December 2018 - mathfun@mailman.xmission.com

[math-fun] Irrational constants
by D J Makin 31 Dec '18

31 Dec '18

Hi all, I don’t normally start a conversation on here basically due to my lack of formal higher math education and a very specific limited math-related skill set, however this is a question that is really just about mathematical preference rather than math itself. I'm writing with the holiday spirit in mind, posing the following question just for pure fun and hoping it won’t cause too much divisive argument ! What is the most important strictly irrational constant ? #pi or #e or sqrt(2) or maybe something else ? (Not including infinity) As a programmer I find it very hard to choose :D (Apologies to those who prefer the term software engineer)

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[math-fun] formulas for producing primes
by Simon Plouffe 31 Dec '18

31 Dec '18

Hello, I have been working on a problem about primes for a while. How to get a formula that would produce primes in any possible way simpler or more direct than the known ones ? Mills formula : simple enough but it grows way too fast, the 4th term is 4932 digits long. Wright formula : simple too but still too fast. Matyasevich expression : very nice but totally useless for computation. Conwa's FRACTAN expression : even more clever but still useless for computation of primes. Is there another way to get out of this deadlock ? Here is one direction, take S(0) = 43.804... see the details here : http://plouffe.fr/NEW/Formula%20for%20primes.pdf If S(0) is 43.804... then S(n+1) = {S(n)^(5/4)} will produce an endless series of primes beginning, { x } is the rounded value of x. 113 367 1607 10177 102217 1827697 67201679 6084503671 1699344564793 1940223714629437 12877001925259260821 771380135526168946568519 722912215706743477640066820689 21079337353575904691781436731789131951 45166994522409258021988187061430676460306223027 20822194129240450122637347266336444580153717439156314146339 ... it is quite smaller than the known ones. Then the next question is : can we find a smaller exponent that will still work ? Answer : yes. If S(0) = 100000000000000000000000000000049.31221074776345 and the exponent being 11/10 then we get the following primes. 100000000000000000000000000000049 158489319246111348520210137339236753 524807460249772597364312157022725894401 3908408957924020300919472370957356345933709 70990461585528724931289825118059422005340095813 3438111840350699188044461057631015443312900908952333 489724690004200094265557071425023036671550364178496540501 ... even smaller ? : yes with this number S(0) = 10^500 + 961 and the exponent being 51/50 (that is 1.02) then the 50'th term of this formula is 2546223984434301126230560405750836844181156799213460578433735627810123608333414\ 6478717920178906592095759064531694091270765292863071164826802108312142369000399\ 0918112200598834059063704371274865057278618525926963160906859256204291253853779\ 0859263106429982680171352298624867958838188610026225438753363257569617236093303\ 5597738787705978929949787991104492121070145202492140062372248031624879195606603\ 9750665982863462494190199174051345073411763460889751017250124325112902369085214\ 0184828179289329015864573760367060446309168404382462687293461046639169729920475\ 3165541783062017852987962447647230150345731151542656737173297014806244172277491\ 6879804546188079063455344339451528745750861580732838099712175302071382677624428\ 4262968338987678773648741141485490879969521079400774328097643567743161861694439\ 2384400589489332204990330011362168526537013264388448543980911860963264776201897\ 0883314652642873229279315244591969596989887077667619942605707704348251758251233\ 1172015772215487384144161658368225114390354140649309995571635019281763656568221\ 8604642413102656623287079740572453580298728150605437498702948770557573781722652\ 8086931990592223405655647742963234918217685490055779977425394611159718143306147\ 7320512686294534342712886388641680379470072208880905199640988562813955595060316\ 87370269455744702618206542285281401557961196762713023289 Since we get here a series of 50 primes in a row : it does brake the 2 known records for a series of primes : namely : Polynomial expression : 46 values (degree 6) Primes in arithmetic progression : 26 values (with a(1) = 17 digits long). The next good question is : can we make simpler , smaller ? I do not kwow. Best regards, ps : Happy new year 2019 = not a prime.

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[math-fun] Count down to the new year
by Tom Duff 31 Dec '18

31 Dec '18

9 + ( 8 + 7 ) * ( 6! / 5 - 4 - 3 - 2 - 1 ) (from Jef Poskanzer on twitter: https://twitter.com/jef_poskanzer/status/1079766212190470144?s=11 )

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[math-fun] 1,409, 481, 24961, 28721, ...?
by Bill Gosper 30 Dec '18

30 Dec '18

*Very* probably spotty list: 1,409, 481, 24961, 28721, 65441, 113241, 121609, 345761, 362401, 384161, 530881, 620321, 854401, 882889, 909321, 1163249,... the squarefree part of integers of the form √(a⁴+b⁴+c⁴). Like the congruent numbers, you can't rule them out with finite searching. 1 doesn't show up until a=414560, b=217519, c=95800 (https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture) Are there really none with four digits? Two? —rwg

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[math-fun] Richard Dedekind, Professor of Polymer Chemistry
by Bill Gosper 30 Dec '18

30 Dec '18

Let p:=(1/2 - Sqrt[23/3]/6)^(1/3) + (1/2 + Sqrt[23/3]/6)^(1/3)] = "the plastic constant", solving a tiny cubic: In[113]:= RootReduce@p Out[113]= Root[-1 - #1 + #1^3 &, 1] http://mathworld.wolfram.com/DedekindEtaFunction.html gives p = i^(-1/12) η(½+i√23/2)/√2/η(1+i√23), suggesting that η(i√23) contains (some power of) p as a factor. I've just now simplified it to DedekindEta[I Sqrt[23]] == (23^(19/24) Sqrt[(3/23)!] (( 13 (2/23)! (4/23)! (6/23)! (8/23)!)/((1/46)! (9/46)! (13/46)!))^( 1/6) ((1/23)! (9/23)!)^(1/3))/(8 2^(1/23) 3^( 2/3) ((1/2 - Sqrt[23/3]/6)^(1/3) + (1/2 + Sqrt[23/3]/6)^(1/3))^( 4/3) \[Pi]^(3/4) ((3/46)!)^(1/3)) with numerous equivalent forms, given the identity (1476395008 2^(21/23) \[Pi]^(3/2) (1/46)! ((3/46)!)^2 (5/46)! ((7/46)!)^4 (9/46)! ((11/46)!)^2 ( 13/46)! ((15/46)!)^3 (17/46)! ((19/46)!)^2 (21/46)!)/( 4853477135 Sqrt[23] ((1/23)!)^2 (2/23)! ((3/23)!)^3 (4/23)! ((5/23)!)^2 (6/ 23)! ((7/23)!)^5 (8/23)! ((9/23)!)^2 (10/23)! ((11/23)!)^3) == 1 Eric, hang on. Those η(½+i√23/2) and η(1+i√23) I sent were undersimplified! —rwg

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Re: [math-fun] New Mersenne; GIMPS Coin? Why not GIMPS for POW?
by Keith F. Lynch 29 Dec '18

29 Dec '18

"Adam P. Goucher" <apgoucher(a)gmx.com> wrote: > The main reason is that proof-of-work solutions should be hard to > find and easy to verify. > Mersenne primes are hard to verify. But what about Mersenne *composites*? It's hard to find the prime divisors of Mersenne numbers (the Lucas-Lehmer primality test won't find them), but easy to confirm a claimed divisor.

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[math-fun] The truth behind the ¼! fraud
by Bill Gosper 28 Dec '18

28 Dec '18

http://gosper.org/¼!.png Just the n=0 term: Out[67]= (1/4)! - √(5GoldenRatio) π^(3/4)/(√√√2 (√√5-1)E^(5π/6)) In[68]:= N[%, 9] Out[68]= -4.67513339*10^-28 —rwg MMachinable series: (1/4)! == π^(3/4) √(5 GoldenRatio) Sum[(-1)^n/E^(30π(n+1/6)^2),{n,-∞,∞}]/(√√√2(√√5 - 1))

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[math-fun] "machine learning" v. "scientific method" ?
by Henry Baker 28 Dec '18

28 Dec '18

I know essentially zero about the current AI/ML hype. I was wondering if there is a trivial example of AI/ML with a very small number of "connections" (or whatever they call them) that is capable of learning some trivial task -- e.g., an AND gate working on 0/1 input data. [I suppose that I could try out some of the current open source ML software on a trivial example like the AND gate to see how difficult this problem is. Perhaps someone here has already done such a "hello world" example? Also, has anyone tried such ML software on the various *integer sequences* ? Not that the ML software would "understand" the sequences, but it might provide some idea of the relative "complexity" of the different sequences.] I'm curious about whether this process can be seen from some perspective as a kind of "scientific method", where some model (which may consist of perhaps 1-4 real or rational numbers) is a "hypothesis" which is "tested" against some training data which then causes some sort of "model refinement". If AI/ML cannot be put into (or forced into) this form, then perhaps it is time to come up with a replacement for the "scientific method" that CAN be put into such an AI/ML form. My point is that if ML really has come far enough to provide a more concrete rationale behind "scientific progress", then perhaps we need to revise what the phrase "scientific progress" should mean.

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[math-fun] Moderately high precision fraud
by Bill Gosper 28 Dec '18

28 Dec '18

Gamma[1/4] == 2^(7/8) Sqrt[ 5] (1 + 5^(1/4)) (E^(-845 \[Pi]/6) + E^(-605 \[Pi]/6) - E^(-245 \[Pi]/6) - E^(-125 \[Pi]/6) + E^(-5 \[Pi]/6)) GoldenRatio^(3/2) \[Pi]^(3/4) N[%, 239] Out[371]= True But when I try to switch to factorial notation, Mathematica catches on! In[377]:= %370/4 /. Gamma@z_ :> z!/z Out[377]= False —rwg

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Re: [math-fun] Amusing Permutation Representations of Group Extensions
by Dan Asimov 28 Dec '18

28 Dec '18

Last line had some numbers wrong but is now fixed below. —Dan ----- Nice pictures in that paper! (I know nothing about group representations, and don't even know the definition of a wreath product.) Funny, just last night I was musing about how for any finite group G the fewest number N = N(G) of points on which it can act faithfully is an interesting invariant. Which is just finding G as a subgroup of the permutation group S_N. For instance, what is this number for the rotation groups of the regular polyhedra: T (order 12), O (order 24), I (order 60) ? Each group acts faithfully on the face centers, edge midpoints, and also the face centers of the dual polyhedron (for O and I). For the octahedron and icosahedron, and for the edges of the tetrahedron, antipodal points get rotated to antipodal points, so we can divide all these numbers by 2 except for the 4 face centers of the tetrahedron. Hence T permutes 4 or 3 = 6/2 points, O permutes 4 = 8/2 or 6 = 12/2 or 3 = 6/2 points, and I permutes 10 = 20/2 or 15 = 30/2 or 6 = 12/2 points. But wait! We can't represent T or O faithfully on 3 points! S_3 is only order 6, while these groups are bigger. Which of these 8 cases are faithful (represent all group elements as distinct permutations)? Can we do better than 4 for T (clearly not), 4 for O, and 6 for I ? -----

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