November 2019 - mathfun@mailman.xmission.com

[math-fun] Find the matrix
by Fred Lunnon 30 Nov '19

30 Nov '19

Given two (unordered) sets of (binary) integer vectors. Considerations of symmetry lead to the conjecture that there exists a linear bijection (indeed, a permutation of the basis) relating them. Design an (efficient) algorithm to construct some such matrix. In the case to hand, the dimension equals 20 (reducible to nearer 10 after some finagling), and vector sets have size 120 and full rank. Fred Lunnon

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[math-fun] nonstandard Heighway Dragon recipe
by Bill Gosper 26 Nov '19

26 Nov '19

Huh? <http://gosper.org/symdrag.gif> —rwg

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[math-fun] Fwd: Re: Hot tea sucks.
by Bill Gosper 25 Nov '19

25 Nov '19

Duct / Duck (sic) tape comes in handy in such situations. Or you could just drink it hot with milk, a la Brit. WFL OMG—I shoulda stood in bed! Next to which sits a huge roll of Gaffer/Gopher (sic) tape! (Way better than duct tape.) —rwg I glossed over the hassle factor of the trip to Draeger's, which has an excessively level parking lot. You might not perceive this as a disadvantage until I reveal that reverse gear on my car is broken. On 11/24/19, Bill Gosper <billgosper(a)gmail.com> wrote: > It just seized again for the first time since this discussion. I ran > out > of nerve trying to > repressurize it in the 𝜇wave, and drove to Draeger's to bum a big > rubber > band from > the produce dept. But their's were too small. Trudging back to my > car, I > found a > bigger one in the parking lot! Which, along with more 𝜇wave courage, > did > the trick. > —rwg > > On Fri, Apr 26, 2019 at 3:54 PM Bill Gosper <billgosper(a)gmail.com> > wrote: > >> I BCCed a lot of people and got a lot of nontrivial suggestions. >> (Restaurateur) Mike Wang's rubber band suggestion >> was probably best. Just encircle the lid with a big one from a >> broccoli >> bunch. It's amazing if you've never tried it. >> But that's not trivial enough! (I forget where I stashed my rubber >> band.) Someone suggested that if I have a mountain >> handy, I could drive up it. Nobody suggested the dicey solution of >> sticking the jar back in the 𝜇wave. Really hot tapwater >> should work, but our water heater is set way low. C'mon >> people—*trivial*! >> >> SPOILER: >> In[532]:= StringReverse@"!no ti pals uoy sa dil eht rednu ria )looc >> ylevitaler( fo llufgnul kciuq a wolb tsuJ" >> —rwg >> >> On Fri, Apr 26, 2019 at 8:12 AM Bill Gosper <billgosper(a)gmail.com> >> wrote: >> >>> I brew it in a 46oz pickle jar that easily tips over, and leaks if >>> the >>> lid isn't snug. >>> But if I forget to momentarily loosen the lid within a few minutes, a >>> powerful vacuum >>> develops that makes it virtually impossible to unscrew. What is the >>> trivial solution? >>> —Bill >>> >> > _______________________________________________ > 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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[math-fun] Hot tea sucks.
by Bill Gosper 25 Nov '19

25 Nov '19

I brew it in a 46oz pickle jar that easily tips over, and leaks if the lid isn't snug. But if I forget to momentarily loosen the lid within a few minutes, a powerful vacuum develops that makes it virtually impossible to unscrew. What is the trivial solution? —Bill

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[math-fun] Nice video NASA
by Brad Klee 24 Nov '19

24 Nov '19

https://m.youtube.com/watch?v=0yNzSwlnQ2Q In case you didn’t see... it takes a lot of math to make this possible, and even more to explain what it is we are seeing. No, mercury isn’t riding a perfect ellipse. Elliptic functions do better to account for precession, but still, nature is messy. I like the music too, great! And no astronauts were harmed, much less involved, another plus. Enjoy —Brad

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[math-fun] Brad's recent interest in sn|cn double angles
by Bill Gosper 21 Nov '19

21 Nov '19

prompts me to t(rot )out these four old 𝜗 half-angle formulas: Out[419]= {EllipticTheta[1, x/2, q] == (1/(2^( 1/4)))((-EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] - EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4)), EllipticTheta[2, x/2, q] == (1/(2^( 1/4)))((EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] - EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4)), EllipticTheta[3, x/2, q] == (1/(2^( 1/4)))((EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] + EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4)), EllipticTheta[4, x/2, q] == (1/(2^( 1/4)))((-EllipticTheta[2, 0, q]^3 EllipticTheta[2, x, q] + EllipticTheta[3, 0, q]^3 EllipticTheta[3, x, q] + EllipticTheta[4, 0, q]^3 EllipticTheta[4, x, q])^(1/4))} Check: In[432]:= Assuming[0 < x < \[Pi]/2 && 0 < q < 1, FullSimplify@Series[%419, {q, 0, 22}, {x, 0, 22}]] // tim During evaluation of In[432]:= 419.850403,4 (*Seven minutes*). Out[432]= {True, True, True, True} —rwg

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[math-fun] Primes with strokes
by Éric Angelini 19 Nov '19

19 Nov '19

Hello Math-Fun, I was wondering if this seq S was sound: "Put a stroke behind every prime digit. Any integer formed by the concatenated digits between two strokes is prime." S = 1,3,2,4,7,5,6,8,9,10,11,14,16,18,83,13,17,,... The same with the strokes: S = 1,3,\2,\4,7\,5,\6,8,9,10,11,14,16,18,83\,13\,17\,... If I'm not wrong, S is the lexicographically first seq of distinct terms with this property. The successive primes are: 13, 2, 47, 5, 689101114161883, 13, 17,... What do you think? Best, É.

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[math-fun] Yellows = non-Yellows
by Éric Angelini 18 Nov '19

18 Nov '19

Hello Math-Fun, S is the lexicographically earliest seq of distinct integers [separated only by a comma, and starting with a(1) = 1] such that the yellow digits appear in the exact same order as the non yellow digits. A yellow digit is a digit in contact with at least one comma of S. Question: For which a(n) will the quantity of yellow digits equal the quantity of non yellow digits [if such an a(n) exists]? S starts like this, if I am not wrong: S = 1,2,3,4,5,6,78,9,10,11,12,13,...,97,98,99, 110,120,130,140,150,160,170,180,190,111, 100,112,113,114,121,115,131,116,141,117,... (S is perhaps already in the OEIS -- but I could not find it; anyway the question remains). Best, É. à+ É. Catapulté de mon aPhone

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[math-fun] Symmetric Angle Doubling Formulas for JacobiSN | CN | DN .
by Brad Klee 16 Nov '19

16 Nov '19

Cross-Post from Wolfram Community: https://community.wolfram.com/groups/-/m/t/1826418 News to me. --Brad

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[math-fun] the impressive speed of primecount
by Simon Plouffe 16 Nov '19

16 Nov '19

Hello, there is a little jewel of program called primecount (easy to find on Windows, mac or linux). It counts the primes. Here is a table of results (see below), from 10^1 to 10^20 with primecount and mathematica : mathematica stops giving answers after 10^14 and takes 26.2656 seconds to find PrimePi[10^14] when primecount takes only : 0.069 seconds (!). This is 431 times faster. I do not mention Maple for decency, it is too slow by far, Pari-Gp is not bad but limited in range too. The -g option stands for the Gourdon method (apparently the fastest). You can download the program here : https://github.com/kimwalisch/primecount it comes with a short but very useful explanation. See for yourself : results are astonishing. The recent record result of Pi(10^27) is mainly due to this program. See http://oeis.org/A006880 for more details. Last thing : the program is 1.52 megs in size and takes all the cpus/threads you may have on one machine automatically. primecount -g 10^1 --time; 4 seconds: 0.000 primecount -g 10^2 --time; 25 seconds: 0.000 primecount -g 10^3 --time; 168 seconds: 0.000 primecount -g 10^4 --time; 1229 seconds: 0.000 primecount -g 10^5 --time; 9592 seconds: 0.000 primecount -g 10^6 --time; 78498 seconds: 0.000 primecount -g 10^7 --time; 664579 seconds: 0.000 primecount -g 10^8 --time; 5761455 seconds: 0.001 primecount -g 10^9 --time; 50847534 seconds: 0.001 primecount -g 10^10 --time; 455052511 seconds: 0.003 primecount -g 10^11 --time; 4118054813 seconds: 0.007 primecount -g 10^12 --time; 37607912018 seconds: 0.015 primecount -g 10^13 --time; 346065536839 seconds: 0.026 primecount -g 10^14 --time; 3204941750802 seconds: 0.069 primecount -g 10^15 --time; 29844570422669 seconds: 0.204 primecount -g 10^16 --time; 279238341033925 seconds: 0.647 primecount -g 10^17 --time; 2623557157654233 seconds: 2.276 primecount -g 10^18 --time; 24739954287740860 seconds: 8.334 primecount -g 10^19 --time; 234057667276344607 seconds: 35.579 primecount -g 10^20 --time; 2220819602560918840 seconds: 155.561 Timing[PrimePi[10^1]] {0., 4} Timing[PrimePi[10^2]] {0., 25} Timing[PrimePi[10^3]] {0., 168} Timing[PrimePi[10^4]] {0., 1229} Timing[PrimePi[10^5]] {0., 9592} Timing[PrimePi[10^6]] {0., 78498} Timing[PrimePi[10^7]] {0., 664579} Timing[PrimePi[10^8]] {0., 5761455} Timing[PrimePi[10^9]] {0., 50847534} Timing[PrimePi[10^10]] {0.046875, 455052511} Timing[PrimePi[10^11]] {0.1875, 4118054813} Timing[PrimePi[10^12]] {1.03125, 37607912018} Timing[PrimePi[10^13]] {5.34375, 346065536839} Timing[PrimePi[10^14]] {26.2656, 3204941750802} Have fun !, Simon Plouffe

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