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[math-fun] In case you hadn't heard, Wolfram has figured out physics.
by Bill Gosper 11 May '20

11 May '20

It turns out General Relativity was a mistake. https://writings.stephenwolfram.com/2020/04/finally-we-may-have-a-path-to-t… —Bill

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[math-fun] 3-smooth numbers #1
by jacobs＠xmission.com 10 May '20

10 May '20

Hi, Sorry this is so long, if you hate 3-smooth numbers, run away now... I'm Steve Jacobs, the new kid on the block. I'm a retired electrical/computer engineer with a passion for Number Theory. In a prior life I spent a lot of time on probability and gambling stuff. Over the last several months, I've been playing with 3-smooth numbers. I've developed some new methods that I think are original, but perhaps I've merely reinvented an existing wheel. But, it has been great fun going down this particular rabbit hole. Sadly, formal proofs are not my strong suit, so I can't characterize my results as more than speculation or conjecture. I'm hoping that some of you will be able to easily (dis-)prove my conjectures, or point to prior work that I've reinvented. I actually started with 5-smooth numbers, and found the algorithm by Dijkstra for generating them in sequence by using queues to store prior results. I adapted this method for 3-smooth numbers, as well as other numbers that I cal {p,q}-smooth (with p,q prime and (p < q)). So {2,5}-smooth numbers and {3,5}-smooth numbers are obvious subsets of the Hamming numbers. It occurred to me that {p,q}-smooth numbers should be closely related to the continued-fraction expansion of log(q)/log(p), so I generated convergents and semi-convergents of log(3)/log(2) and started looking for them to show up when generating 3-smooth numbers. I like to reduce a(n) = A003586(n) = (2^i * 3^j) to the (i,j) pair of exponents, which corresponds to (A022328(n), A022329(n)). So I listed (i,j) pairs for 3-smooth numbers, and looked at how the exponents change when advancing from a(n) to a(n+1). So I generated the first million and looked for patterns in the results: Note: form [di dj]h is the change in (i,j) to go to a(n+1), followed by the "hop index" that I use to track how a(n) "hops around" the I-J plane. n i j [di dj]h ------------------------------------- 1: (0 0) [1 0]0 2: (1 0) [-1 1]1 3: (0 1) [2 -1]2 4: (2 0) [-1 1]1 5: (1 1) [2 -1]2 6: (3 0) [-3 2]4 7: (0 2) [2 -1]2 8: (2 1) [2 -1]2 9: (4 0) [-3 2]4 ... (lots deleted) 999991: (844 590) [485 -306]17 999992: (1329 284) [-1054 665]40 999993: (275 949) [485 -306]17 999994: (760 643) [485 -306]17 999995: (1245 337) [-1054 665]40 999996: (191 1002) [1539 -971]18 999997: (1730 31) [-1054 665]40 999998: (676 696) [485 -306]17 999999: (1161 390) [-1054 665]40 1000000: (107 1055) [1539 -971]18 So the hop from a(n) to a(n+1) is always either (di, -dj) or (-di, dj) corresponding the rational factors (2^di / 3^dj) or (3^dj / 2^di), which is sorta obvious, advancing always exchanges some power of 2 for some power of 3, or vice-versa. Perhaps less obvious is (conjecture) that |di / dj| is always a (semi-)convergent of CF of log(3)/log(2). Digging deeper, I found simple patterns in the points in I-J space that share the same hop when going from (i,j)=(A022328(n), A022329(n)) to (i,j)=(A022328(n+1), A022329(n+1)). For a given hop(h)==[di,dj] the sites that use hop(h) are covered by a rectangle in the I-J plane. Other than the trivial hop(0)=[1 0] case, each of these rectangles have one side that overlaps with the I axis. The other thing that jumps out is that these hop rectangles grow pretty quickly, so it doesn't take very many to cover huge chunks of the I-J plane. In fact, if all my calculations pan out, then a mere 569 distinct hops are sufficient to cover enough of the I-J plane to include (i,j) points up to and beyond a(10^100). It then dawned on me that a little bit of number crunching on the list of hops could create a lookup table to take (i,j) and find the right hop to propagate to the next 3-smooth number. So, I implemented a function that computes (di, dj) = hop(i, j) using a quick table lookup, to give the exponents of the next 3-smooth number (i+di, j+dj). This has obvious advantage over using the queues associated with the Dijkstra method, since those queues grow large the further one goes. Another obvious advantage is that no prior history is needed to compute the next 3-smooth number in the sequence. A small lookup table based on 569 (semi-)convergents is sufficient to hop from one (i,j) to the next, so the first googol or so 3-smooth numbers. The really fun part concerns A022330 and A022331. Can anyone confirm (or refute) any of the numbers below? I've done numerous Google searchs with no luck. Note: I use the "usual" est_count(n) = (log(sqrt(6)*N))^2 / (2*log(2)*log(3) to compare with my results. The error of this estimate (presuming my numbers are correct) is listed in the "err" column: err = est_count(n) - i_indx(n) k err i_indx(10^k) [speculative A022331(10^k) numbers] ------------------------------------------------------------------------- 1 -1 41 * '*' means confirmed by A022331 or comments/links 2 -1 3237 * 3 -1 316281 * 4 1 31554641 5 -1 3154730315 6 1 315465692250 7 -5 31546495833227 8 -1 3154648849403776 9 -5 315464877601193600 10 -3 31546487686727520626 11 -5 3154648767938833673181 12 -7 315464876786544183426556 13 -7 31546487678581026503744220 14 -2 3154648767857368731985314147 15 -5 315464876785729534014640342905 16 -7 31546487678572880009625123574432 17 -3 3154648767857287267044123250286679 18 -3 315464876785728719365228433957109149 19 -2 31546487678572871863131004484995329903 20 -3 3154648767857287185579182059392377143356 ------------------------------------------------------------------------- If my numbers hold up, then est_count(n) is incredibly accurate for an estimate. k err j_indx(10^k) [speculative A022330(10^k) numbers] ------------------------------------------------------------------------- 1 -1 93 * '*' means confirmed by A022330 or comments/links 2 -1 8055 * 3 -1 793775 * 4 -4 79261054 * 5 -3 7924941755 * 6 1 792482542841 * 7 -2 79248137960872 (failed Google search) 8 4 7924812632853902 9 -1 792481251653059342 10 4 79248125048982621572 11 7 7924812503735029032298 12 6 792481250361870571977224 13 2 79248125036070733885190551 14 6 7924812503605910155393730772 15 3 792481250360579383208119832549 16 3 79248125036057821997499450803169 17 -1 7924812503605781036516819755796893 18 -3 792481250360578092019350722334486348 19 1 79248125036057809085611759700996606344 20 0 7924812503605780907397942844775140352872 ------------------------------------------------------------------------- I'm perfectly willing to share my methods for computing these numbers, but this post is already too long, so I didn't want to get too bogged down in details, especially since I could be sadly but completely wrong. Any help is greatly appreciated.

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[math-fun] Conway's "Life" & Covid rule
by Henry Baker 10 May '20

10 May '20

According to wikipedia, "3. Any live cell with more than three live neighbors dies, as if by overpopulation." In retrospect, this appears to be a "social distancing" rule. In various generalizations of "Life" -- e.g., with different sets of rules, on different grid patterns -- e.g., hex, etc. -- are there other interesting versions of "Life" with (and without) "social distancing" rules?

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Re: [math-fun] Sum of initial nth powers equal to a higher power?
by Dan Asimov 10 May '20

10 May '20

I get that 41616 = (288 x 289)/2 = (204)^2 is a square triangular number. Is that special enough, or is it even specialer? * * * I imagine a square triangular numbers — say 36 — represented as a set of points in 4-space {(s_n, t_n) in C^2} where C is the complex plane, such that {s_n} forms a 6x6 square in C, and {t_n} forms the triangle of side 8, each centered at the origin. And then computer graphically, this set of 36 points in 4-space rotates around, perhaps randomly, at each step being projected onto the first 2 (real) coordinates and displayed on a computer screen. It should appear to be a square morphing into a triangle and back again. By varying the one-to-one correspondence between the 36 indices of {s_n} and {t_n} (i.e., considering {(s_n, t_𝜎(n)} instead of {(s_n, t_n)} where 𝜎 is a permutation of {1,2,...,36}, the exact set of points in 4-space will vary. I wonder if there are criteria for an "optimal" one-to-one correspondence. This suggests considering a homeomorphism from a square S = [-1,1]^2 in R^2 to a triangle T = C * convex hull of {(2,-1-1), (-1,2,-1), -1,-1,2)} in R^3, where C = (64/243)^(1/4) is a linear scaling factor to make their areas equal. Assuming that a homeomorphism h : S —> T is angle-preserving (conformal) on their interiors and takes the center of the square to the center of the triangle, there is only one real parameter remaining — one degree of freedom. That can be specified by the image of the positive x-direction in R^3, i.e., Dh(d/dx). What should it be for the optimal homeomorphism? —Dan ----- If we allow the sum to be a different power ... The sum of the first N cubes is the square of a triangle. 1^3 + 2^3 + 3^3 + 4^3 = 10^2 Sometimes a triangle is also a square: T(8) = 36 = 6^2. The square of a square is a 4th power: 1^3 + 2^3 + ... + 8^3 = 1296 = 36^2 = 6^4 The next triangle-square is 1225. 1^3 + 2^3 + ... + 49^3 = 1225^2 = 35^4 There's a hidden fact in the Ramanujan-Hardy 1729 story, but it's never mentioned. 1729 is a (base 2) pseudoprime: 2^1728 (mod 1729) = 1. Here's how I flunked my Ramanujan test. My University of Arizona employee number was 41616. I didn't look at it very often-- mainly when I used my ID card to check out a library book. Years later, I noticed the number was special. In fact, very special. Puzzle: Why is it special? Puzzle2: Why is it very special? -----

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[math-fun] Looking for Feller Probability Theory" volume 2
by Dan Asimov 10 May '20

10 May '20

Does anyone have a copy of *[volume 2]* of "Introduction to Probability Theory and Its Applications" — 2nd edition, by William Feller — *[copyright 1971]* in *very good condition* that they'd be willing to sell (or donate) to me? I recently bought a copy online that turned out to be in much worse physical condition than advertised plus strongly smelling of mildew, so I need another copy to replace it. If you do, please write me directly at <asimov(a)msri.org>. Many thanks, Dan

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[math-fun] Comment on tian-tsui
by Brad Klee 09 May '20

09 May '20

Apparently, for over 2000 years, the Chinese have used iridescent kingfisher feathers in the fabrication of beautiful adornments: https://en.wikipedia.org/wiki/Tian-tsui (But this came to an end during the Mao era.) The good analogy is between kingfishers and dung beetles, because both species are examples of beauty born from shit, or at least from a "fetid" mess. If you are trying to understand Olson's famous poem... There are reports from the 19th century of maggots being found in kingfisher nests, and I'm sure that similar observations could be made today, see also: https://www.inaturalist.org/taxa/2314-Alcedinidae Maggots are the larvae of flies, and from what I've seen in the field lately, beetles have to share dung pats with species of Diptera--yes it is disgusting, and nothing like honey. Too bad Charles Olson is not alive today, because I think he would appreciate the special symmetry of the dung beetles, even if no one else gives a damn. Yet it is only a part of the epic poem of the anthropocene. Sorry again, I will shut up for a while. --Brad

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[math-fun] Pattern in Poetry
by Brad Klee 09 May '20

09 May '20

This is the fourth or fifth draft of a poem I am somewhat happy with: https://0x0.st/i__e.pdf One fun part was realizing that an odd spacing pattern on the first three stanzas could be appropriate. Deciding which pattern to use exactly is something of a mathematical question. I wanted it to zig-zag as butterfly flight patterns often do. Perhaps I don't have it right just yet? If anyone wants to engage in constructive criticism, on line spacing or otherwise, comments welcome! --Brad

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Re: [math-fun] Dodecahedron packing conjecture
by Steve Witham 09 May '20

09 May '20

> From: Bill Gosper <billgosper(a)gmail.com> > Date:Â 5/9/20, 2:46 PM > > Ages ago I mentioned that regular dodex can interlock in an "airtight" sheet > <http://gosper.org/dodex.gif>, > analogous to a sheet of cubes ("Martin's Marbles") > <http://gosper.org/martinsmarbles.png>.Â Isn't this just a plane > section perpendicular to (1,1,1) through the 3D endo-dodec checkerboard... I "see" the cross-section that's a regular hexagonal tiling, I see it as cutting through the "equators" in dodex.gif. I'm pretty convinced the dodecahedrons don't intersect. I see an analogy to martinsmarbles. My brain can't grasp the endo-dodec picture well enough to answer the question. On both the dodex and martinsmarbles, it seems to me that the airtightness is just a 2D slice. There are leaks "of measure zero." Both arrangements have corner points with air on both sides. On dodex it's clearer to me anyway: some of the cutaways show pyramidal air-pockets with their tips touching the "equator". I'm pretty sure the air-pyramids come in pairs with tips (almost?) touching. Is that why you put scare-quotes on "airtight"? On May 9, 2020, at 4:00 PM, James Propp <jamespropp(a)gmail.com> wrote: > I don't understand this [endo-dodec] question. The airtight sheet is a 3-dimensional > structure; a plane section of a 3-d honeycomb is a 2-dimensional structure. From: Veit Elser <ve10(a)cornell.edu> > Jim, look at a the cut-away dodec. on the left of <http://gosper.org/dodex.gif>. > ThereÂ’s a special plane orthogonal to the 3-fold axis where the cross-sections > are regular hexagons. You need to check that when extended above and below > this plane the dodecÂ’s donÂ’t interpenetrate, but it seems to me that part of the > construction also holds water. Heh. --Steve

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[math-fun] Sum of initial nth powers equal to another nth power?
by Dan Asimov 09 May '20

09 May '20

Famously, 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2 is the only case where the sum of an initial sequence of squares equals another square (ignoring 1^2 = 1^2). Are there any non-trivial examples of this for powers higher than 2 ? (For non-initial cases, 3^3 + 4^3 + 5^3 = 6^3 is the first example, and this website <https://www.mathpages.com/home/kmath147.htm> gives many other examples for cubes.) But I'm interested in when 1^k + 2^k + ... + n^k is an exact kth power for k > 2 and of course n > 1. —Dan

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[math-fun] Playing the numbers: another miraculous hit!
by Brad Klee 09 May '20

09 May '20

During a combinatorial search for more well-integrable geometries (ha ha), the following differential equation turned up: 8*(2 - 3*x)*T - d/dx( x*(1 - x)*(25 - 27*x)*T' ) = 0. As the period constraint of a Hamiltonian: 2H = r^2 - (8/25)*r^4 + (18/625)* r^6 - (2/625)* r^6 *Cos[6*z] . ( in (r,z) polar coordinates ) The integer coefficients of the period expansions are: T@0=1, 16, 330, 7360, 170650, 4050816, 97699056, 2383814400,... T@(25/27)=1, 84, 16830, 3971640, 1030948650, 282553385184, 80264956143456,... T@1=1, 4, 30, 280, 2890, 31584, 358176, 4168560, 49455450, 595480600,... and the integer coefficients of nome expansions are: q@0=0, 1, 40, 1746, 79632, 3730134, 177917808, 8598649428, 419770317216,... q@(25/27)=0, 1, 285, 122346, 59567918, 31213994709, 17129120744817,... q@(1)=0, 1, 65, 5550, 540250, 56586875, 6208576875, 703359012500, ... Only T@1 is in OEIS: https://oeis.org/A274665, and it comes from a list by Bostan et al. see ref. [1,2] entry SS[16]. The authors are interested in connections to physics, so I think it's right in line to mention this alternative interpretation from Hamiltonian mechanics. Entries SS[2] and SS[15] have similar interpretations in terms of Hamiltonian functions with elliptic level curves. I checked the list given on J.A. Weil's website, and didn't find any other matches with my own records. It would be nice to have a more complete list of similar geometries, but so far it has not been too easy to make progress (stronger, faster algorithms are always needed). --Brad [1] https://arxiv.org/pdf/1507.03227.pdf [2] http://www.unilim.fr/pages_perso/jacques-arthur.weil/diagonals/3var/3var_OR…

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