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[math-fun] stupid number theory problem
by Henry Baker 26 Dec '19

26 Dec '19

Consider the sequence of integers generated by: Given a prime p, Ap_0 = p Ap_(n+1) = 2*Ap_n + 1 This sequence *terminates* when Ap_n *isn't* prime. Finally, define the function F(p), p prime, as the *length* of this sequence Ap_n. Informally, F(p) is the length of an all-prime sequence consisting of p, 2p+1, 2(2p+1)+1, ... These sequences of primes seem to be common enough, that I'm guessing that for *every* n, there exists a prime p such that F(p)=n. Here's the real kicker: there probably isn't a prime p s.t. F(p)=oo, is there ?

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[math-fun] Boustrophedon primes
by Walter Trump 23 Dec '19

23 Dec '19

The next Boustrophedon primes are 3821 and 3989. Here is a continuation of Eric's arrangement of numbers (you cannot read the digits, they are too small): https//www.trump.de/Boustrophedon-Primes.png Walter

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[math-fun] Hacking RSA/DH in Lisp
by Henry Baker 23 Dec '19

23 Dec '19

The Linux bash script at the end of this message might be useful for anyone who wants to hack RSA or DH crypto in Lisp. The script converts an OpenSSL .pem file into a .lsp (Lisp) file. YMMV... cat key512.pem -----BEGIN RSA PRIVATE KEY----- MIIBOwIBAAJBALgG3dbTsWaMMd6l5KYLCVpP4O0xZ9OcMqWJHvPQ050BaFdgLRsj DOECyWb1HzWJ25E5EmwVV7xv6RTe/JzVcscCAwEAAQJBAKpFQDErQfMeOG5HlF3F zGuRiQ4uLmDvHzVmwArb51BWkQK/rYs2FRNdHwFVbECzn2FmakjokX17qJ4NiFFO QZECIQDjAnODewWyLr77b8QIFLz+Apo0tPcxWe9AW2frlsEp/QIhAM+HMce/j13H 1DANadMFnNAYAaYEHGRaDd5b722AnjkTAiAYQPjKgWhYYzowbv6Ixn5yvSBwpS0L 7snrHFj8CL1WmQIhALTx1e+qEKDeOR5ks55wyFeQdu9/qhm9g1Ybptk9SbfNAiBK 1COirtJRWGD4cWY4doI+qpR1/kGZ/w6mo6GTaQc4Vg== -----END RSA PRIVATE KEY----- cat key512.pem | ./rsa2lsp.sh ;;; RSA Private-Key (512 bit, 2 primes) (setq modulus #x00b806ddd6d3b1668c31dea5e4a60b095a4fe0ed3167d39c32a5891ef3d0d39d016857602d1b230ce102c966f51f3589db9139126c1557bc6fe914defc9cd572c7) (setq publicExponent 65537) (setq privateExponent #x00aa4540312b41f31e386e47945dc5cc6b91890e2e2e60ef1f3566c00adbe750569102bfad8b3615135d1f01556c40b39f61666a48e8917d7ba89e0d88514e4191) (setq prime1 #x00e30273837b05b22ebefb6fc40814bcfe029a34b4f73159ef405b67eb96c129fd) (setq prime2 #x00cf8731c7bf8f5dc7d4300d69d3059cd01801a6041c645a0dde5bef6d809e3913) (setq exponent1 #x1840f8ca816858633a306efe88c67e72bd2070a52d0beec9eb1c58fc08bd5699) (setq exponent2 #x00b4f1d5efaa10a0de391e64b39e70c8579076ef7faa19bd83561ba6d93d49b7cd) (setq coefficient #x4ad423a2aed2515860f871663876823eaa9475fe4199ff0ea6a3a19369073856) cat dhkey1.pem -----BEGIN PRIVATE KEY----- MFwCAQAwMwYJKoZIhvcNAQMBMCYCIQDOxpswAa6BOww02q1Y3OdzFblZJkIiQoxF FbUZgNHliwIBAgQiAiBCXWiQd0cpVXcNKj6j+knJWnLX6mbx/9o8oa99XPr4fA== -----END PRIVATE KEY----- cat dhkey1.pem | ./rsa2lsp.sh ;;; DH Private-Key (256 bit) (setq private-key #x425d689077472955770d2a3ea3fa49c95a72d7ea66f1ffda3ca1af7d5cfaf87c) (setq public-key #x4ac901eff14eb826b1c01d17408913f73fd55a30eb6d98d110f05ec07424a28a) (setq prime #x00cec69b3001ae813b0c34daad58dce77315b959264222428c4515b51980d1e58b) (setq generator 2) cat rsa2lsp.sh # Use "sed" to transform .pem keys into something readable by Common Lisp. openssl pkey -text -noout | \ tr '\n' '|' | \ sed 's/$/)\n/g' | \ sed 's/[[:space:]]*(0x10001)[[:space:]]*//g' | \ sed 's/[[:space:]]*(0x2)[[:space:]]*//g' | \ sed 's/|[[:space:]]*/|/g' | \ sed 's/RSA/\;\;\; RSA/g' | \ sed 's/DH/\;\;\; DH/g' | \ sed 's/|modulus:|/\n(setq modulus #x/g' | \ sed 's/|private-key:|/\n(setq private-key #x/g' | \ sed 's/|public-key:|/)\n(setq public-key #x/g' | \ sed 's/|prime:|/)\n(setq prime #x/g' | \ sed 's/|publicExponent: /)\n(setq publicExponent /g' | \ sed 's/|generator: /)\n(setq generator /g' | \ sed 's/|privateExponent:|/)\n(setq privateExponent #x/g' | \ sed 's/|prime1:|/)\n(setq prime1 #x/g' | \ sed 's/|prime2:|/)\n(setq prime2 #x/g' | \ sed 's/|exponent1:|/)\n(setq exponent1 #x/g' | \ sed 's/|exponent2:|/)\n(setq exponent2 #x/g' | \ sed 's/|coefficient:|/)\n(setq coefficient #x/g' | \ sed 's/|//g' | \ sed 's/://g'

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Re: [math-fun] Boustrophedon primes
by Walter Trump 20 Dec '19

20 Dec '19

Sorry for the error in my last contribution. Of course A282178 does not start with number 1. I wanted to say that the elements of A282178 are in the column with number 1. The primes of A330339 are equivalent to 1 modulo 4. The primes of A282178 are equivalent to 3 modulo 4. If n is odd p = prime(n) belongs to A282178 if 2*(s1(n) - s2(n)) + p = 1 p = prime(n) does not belong to A330339 Where s1(n) = sum(prime(k) for k=1,3,5,...,n-2) and s2(n) = sum(prime(k) for k=2,4,6,...,n-1) If n is even p = prime(n) belongs to A282178 if 2*(s1(n) - s2(n)) - p = 1 p = prime(n) belongs to A330339 if 2*(s1(n) - s2(n)) - p = -1 Where s1(n) = sum(prime(k) for k=1,3,5,...,n-1) and s2(n) = sum(prime(k) for k=2,4,6,...,n-2) Walter

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Re: [math-fun] walking every road in a city - math-fun Digest, Vol 202, Issue 28
by Stuart Anderson 17 Dec '19

17 Dec '19

One method to walk every road in a city, especially if, like a postman you must walk both sides of every street, is to treat the road network as an electrical network with unit resistors in each street , then designating one street as the battery, calculate the currents and voltages using node analysis. If the voltage is chosen as the number of spanning trees in the network, the currents and voltages should come out as integers. The network of streets then becomes a squared rectangle in the squared rectangle theory of Brooks, Stone, Smith and Tutte. To walk both sides of every street exactly once, starting at an outside corner of the rectangle, walk the diagonal to the opposite corner of that square, then at the T-intersection 'bounce' at 45 degrees like a reflected laser beam into the adjacent square, heading for it's opposite corner, and so on. The diagonal walk will eventually exit at one of the other corners. Then enter at one of the other unused corners and diagonally walk until you exit the final corner. It may happen that these two walks exhaust all the squares, and hence traverse all the streets in both directions, but it may not, there may be intranversed squares. If there are, enter one the intraversed squares at the boundary of the rectangle, and diagonal walk from there until you return to where you started that particular walk. Keep doing this until all squares are traversed. Some complications, the network may have bridges and overpasses so it is not planar. Partition the network into planar and K5 or K2-3 components and traverse those separately. One way streets and dead ends shouldn't be a problem as the can be taken out of the network and added back to the network later. Another complication is that I stead of a T-intersection in the interior of the squares rectangle you may encounter a cross, in which you need to decide a convention on which way to turn at the corner. > > Message: 4 > Date: Tue, 17 Dec 2019 22:30:17 -0500 (EST) > From: "Keith F. Lynch" <kfl(a)KeithLynch.net> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] Walking every road in a city > Message-ID: <47d0rT2pbFz1ZVx(a)panix2.panix.com> > > With a map in hand and an app on his phone, Jarad Schofer is taking > the concept of a jog through the city to a whole new level: He plans > to run every road and alley in the District. > > "You know I've lived here a long time and I kind of thought I'd run > everywhere and now I can see that's really nowhere near the case," > he said. > > He's a teacher with a doctorate in mathematics and only recently got > into running. He started with 5Ks, then moved to 10K races. > > Now with a few big races under his belt, Schofer has knocked out > 860 miles on his mission to tackle the streets of D.C. In all, he > intends to run around 2,500 miles. > > . . . > > He said the biggest challenge has been overlapping certain streets > but according to his calculations, it's impossible to avoid. > > > https://wtop.com/dc/2019/12/dc-runner-is-on-a-mission-to-cover-every-inch-o… > > What's the best algorithm for walking (or running) every road in a > city while minimizing duplication? > > Only a small part of DC is on a regular grid. Mostly its road network > is pretty much random. > > > > ------------------------------ > > Message: 5 > Date: Tue, 17 Dec 2019 23:21:13 -0500 > From: Neil Sloane <njasloane(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Boustrophedon primes > Message-ID: > < > CAAOnSgTeWCo3smnjuCWL94xwj-TYX+z0Qi_1+f63rzUiKrhQYQ(a)mail.gmail.com> > Content-Type: text/plain; charset="UTF-8" > > Allan, I had exactly the same idea, and this afternoon I created A330545! > It is indeed the key to the whole problem - and of course the recurrence > makes it a lot easier to compute. > > > > > On Tue, Dec 17, 2019 at 5:27 PM Allan Wechsler <acwacw(a)gmail.com> wrote: > > > There is an underlying sequence, which tells what columns successive > prime > > numbers land in. This is the alternating sum of the decremented first > > differences of the primes, with the sequence starting at 2. Just reading > > off Eric's picture, I get > > > > 2,2,3,2,5,4,7,6,9,4,5,0,3,2,5,0,.... > > > > A330339 is then the primes indexed by occurrences of zero in this > > alternating sum. > > > > On Tue, Dec 17, 2019 at 5:14 PM Neil Sloane <njasloane(a)gmail.com> wrote: > > > > > See https://oeis.org/A330339 > > > > > > > > > > > > On Tue, Dec 17, 2019 at 4:24 PM ?ric Angelini <eric.angelini(a)skynet.be > > > > > wrote: > > > > > > > Thank you Walter! This is an incentive > > > > for ski slopes! > > > > Best, > > > > ?. > > > > > > > > > > > > > Le 17 d?c. 2019 ? 20:05, Walter Trump <w(a)trump.de> a ?crit : > > > > > > > > > > The next Boustrophedon primes are 3821 and 3989. > > > > > Here is a continuation of Eric's arrangement of numbers (you cannot > > > read > > > > the > > > > > digits, they are too small): > > > > > https//www.trump.de/Boustrophedon-Primes.png > > > > > > > > > > Walter > > > > > > > > > > > > > > > > > > > > _______________________________________________ > > > > > 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 > > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > ------------------------------ > > Subject: Digest Footer > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > ------------------------------ > > End of math-fun Digest, Vol 202, Issue 28 > ***************************************** >

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[math-fun] Walking every road in a city
by Keith F. Lynch 17 Dec '19

17 Dec '19

With a map in hand and an app on his phone, Jarad Schofer is taking the concept of a jog through the city to a whole new level: He plans to run every road and alley in the District. "You know I've lived here a long time and I kind of thought I'd run everywhere and now I can see that's really nowhere near the case," he said. He's a teacher with a doctorate in mathematics and only recently got into running. He started with 5Ks, then moved to 10K races. Now with a few big races under his belt, Schofer has knocked out 860 miles on his mission to tackle the streets of D.C. In all, he intends to run around 2,500 miles. . . . He said the biggest challenge has been overlapping certain streets but according to his calculations, it's impossible to avoid. https://wtop.com/dc/2019/12/dc-runner-is-on-a-mission-to-cover-every-inch-o… What's the best algorithm for walking (or running) every road in a city while minimizing duplication? Only a small part of DC is on a regular grid. Mostly its road network is pretty much random.

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[math-fun] Tetranacci Tetrahedron
by ed pegg 17 Dec '19

17 Dec '19

Let t^2 =1.92756 be the tetranacci constant. Build a tetrahedron with edgelengths t^0 opposite t^3 t^1 opposite t^1 t^2 opposite t^2 This has 4 similar triangles of 2 sizes. If this tetrahedron is rolled on its edges, it only reaches 12 different orientations. -Ed Pegg Jr

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[math-fun] More On Ellipse Area Integrals
by Brad Klee 17 Dec '19

17 Dec '19

Define an ellipse of eccentricity "e" as the real points {x,y} satisfying: (1-e^2)*(x+e)^2 + y^2 = (1-e^2), y[x] = sqrt((1-e^2)*(1-(x+e)^2)). Notice that (e^2-1)*d/de*y[x-e] - e*y[x-e] = 0, and that: (e^2-1)*d/de*y[x] - e*y[x] = d/dx*((e^2-1)*y[x]) = d/dx*Cy[x] . Action angle coordinates {p,q} satisfy {x,y}=sqrt(2*p)*{cos[q],sin[q]}, with y[x] replaced by p[q]=(1/2)*((1-e^2)/(1+e*cos[q]))^2, and then: (e^2-1)*d/de*p[q] - e*p[q] = d/dq*((2*sin[q]+e*cos[q]*sin[q])*p[q]) = d/dq*Cp[q] . Subtracting both equations, and integrating between points on the ellipse circumference, obtains Int_{P1}^{P2} e*( p[q]*dq - y[x]*dx ) = e*(Apq - Axy) = Cy[x_P2] - Cy[x_P1] - Cp[q_P2] + Cp[q_P1], which says that the difference of sectorial area Apq and Cartesian area Axy is a difference of certificates evaluated at P1 and P2. ( I tested this equation numerically and it works, no problem. ) The difference Apq - Axy is also the area difference of a triangle and a quadrilateral, so can alternatively be calculated by trigonometry. When this is done separately, it should then be possible to reverse the derivation and conclude that p[q] is the desired function "1/2*r^2". It satisfies a certain differential equation, with a known, unique certificate. --Brad PS. In case there are any typos in the above, here are the computes: Factor@Dot[ D[Sqrt[(1 - e^2) (1 - x^2)], {e, #}] & /@ {0, 1}, {-e, e^2 - 1}] With[{ye = Sqrt[(1 - e^2) (1 - (x + e)^2)]}, Factor@Plus[Factor@Dot[D[ye, {e, #}] & /@ {0, 1}, {-e, e^2 - 1}], D[(1 - e^2) ye, x]]] With[{r2e = (1/2)*((1 - e^2)/(1 + e Cos[q]))^2}, ReplaceAll[ Factor@Subtract[Dot[D[r2e, {e, #}] & /@ {0, 1}, {-e, e^2 - 1}], D[(2*Sin[q] + e*Cos[q] Sin[q]) r2e, q]], Sin[q]^2 -> 1 - Cos[q]^2]]

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Re: [math-fun] Heat leaps through empty space -- quantum weirdness
by Henry Baker 17 Dec '19

17 Dec '19

Re: "Moving them apart makes the tunneling probability drop off exponentially" Hmmmm... At the vanishingly small absolute temperatures of current quantum computers, the Compton wavelength grows to become not very small; analogy to "slowing (i.e., 'cooling') neutrons to enhance probability of interaction in an atomic reactor". At 07:37 AM 12/17/2019, Mike Stay wrote: >On Tue, Dec 17, 2019 at 8:28 AM Henry Baker <hbaker1(a)pipeline.com> wrote: >> Remind me again how we're planning to insulate qubits from the environment ? > >In this particular case, the phonons tunnel from one resonator to the >other. Moving them apart makes the tunneling probability drop off >exponentially. > >For the more general case, topological qubits like nonabelian anyons >(think "smoke rings of charge that pass through each other") are >likely to be very resilient to environmental noise. >-- >Mike Stay - metaweta(a)gmail.com >http://math.ucr.edu/~mike >https://reperiendi.wordpress.com

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[math-fun] Prime modulo prime
by Éric Angelini 17 Dec '19

17 Dec '19

Hello Math-Fun, I was playing with a(n) being the remainder of the division by a(n+1) of a(n+2), with all terms prime. I've found this succession of primes, but I don't know if this is the lexico first such seq: S = 2,3,5,23,97,1187,14341,57073,... I found empirically by hand the last three terms because of this: the biggest factor that multiplies 5 to produce 3 as remainder is 4 (indeed, 23-4*5=3); the biggest factor that multiplies 23 to produce 97 as remainder is 4 (indeed, 97-4*23=5); the biggest factor that multiplies 97 to produce 23 as remainder is 12 (indeed, 1187-12*97=23); the biggest factor that multiplies 1187 to produce 97 as remainder is 12 (indeed, 14341-12*1187=97). Then, seing the pattern 4, 4, 12, 12,... I tried to multiply by 36... and got a hit: indeed 14341*36+1187 = 57073 which is prime. Any thoughts? à+ É. Catapulté de mon aPhone

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