June 2020 - mathfun@mailman.xmission.com

[math-fun] Read by pairs
by Éric Angelini 22 Jun '20

22 Jun '20

Hello Math-Fun, there are only 8 terms here — then the seq stops. Can you do better (more terms with another start)? S = 22,55,628,5527,54140,52707,539525,5296736. The rule is simple: read by pairs. The first pair reads « 22,55 is 2 x (2.5) = 5 » The second pair reads « 55,628 is 5 x (5.6) = 28 » The third pair reads « 628,5527 is 62 x (8.5) = 527 » etc. Note that inside the parentheses you have only 3 elements: — the rightmost digit of a(n) — a dot — the leftmost digit of a(n+1) Note also that the first element can be zero: « 54140,52707 is 5414 x (0.5) = 2707 ». Best, É.

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[math-fun] Fwd: Decently convergent gamma (factorial) formula
by Bill Gosper 19 Jun '20

19 Jun '20

---------- Forwarded message --------- From: Bill Gosper <billgosper(a)gmail.com> Date: Fri, Jun 19, 2020 at 10:41 AM Subject: Decently convergent gamma (factorial) formula To: Wolfram Tech Support Assuming you precompute the small Zetas, z! == E^((-EulerGamma)*z + Sum[((-z)^j*Zeta[j])/j, {j, 2, k}])*Product[ 1/(E^Inactive[Sum][(-(z/n))^j/j, {j, k}]*(1 + z/n)), {n, k}] Error ~ (z/k)^(1 + k)/(1 + k) , E.g., with k=9 terms, In[349]:= List @@ %304 /. z -> 1/4 /. k -> 9 // Activate Out[349]= {(1/4)!, (1/65211575) 33554432 E^( 6424827001954153772735623651396190479/ 9669127754221773701410455552000000000 - EulerGamma/4 + \[Pi]^2/ 192 + \[Pi]^4/92160 + \[Pi]^6/23224320 + \[Pi]^8/4954521600 - Zeta[3]/192 - Zeta[5]/5120 - Zeta[7]/114688 - Zeta[9]/2359296)} In[353]:= N@%349 Out[353]= {0.906402477055477, 0.906402477055477} The Inactive was to suppress the LerchPhi junk function. In[360]:= %304 /. k -> ∞ // Activate Out[360]= z! == Gamma[1 + z] Not great, but usable for complex z. I should investigate why it gets as many digits for z = ½ as for z = ¼. This is a good time to mention the (generalized) Remez algoritthm—an ingenious way to optimize an approximation over a fixed interval. E.g., for real Factorial you only need z ϵ [0,1] or [-½,½], e.g.. Introduce a new unknown, traditionally named 𝜇, and use all your numerical coefficients as first approximations to the solution of a system of simultaneous nonlinear equations stating that the error (absolute or relative, your choice) is alternately 𝜇 and -𝜇 at a bunch of turning points that you inititally imagine are distributed along the interval like the turning points of a Chebychev polynomial mapped onto that same interval. Notice that nowhere do we minimize anything. Just equalize the ripple amplitude. Depending on the form of the approximation, it can be hard as heck to get this process started. But once you get a solution, compute where the ripples really are, and solve the new system for new coefficients. Then the process will converge like a beast (quadratically). It may happen that the form of your approximation fortuitously has an extra turning point, and you have to manually choose which ripples to equalize. But when you succeed, the final |𝜇| will probably be dramatically small. And then explode as soon as you step ouside the interval. —rwg

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Re: [math-fun] "Density" of a regular polytope
by Dan Asimov 19 Jun '20

19 Jun '20

PS An extremely kind math-fun member has forwarded to me copies of these papers — so I no longer need to acquire them. —Dan ----- In what I can read of his papers "Densities of regular polytopes, I, II, III" Coxeter defines the "density" of a stellated polytope to be the number of times that it covers (almost) each point of the sphere (if it's centrally projected to the unit sphere). Among all star polygons like the pentagram {5/2}, or {19/8}, density can be any positive integer.* But then things get very interesting: In 3D, the density can be 1, 3, 7. In 4D, the density can be 1, 4, 6, 20, 66, 76, 191. Strange numbers for these highly symmetric situations! -----

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[math-fun] "Density" of a regular polytope
by Dan Asimov 19 Jun '20

19 Jun '20

In what I can read of his papers "Densities of regular polytopes, I, II, III" Coxeter defines the "density" of a stellated polytope to be the number of times that it covers (almost) each point of the sphere (if it's centrally projected to the unit sphere). Among all star polygons like the pentagram {5/2}, or {19/8}, density can be any positive integer.* But then things get very interesting: In 3D, the density can be 1, 3, 7. In 4D, the density can be 1, 4, 6, 20, 66, 76, 191. Strange numbers for these highly symmetric situations! —Dan PS I don't have access to the 3 papers of Coxeter mentioned above. If anyone can send (any of) them to me I'd greatly appreciate it. They appear in Proceedings of the Cambridge Philosophical Society v27, 1931, p 201ff; v28, 1932 p 509ff; v29, 1933, p 1ff; respectively. * {p/q} (in lowest terms) denotes the star polygon obtained from dividing the circle into p equal parts and then connecting them with edges between pairs of vertices separated by q of these parts.

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Re: [math-fun] Schinzel-Zassenhaus conjecture
by Dan Asimov 17 Jun '20

17 Jun '20

I glanced at that yesterday and immediately didn't understand a number of terms that were used, like "the compositum of all abelian and all totally p-adic extensions of Q" for a fixed prime p. The math articles in Quanta explain things at an extremely low level and never seem to rise to mention things at a higher level. The Quanta article on this research seemed to barely mention that all the polynomials being considered had integer coefficients. It *was* mentioned, but very casually, as if it hardly mattered. —Dan Mike Stay wrote: ----- Here's numthe paper: https://arxiv.org/abs/1912.12545 -----

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Re: [math-fun] Quantifiers
by Dan Asimov 16 Jun '20

16 Jun '20

Just to be clear, that was distinctly *not* my point. I *do* believe that "to deceive" need not imply intent. But it always means that the deceivee was misled, intentionally or not. —Dan ----- as Dan Asimov (and you, and others pointed out), "deceive includes intent". -----

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[math-fun] Number of ways to pick 4 points from an m X n grid?
by Neil Sloane 16 Jun '20

16 Jun '20

If we take an m X n rectangular grid of points, and choose four of them (in binomial(m*n,4) ways), there are four possibilities: (i) the 4 points are collinear, (ii) 3 points are collinear and one point is not on that line, (iii) 3 points form a triangle of positive area and the other point is in the interior of that triangle, (iv) the 4 points form a convex quadrilateral of nonzero area. As a function of m and n, what are these numbers? This must be a classical problem, but I don't have enough data to check in the OEIS. (It might be called the "moduli space" of 4-tuples from an m X n grid.) I would like the 4 tables of numbers, and exact and/or asymptotic formulas. Which is more likely, (iii) or (iv)? I just looked at what happens if we choose three rather than four points. Now there are only two cases, collinear or not. This produces two triangles of numbers, which are now A334704 and A334705. E.g. A334704(m,n) is the number of ways to choose 3 collinear points from an m X n grid, for m >= n >= 1. Both triangles need more terms, and are lacking exact formulas (or generating functions). Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com

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Re: [math-fun] Quantifiers
by Hilarie Orman 16 Jun '20

16 Jun '20

Even with mathematics, stating the pigeon/person attack distribution is a challenge. Indeed, it could be the same pigeon attacking the same person. Every 30 seconds exactly? "The Seven Types of Ambiguity" by William Empson describes how English literature utilizes ambiguity. Nothing will come of nothing. Hilarie

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[math-fun] primes in geometric progressionm new record.
by Simon Plouffe 15 Jun '20

15 Jun '20

Hello, I continued the calculation of large primes with the function { c^k } , and found that if c = 999982.6807... then, 899 primes are generated. http://plouffe.fr/NEW/899%20primes%20in%20geometric%20progression.txt By the way, I compared the speed of calculation for <isprime> in Maple, Mathematica, Pari-Gp and PFGW. PFGW is the fastest for testing if a prime is probable, well at least for large values. But if you want from a given n to find the next prime after n then Maple is the fastest. I could not use PFGW for doing that at all, too slow. The function nextprime() within PFGW is <mickey mouse> , it works for small values but fails miserably for n > 10^20. I tried an intelligent loop on it and it is by far too slow. I believe that for once, Maple is the best. Which brings up the next question. Emile Borel once calculated what could be what is considered impossible on a human, terrestrial or cosmic scale. If I remember correctly, a probability of 1/1000000 is on a human scale impossible. It was 10 ^ (- 12) for the terrestrial scale and 10 ^ (- 50) for the cosmic scale. In other words, if you calculate that there is less than 1 in 1 million chance that an unfortunate event can happen to you, then you can rest easy and if it is on a cosmic scale then all of humanity can also sleep peacefully. Which brings me to the next question. Does anyone know what is the probability that a probable prime number of 1000 digits can in fact be composed like a Carmichael number? Best regards (and chance) to you, Simon Plouffe

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Re: [math-fun] Quantifiers
by Dan Asimov 15 Jun '20

15 Jun '20

You're right, there can be no precise definition of probability for real-world events. But in practice this means that among situations on record having approximately the same parameters as the current one, about 30% of them have resulted in (greater-than-threshold) rain (during the period in question). —Dan ----- On Mon, Jun 15, 2020 at 4:48 PM Brent Meeker via math-fun < math-fun(a)mailman.xmission.com> wrote: > The ambiguity that most often bothers me is announcements of the form, > "Thirty per cent chance of rain today." I know rain is defined as a > measure greater than 0.1" But what does the 30% mean? Does it mean > there will be at least 0.1" of rain on 30% of the people within range of > the radio station? Or does it mean that if you hear this a hundred > times, all the people in range will experience rain 30 times? -----

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