. >> >> >> >> Can the upper bound of 8 for the square torus be improved on? And what >> >> about a non-square rectangular torus? >> >> >> >> —Dan >> >> >> >> _______________________________________________ >> >> math-fun mailing list >> >> 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 >> > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> >>

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