September 2019 - mathfun@mailman.xmission.com

Re: [math-fun] Three equal sums
by Keith F. Lynch 19 Sep '19

19 Sep '19

Neil Sloane <njasloane(a)gmail.com> wrote: > Keith, Bearing in mind your remark about tuits, I created A327448 and > A327449 for cubes and primes. > If you have others (squares?), please post them here - or email them to me. Thanks. I'd be more comfortable if someone else confirmed them. I lost a lot of confidence due to my screwup with the areas of plane-cube intersection polygons. I should really be trying to figure out where I went wrong on that rather than doing new stuff. (But new stuff is more fun.) Plus, I've been making a lot of careless errors this week due to being very short on sleep. (I'm short on sleep because a company has been dumping tons of dirt right across the street literally all night long every night, and slamming the dump truck tailgates, which is louder than a shotgun blast. I may have to hire a lawyer.) Allan Wechsler <acwacw(a)gmail.com> wrote: > I just spent ten minutes trying to partition the first n squares > into three parts with equal sums. I can plug any sequence whatsoever into my program. Maybe I can triple the number of OEIS sequences by plugging every OEIS sequence into it. :-) Bearing in mind the caveat above, here are my results for squares: Starting with n = 13: 1,0,0,0,137,211,0,0,0,3035,0,0,0,120465,259383 The unique solution with 13 is: 1 + 9 + 25 + 36 + 81 + 121 = 16 + 49 + 64 + 144 = 4 + 100 + 169

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[math-fun] Three equal sums (was Re: Returns to zero when subtracting and adding squares)
by Keith F. Lynch 19 Sep '19

19 Sep '19

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > Interestingly, the solution for cubes, A113263, is phrased in OEIS > as "a(n) is the number of ways the set {1^3, 2^3, ..., n^3} can > be partitioned into two sets of equal sums." I think I like that > phrasing better than the "sum to zero" phrasing, as it leads to > further questions, such as how many ways a set can be partitioned > into *three* sets of equal sums. Surprisingly, that isn't in OEIS, nor for primes, nor for squares. I'll add them when I get a tuit with fewer corners. Here's how many ways the first n cubes can be partitioned into three sets of equal sums, starting with n=23: 1,0,0,691,3416,0,233,1168,0,17714,30601,0,3087050,... Here's the unique first solution: 27 + 216 + 1000 + 2197 + 5832 + 6859 + 9261 = 1 + 64 + 343 + 512 + 1728 + 4096 + 8000 + 10648 = 8 + 125 + 729 + 1331 + 2744 + 3375 + 4913 + 12167 And here's the sequence for primes, starting with n=10: 3,0,0,0,0,0,423,0,2624,0,13474,0,611736,0,4169165,0,30926812,0,... (No, every other term isn't always zero.) And here's one of the first solutions for primes: 3 + 17 + 23 = 2 + 5 + 7 + 29 = 11 + 13 + 19

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[math-fun] self-avoiding dragons
by Bill Gosper 19 Sep '19

19 Sep '19

It's a bit hard to get a self-avoiding polygon by periodically sampling the Dragon function. Here's one. <http://gosper.org/dragavoid.png> The usual recipe for a self-avoiding Dragon polygon is the "median curve <http://gosper.org/dragmed.png>", which, for fixed n, samples points m/2^n, but then averages pairs for adjacent m. Notice how this "median curve" (barely) stays inside the Dragon image. So for each n, there's a sequence of 2^n (unequally spaced) points in [0,1] that draw a median curve. Julian has found a remarkably simple formula <http://gosper.org/drag samp=med.png> (i.e., periodic sampling except for some phase modulation). This wasn't trivial—many median curve vertices have multiple preimages. Which are found by his function undrag, which of course inverts the function dragun. Self avoiding polygons are attractive because they're easy to follow, but I object to calling them "spacefilling curves", which anything but self avoid. They hit uncountably many points twice, and densely many thrice! —rwg

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[math-fun] A bbp for the numbers different from 2 and 3 is it possible??
by françois mendzina essomba2 17 Sep '19

17 Sep '19

I came very close to a generalized BBP formula for all numbers. But that's a pity ... I still found some formulas that I suppose new... x=sum(sum(((-1)^(j+1)*(binomial(n,j))*sinh(j*x))/k^j,j=1..n)/n,n=1..inf); by variable change x=sum(sum(((-1)^(j+1)*(binomial(n,j))*sin(j*x))/k^j,j=1..n)/n,n=1..inf); we cant get: Pi/4=sum(sum(((-1)^(j+1)*(binomial(n,j))*sin(j*Pi/4))/k^j,j=1..n)/n,n=1..inf); Pi/8=sum(sum(((-1)^(j+1)*(binomial(n,j))*sin(j*Pi/8))/k^j,j=1..n)/n,n=1..inf); true result with some exceptions on integers k Pi/4=sum(sum(((-1)^(j+1)*(binomial(n,j))*sin(j*Pi/4))/2^j,j=1..n)/n,n=1..inf); Pi/4=sum(sum(((-1)^(j+1)*(binomial(n,j))*sin(j*Pi/4))/3^j,j=1..n)/n,n=1..inf); Pi/4=sum(sum(((-1)^(j+1)*(binomial(n,j))*sin(j*Pi/4))/5^j,j=1..n)/n,n=1..inf);

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[math-fun] Youtube, also good for experiments.
by Brad Klee 16 Sep '19

16 Sep '19

Last week, a //Limenitis Arthemis// happened to be in the wrong place at the wrong time, sunning itself near a walking trail between my house and the Physics laboratory. This species is not currently threatened by extinction. There are plenty to observe all around town, so I did not feel too bad about capturing it with my bare hands and paralysing it by the method of thorax pinching. Later I put it into a plastic bag, and killed it by freezing. ( Though it does go against Buddhist vows. ) Students should not be recommend to become "copycat killers". That is a very bad idea. The purpose of collecting the specimen was to take a closer look at the scales, including an exploratory foray into measuring reflection and transmission. The videos turned out okay: https://www.youtube.com/watch?v=FrRozvSffqc https://www.youtube.com/watch?v=ldpUEeWwhFE A scanning procedure along a distal wing cell goes through a reflection sequence--light-dark-light--which supports what we already know, that the iridescent photonic structures are more reflective than the dyed-black scales. The body and tracheal tubes are also very reflective, leading to quite a nice light show. Comparatively, scattering from iridescent scales is more dispersive, and scattering from tubes produces tight and interesting curvilinear forms. We also noticed transmission and diffraction through the wings, so made a second recording. The spot pattern does appear at first glance to be same or similar to the typical sinc-squared function. The data shows a directional bias at about a right angle to the tracheal tubes bounding any particular wing cell. One more hypothesis is that the directional bias *is not* due to bias in damage done to the wings. Though it could be, so we will probably test more later on a few of the roadkill samples. There are a number of ways this project could be improved: better quantification, more colors, standard intensity, LED illumination, spectrometry, theory and calculation, etc... "Light from Orion's stars also reflects off the big fat moon of September, barely noticeable photons, unless you've already killed a little Arthemis, on the battlefield where the sun says I hate you." --Brad

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Re: [math-fun] Flat earth
by Keith F. Lynch 15 Sep '19

15 Sep '19

You just need to replace Einstein's physics with Mercator's. Space repeats in the east-west direction with a period of Eratosthenes' constant, epsilon (about 40,000 kilometers). The vacuum speed of light isn't constant, but varies with both altitude and with so-called latitude. Where c is the local speed of light, c0 is the speed of light at sea level at the equator, l is the latitude, a is the altitude, and epsilon-bar is Eratosthenes' reduced constant, (about 6366 kilometers, epsilon divided by 2 pi), I think the correct equation is c = c0 / cos(lat) exp(a/epsilon-bar). The size of atoms, hence of objects made of atoms, are in proportion to c. So objects change in size with latitude and altitude. They change in density by the cube of this factor, so their mass remains constant. Because of the size increase, they can reach points at infinity (the so-called north pole, south pole, and center of the Earth) in finite time. The varying c results in light curving upwards. That's what results in the illusion of a horizon, and in Earth appearing round when it's viewed from a high altitude. Someone please double-check my math. For extra credit, figure out how to make Earth a sphere again, but wrongside out, with the stars and galaxies on the inside, and solid rock extending to infinity in all directions.

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[math-fun] Math synesthesia
by Henry Baker 14 Sep '19

14 Sep '19

Is it just me, or do others on this list experience numbers, formulae, etc., with more senses than just eyesight? For example, I find formulae that have polarity -- e.g., determinants -- to feel "sharp" and/or have a sharp taste, whereas formulae that are non-negative to have a rounded shape and/or have a smooth taste. Thus, a squared determinant loses its sharpness. By analogy, quadratic residues feel "smoother" than do non-residues. Perhaps these feelings come from formulae such as: abs(x)^2 = x^2 (abs(x) has a *sharp* edge at x=0, whereas abs(x)^2=x^2 does not.) Perhaps numbers & formulae could also come with *sound effects* ? --- Should math education and/or math museum exhibits attempt to capitalize on math synesthesia?

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[math-fun] Flat earth [was: math-fun How many 5 digit #s are (US) zip codes?]
by Bill Gosper 14 Sep '19

14 Sep '19

> From: Thane Plambeck <tplambeck(a)gmail.com> > To: Andres Valloud <avalloud(a)smalltalk.comcastbiz.net> > Cc: math-fun <math-fun(a)mailman.xmission.com> > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > The G4Ger who did the zip code act is named David Rosdeitcher. He lives > near Boulder, CO. > > He has several flat Earth videos on YouTube. > Until now, I've managed to ignore flat earth advocates as crackpots amusing each other by weaving tangled webs: Wikipedia: When satellite images showed Earth as a sphere, Shenton remarked: "It's easy to see how a photograph like that could fool the untrained eye".[15] <https://en.wikipedia.org/wiki/Modern_flat_Earth_societies#cite_note-15> Later asked about similar photographs taken by astronauts, he attributed curvature to the use of wide-angle lens <https://en.wikipedia.org/wiki/Wide-angle_lens>, adding, "It's a deception of the public and it isn't right".[13] <https://en.wikipedia.org/wiki/Modern_flat_Earth_societies#cite_note-Gilmore…> But if I met an advocate, morbid curiosity would compel me to ask: "Forget NASA. Why do air routes from Atlanta to Tokyo "detour" north of the Aleutians?" Great Circles were long known to old time mariners. But I'd just get sprayed in the face by Occam's Lawnmower. —rwg Yup, the Titanic was lost to the foolish belief that veering northward is actually shorter. > > On Fri, Sep 13, 2019 at 11:52 AM Andres Valloud < > avalloud(a)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 > > > 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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Re: [math-fun] Returns to zero when subtracting and adding squares
by Keith F. Lynch 11 Sep '19

11 Sep '19

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > Powers of 2? No solutions, unsurprisingly. Fibonacci series? > It's past my bedtime so I'll check tomorrow. The number of solutions for the first n terms of the Fibonacci series is: 0,0,1,0,0,2,0,0,4,0,0,8,0,0,16,0,0,32,0,0,64,... Interestingly, it's the same no matter how far into the Fibonacci series I start. As I said last night, there are no solutions for the powers of 2. That's because each one is more than the sum of all the previous. But suppose I moderate it? 2^n - 1? Still no solutions, at least for 20 or fewer terms. 2^n - 2? Nope. 2^n - n? Still nothing. 2^n - n^2. No joy. 2^n - n^3. I finally found a solution -- but only one: 1 1 -4 -19 -48 93 152 -215 256 -217 Note that all but the first two of those terms were "born negative": n^3 quickly pulls ahead of 2^n, but of course exponential fun always outruns polynomial fun in the long run. See A024013. Since the numbers rapidly get larger after that, I doubt there are any other solutions. How about primes? For the 2n+1th term, the number of solutions is: 1,0,2,1,4,25,47,237,562,1965,... (A261059). (For an even number of terms, there are of course no solutions.) The first solution is: 2 3 -5. How about just the odd primes? For the 2nth term, the number of solutions is: 0,0,1,2,4,7,33,88,344,1160,3516,13225,45876,161871,573749 (For an odd number of terms, there are of course no solutions.) The first solution is: 3 5 7 -11 13 -17 Surprisingly, it's not in OEIS. Before I submit it, would anyone care to check my numbers? Thanks. (In every case, I make the first non-zero term positive, so I don't also get mirror-images of every solution.) Interestingly, the solution for cubes, A113263, is phrased in OEIS as "a(n) is the number of ways the set {1^3, 2^3, ..., n^3} can be partitioned into two sets of equal sums." I think I like that phrasing better than the "sum to zero" phrasing, as it leads to further questions, such as how many ways a set can be partitioned into *three* sets of equal sums.

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Re: [math-fun] Returns to zero when subtracting and adding squares
by Keith F. Lynch 10 Sep '19

10 Sep '19

Allan Wechsler <acwacw(a)gmail.com> wrote: > Michael Collins <mjcollins10(a)gmail.com> wrote: >> I suspect that, more generally, a sum a_1 + ... + a_n with each a_k >> being either k^2 or -k^2 can never be zero? > Ah, no, I thought of that, and there are examples! > 1 + 4 - 9 + 16 - 25 - 36 + 49 = 0 > Here is another: 1 - 4 - 9 + 16 - 25 + 36 + 49 - 64 = 0 Apparently these have to end with the square of a number that's congruent to 0 or 3 mod 4. Can anyone find a counterexample? The number of solutions, starting with 0, with the 1 positive, are: 1,0,0,0,0,0,0,1,1,0,0,1,5,0,0,43,57,0,0,239,430,0,0,2904,5419 I see that it's already in OEIS. A083527. Then I tried it with cubes. Here's the first solution: > 1 + 8 - 27 + 64 - 125 - 216 - 343 + 512 + 729 - 1000 - 1331 + 1728 Again, apparently these have to end with the cube of a number that's congruent to 0 or 3 mod 4. The numbers of solutions, starting with 12, with the 1 positive, are: 1,0,0,2,1,0,0,2,62,0,0,268,356. A113263. I also tried 4th powers. And 5th. And 1st. A058377. Again, congruent to 0 or 3 mod 4. Powers of 2? No solutions, unsurprisingly. Fibonacci series? It's past my bedtime so I'll check tomorrow.

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