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Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Andy Latto 08 Aug '17

08 Aug '17

On Aug 9, 2017 04:32, "Dan Asimov" <dasimov(a)earthlink.net> wrote: And paradoxically, the conformal bijection z |—> -1/(1+z) of the Riemann sphere S^2 = C u {oo} to itself that preserves the integer points Z^2 ... In what sense does this preserve the integer points? F(17) = -1/18 Andy (i.e., a fractional linear transformation of form z |—> (az + b)/(cz + d), a,b,c,d in Z, ad - bc = 1, i.e., a member of the group PSL(2,Z) = SL(2,Z) / {+-I}) ... is of order three: f(f(f(z))) == z (for example). —Dan >From: Eugene Salamin via math-fun <math-fun(a)mailman.xmission.com> >Sent: Aug 8, 2017 5:17 PM Suppose, to the contrary, that such a triangle exists. We may translate one vertex to the origin, the remaining vertices being (x1, y1) and (x2, y2). If all 4 of these numbers are even, there exists a half-size triangle. So we may assume that one of these numbers is odd, and by rotation and reflection, we can take x1 to be odd. We have x1^2 + y1^2 = x2^2 + y2^2 = (x1 - x2)^2 + (y1 - y2)^2 = s^2, and thus that 2 (x1 x2 + y1 y2) = s^2. Since s^2 is even, x1 and y1 have the same parity, and likewise for x2 and y2. So y1 is odd. But then s^2 = x1^2 + y1^2 = 2 mod 4. Hence x2 and y2 must also be odd. With all 4 numbers odd, x1 x2 + y1 y2 = 2 mod 4. But s^2 is twice this, so s^2 = 0 mod 4, and we have a contradiction. On Monday, August 7, 2017, 4:04:20 AM PDT, Bill Gosper <billgosper(a)gmail.com> wrote: More generally, there is no threefold rotational symmetry. Julian's function can produce only gridpoints: ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@ {#[[1]], #[[2]] + Floor[381 #[[1]]/128]} &@ {#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {2, 1}, # != -{1, 4} &], AspectRatio -> 1, Axes -> False, Frame -> True] gosper.org/trinsky3spiral.png And all the arguments to the Floors are merely rational. --rwg _______________________________________________ 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] Generalized Fibonacci sequences and arithmetic progressions
by James Propp 08 Aug '17

08 Aug '17

Given a two-sided generalized Fibonacci sequence ...,n-m,m,n,m+n,... (with m,n in Z and not both zero), must there exist a two-sided arithmetic progression ...,a-d,a,a+d,... (with a,d in Z and d nonzero) that is disjoint from it? Jim Propp

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Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Dan Asimov 08 Aug '17

08 Aug '17

And paradoxically, the conformal bijection z |—> -1/(1+z) of the Riemann sphere S^2 = C u {oo} to itself that preserves the integer points Z^2 ... (i.e., a fractional linear transformation of form z |—> (az + b)/(cz + d), a,b,c,d in Z, ad - bc = 1, i.e., a member of the group PSL(2,Z) = SL(2,Z) / {+-I}) ... is of order three: f(f(f(z))) == z (for example). —Dan >From: Eugene Salamin via math-fun <math-fun(a)mailman.xmission.com> >Sent: Aug 8, 2017 5:17 PM Suppose, to the contrary, that such a triangle exists. We may translate one vertex to the origin, the remaining vertices being (x1, y1) and (x2, y2). If all 4 of these numbers are even, there exists a half-size triangle. So we may assume that one of these numbers is odd, and by rotation and reflection, we can take x1 to be odd. We have x1^2 + y1^2 = x2^2 + y2^2 = (x1 - x2)^2 + (y1 - y2)^2 = s^2, and thus that 2 (x1 x2 + y1 y2) = s^2. Since s^2 is even, x1 and y1 have the same parity, and likewise for x2 and y2. So y1 is odd. But then s^2 = x1^2 + y1^2 = 2 mod 4. Hence x2 and y2 must also be odd. With all 4 numbers odd, x1 x2 + y1 y2 = 2 mod 4. But s^2 is twice this, so s^2 = 0 mod 4, and we have a contradiction. On Monday, August 7, 2017, 4:04:20 AM PDT, Bill Gosper <billgosper(a)gmail.com> wrote: More generally, there is no threefold rotational symmetry. Julian's function can produce only gridpoints: ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@ {#[[1]], #[[2]] + Floor[381 #[[1]]/128]} &@ {#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {2, 1}, # != -{1, 4} &], AspectRatio -> 1, Axes -> False, Frame -> True] gosper.org/trinsky3spiral.png And all the arguments to the Floors are merely rational. --rwg

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[math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Bill Gosper 08 Aug '17

08 Aug '17

More generally, there is no threefold rotational symmetry. Julian's function can produce only gridpoints: ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@ {#[[1]], #[[2]] + Floor[381 #[[1]]/128]} &@ {#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {2, 1}, # != -{1, 4} &], AspectRatio -> 1, Axes -> False, Frame -> True] gosper.org/trinsky3spiral.png And all the arguments to the Floors are merely rational. --rwg

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[math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Bill Gosper 08 Aug '17

08 Aug '17

Six pngs deleted in the name of "fun": gosper.org/3fold1.png, . . ., 3fold6.png plus gosper.org/minskytrispiral.png . ---------- Forwarded message ---------- From: Bill Gosper <billgosper(a)gmail.com> Date: Tue, Aug 8, 2017 at 6:53 AM To: mathfuneavesdroppers(a)googlegroups.com After finding this remarkable orbit, I'm sure Julian tried some obvious parameter tweaks, although he must have been using "circularized" Minsky rather than "tripled Trinsky". Again, the latter produces only integer coordinates. I just found these: ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@{#[[ 1]], #[[2]] + Floor[381 #[[1]]/128]} &@{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {1, 0}, #2 != {1, 0} &, 2], AspectRatio -> 1, Axes -> False, Frame -> True] ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/(128 - 1/2)], #[[2]]} &@{#[[ 1]], #[[2]] + Floor[(384 - 3/2) #[[1]]/128]} &@{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {1, 0}, #2 != {1, 0} &, 2], AspectRatio -> 1, Axes -> False, Frame -> True] ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/(128 - 1/2)], #[[2]]} &@{#[[ 1]], #[[2]] + Floor[(384 - 3/2) #[[1]]/128]} &@{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {1, 2}, #2 != {1, 2} &, 2], AspectRatio -> 1, Axes -> False, Frame -> True] ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/(128 - 1/3)], #[[2]]} &@{#[[ 1]], #[[2]] + Floor[(384 - 1) #[[1]]/128]} &@{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {6, 9}, #2 != {6, 9} &, 2], AspectRatio -> 1, Axes -> False, Frame -> True, PlotStyle -> PointSize[.001]] (Period 658047. Note increasing circularity with increasing radius, where the effects of the Floor functions diminish.) ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/(128 - 1/3)], #[[2]]} &@{#[[ 1]], #[[2]] + Floor[(384 - 1) #[[1]]/128]} &@{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {5, 0}, #2 != {5, 0} &, 2], AspectRatio -> 1, Axes -> False, Frame -> True] (The vexing white vertical stripes are some artifact of SaveAs and are not on my screen.) ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/(128 - 1/4)], #[[2]]} &@{#[[ 1]], #[[2]] + Floor[(384 - 3/4) #[[1]]/128]} &@{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {1, 2}, #2 != {1, 2} &, 2], AspectRatio -> 1, Axes -> False, Frame -> True] --rwg On Mon, Aug 7, 2017 at 3:56 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > More generally, there is no threefold rotational symmetry. > Julian's function can produce only gridpoints: > ListPlot[{3, 1} # & /@ > NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@ > {#[[1]], #[[2]] + Floor[381 #[[1]]/128]} &@ > {#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, > {2, 1}, # != -{1, 4} &], > AspectRatio -> 1, Axes -> False, Frame -> True] > > And all the arguments to the Floors are merely rational. --Bill Gosper >

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Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Dan Asimov 07 Aug '17

07 Aug '17

In this vein, it's interesting that there are plenty of tilted cubic sublattices in of the integer points Z^3 in 3-space, perhaps the simplest generated by the columns of (-1 2 2) ( 2 -1 2) ( 2 2 -1) . Cayley noticed that if F is any subfield of the reals and S is any n x n skew-symmetric (real) matrix over F, then the matrix A = (S - I)^(-1) (S + I) is an orthogonal matrix over F. Letting F = Q and applying this formula, A will be a rotation matrix over Q. Hence the columns of A are orthonormal. We can then multiply by an integer den(A) to clear the denominators, to get B = den(A) A whose columns will then generate a cubic lattice (i.e., their Z-linear combinations are the lattice points forming a set similar to the set Z^n of integer lattice points. E.g., in R^3 we could have ( 2 6 9) (-9 6 -2) ( 6 7 -6) (Is there an example where no two rows have the same set of absolute values?) —Dan Bill Gosper wrote: --- More generally, there is no threefold rotational symmetry. Julian's function can produce only gridpoints: ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@ {#[[1]], #[[2]] + Floor[381 #[[1]]/128]} &@ {#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {2, 1}, # != -{1, 4} &], AspectRatio -> 1, Axes -> False, Frame -> True] gosper.org/trinsky3spiral.png And all the arguments to the Floors are merely rational. ---

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[math-fun] Cool fact about certain matrices over GF(2)
by Henry Baker 06 Aug '17

06 Aug '17

Compute the inverse of the 3x3 companion matrix over an arbitrary field: (%i1) [M,M^^-1]; [ f ] [ - - 1 0 ] [ c ] [ 0 0 -c ] [ ] [ ] [ i ] (%o1) [[ 1 0 -f ], [ - - 0 1 ]] [ ] [ c ] [ 0 1 -i ] [ ] [ 1 ] [ - 0 0 ] [ c ] Charpoly can't be irreducible unless c=1, so choose i,f,c for x^3+i*x^2+f*x+c = x^3+x+1, which is irreducible (mod 2): (%i2) %,i=0,f=1,c=1; [ 0 0 -1 ] [ -1 1 0 ] [ ] [ ] (%o2) [[ 1 0 -1 ], [ 0 0 1 ]] [ ] [ ] [ 0 1 0 ] [ 1 0 0 ] Reduce mod 2: (%i3) rat(%); [ 0 0 1 ] [ 1 1 0 ] [ ] [ ] (%o3)/R/ [[ 1 0 1 ], [ 0 0 1 ]] [ ] [ ] [ 0 1 0 ] [ 1 0 0 ] (Note that M^^-1 = M^^6 and M^^7=I.] Choose i,f,c for x^3+x^2+1, also irreducible (in fact, the *only* other irreducible poly of deg 3): (%i4) [M,M^^-1],i=1,f=0,c=1; [ 0 0 -1 ] [ 0 1 0 ] [ ] [ ] (%o4) [[ 1 0 0 ], [ -1 0 1 ]] [ ] [ ] [ 0 1 -1 ] [ 1 0 0 ] Again, reduce mod 2: (%i5) rat(%); [ 0 0 1 ] [ 0 1 0 ] [ ] [ ] (%o5)/R/ [[ 1 0 0 ], [ 1 0 1 ]] [ ] [ ] [ 0 1 1 ] [ 1 0 0 ] Note that the inverse of the second matrix (%o5[2]) is the first matrix (%o3[1]) reflected about the central vertical axis and then cyclicly rotated vertically by one, and ditto for the inverse of the second matrix. (Due to the structure of the inverse of the companion matrix, this should be true for *every* nxn binary companion matrix.) I.e., %o3[2] is the reflection+cyclic of %o5[1], and %o3[1] is the reflection+cyclic of %o5[2]. How's that for a cool proof that these two extension fields are isomorphic! Thus, these two matrices/irreducible polynomials are in some sense "conjugates". Presumably, there are other types of conjugation that make *all* irreducible polynomials of the same degree over GF(p) in some sense "conjugate" ??

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[math-fun] Eulerian triangle mod 6, 7
by Bill Gosper 05 Aug '17

05 Aug '17

gosper.org/eulerian67.png --rwg

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[math-fun] What's the proof?
by Bill Gosper 05 Aug '17

05 Aug '17

On 2015-04-20 01:58, Gareth McCaughan wrote: <omitted praise for Kac monograph> > On 18/04/2015 10:16, Bill Gosper wrote: > >> But the cover art >> http://img2.imagesbn.com/p/9780883850251_p0_v1_s260x420.jpg >> is really stupid. --rwg > > Ha. (Note in back of book: "These pictures were provided by > R. William Gosper, Jr. of SYMBOLICS.") That was back before we even had a color display, let alone a color printer. So I just sent the publisher some random color separations of Pascal's Triangle mod a couple of things. When I saw the cover, I just assumed the book designer slapped together whatever (s)he thought looked nice. But Rohan urged me to figure out what it actually was. Surprisingly, it's a coherent picture of Pascal's Triangle mod 6 (with some residues indistinguished): Graphics[Table[{EdgeForm[#], #, RegularPolygon[{4√3 (k - n/2), -6 n}, {4,π/6}, 6]} &[<|0 -> LightGray, 1 -> Red, 2 -> White, 4 -> RGBColor @@ ({74, 131, 231}/256), 3 -> Red, 5 -> Red|>@ Mod[Binomial[n, k], 6]] /. {Red ->RGBColor @@ ({250, 46, 121}/256)}, {n, 91, 161}, {k,0, n}]] Then Set Background Color to LightGray. (Not even Rohan could figure out the textual command for this.) gosper.org/kac.png --rwg

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[math-fun] Torus puzzle
by Dan Asimov 04 Aug '17

04 Aug '17

The distant planet of Torus is basically the unit cube C = [0,1]^3 with corresponding points on opposite faces identified: (0,y,z) ~ (1,y,z), (x,0,z) ~ (x,1,z), (x,y,0) ~ (x,y,1). Or if you prefer, it's the quotient group of R^3 as an abelian group by the subgroup Z^3. ----- ----- It is desired to create a road network through Torus with two properties: I. It must be a connected network, and (When the torus is viewed as a cube:) II. There must be a circuit all around the torus from any point in the plane x = 0 to the corresponding point in x = 1, from any point in the plane y = 0 to the corresponding point in y = 1, and from any point in z = 0 to the corresponding point in z = 1. ----- For example, the 3D plus-sign P ("+") — consisting of the 3 segments connecting each face-midpoint of the cube to the corresponding face-midpoint on the opposite face — will do the job. The total length of the plus-sign is 3. However, it is not the shortest solution. Can you find a network satisfying I. and II., that also is shorter than P ??? How short can it be ??? —Dan

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