July 2018 - mathfun@mailman.xmission.com

Re: [math-fun] Looks fishy; Smells OK.
by Bill Gosper 23 Jul '18

23 Jul '18

I probably shouldn't've rationalized the denominators. Out[971]= {(3 (1 + 2 Sqrt[2] - Sqrt[10]) (1 + 3 Sqrt[2] - Sqrt[ 10]) (1 + Sqrt[2] + 3 Sqrt[5] - Sqrt[10])^3)/((2 - Sqrt[2] + 2 Sqrt[5] - Sqrt[10]) (-1 + 3 Sqrt[5] - Sqrt[10]) (1 + 3 Sqrt[2] + 3 Sqrt[5] - Sqrt[10]) (1 - Sqrt[2] + Sqrt[10])^3), ((1 + 2 Sqrt[2]) (-1 + 3 Sqrt[2] - Sqrt[10]) (-1 + 3 Sqrt[5] - Sqrt[10])^3 (1 - Sqrt[2] + Sqrt[10]) (1 - Sqrt[5] + Sqrt[10]))/((1 + 2 Sqrt[2] - Sqrt[10])^3 (-1 + Sqrt[2] + 2 Sqrt[5] - Sqrt[10])^2 (3 - Sqrt[2] + 3 Sqrt[5] - Sqrt[ 10]) (1 + Sqrt[2] + 3 Sqrt[5] - Sqrt[10])), ((1 + 3 Sqrt[2] - Sqrt[10]) (-1 + 2 Sqrt[5] - Sqrt[10]) (-1 + 3 Sqrt[5] - Sqrt[ 10])^3 (3 + Sqrt[2] - 3 Sqrt[5] + Sqrt[10]))/((1 + 2 Sqrt[2] - Sqrt[10])^3 (-1 + 3 Sqrt[2] - Sqrt[10]) (1 + 2 Sqrt[5] - Sqrt[ 10]) (3 + Sqrt[2] + 3 Sqrt[5] - Sqrt[10]) (2 + Sqrt[2] - 2 Sqrt[5] + Sqrt[10]) (1 + Sqrt[2] + 2 Sqrt[5] + Sqrt[10])), ((1 + 2 Sqrt[2] + 2 Sqrt[5]) (1 + 2 Sqrt[2] - Sqrt[ 10])^2 (-1 + 3 Sqrt[2] - Sqrt[10])^2 (1 + 2 Sqrt[5] - Sqrt[ 10]) (3 + Sqrt[2] + 3 Sqrt[5] - Sqrt[10]) (2 + Sqrt[2] - 2 Sqrt[5] + Sqrt[10]))/((1 + 3 Sqrt[2] - Sqrt[10]) (-1 + 2 Sqrt[5] - Sqrt[10]) (-1 + 3 Sqrt[5] - Sqrt[10])^2 (-2 - Sqrt[ 2] + 3 Sqrt[5] - Sqrt[10]) (3 + Sqrt[2] - 3 Sqrt[5] + Sqrt[ 10]) (1 - Sqrt[5] + Sqrt[10]))} In[972]:= RootReduce@% Out[972]= {1, 1, 1, 1} I wonder if this calculation was humanly possible. --rwg On 2018-07-22 07:04, Fred Lunnon wrote: > But you are not comparing like with like --- the LLL products do have > more than single digit coefficients; 43267971744 is (quite) large; > and output from LLL can hardly be considered uniformly random anyway! > > > On 7/22/18, Bill Gosper <billgosper(a)gmail.com> wrote: >> FindIntegerNullVector constructed 47 products of linear combinations of >> √2, √5, and √10, all = 1. The very first thing it found was >> >> (3 + 3 √2 - 2 √5 - √10) (3 - 3 √2 - 2 √5 + √10). >> >> Is this even obviously rational? It's 1 iff >> (3 - 3 √2 + 2 √5 - √10) (3 + 3 √2 + 2 √5 + √10) is. >> >> Things got rapidly worse: >> >> (√2 - 1) (3 + √2 - √10) (43 + 26 √2 + 10 √5 + 11 √10)/71, >> >> until bewilderment: replaced doubt: >> >> (51 + 38 √2 - 15 √5 - 16 √10) (1 + 3 √2 - √10) (-1 + 2 √5 - √10) >> (1 - √2 + 2 √5 - √10) (-1 + 3 √5 - √10)^3 (4 + 3 √2 + √10) >> (3 + √2 - 3 √5 + √10) (3 + 7√2/3 + 4√5/3 + √10) >> (17 + 6 √2 + 8 √5 + √10)^3 (-29 - 20 √2 + 18 √5 + 11 √10)/43267971744 >> >> In:= RootReduce@% >> Out= 1 >> >> Such things ought to be trivial to construct: >> Just multiply a random pile of surds by their reciprocal, with denominator >> rationalized. But I'd expect these reciprocals to have coefficients a lot >> bigger than 51: >> >> In[707]:= RandomInteger[{-9, 9}, {4, 4, 4}].Sqrt[{1, 2, 5, 10}] >> >> Out[707]= {{6 + 5 Sqrt[2] + 8 Sqrt[10], >> 6 + 7 Sqrt[2] + 8 Sqrt[5] - 4 Sqrt[10], -8 + 5 Sqrt[2] - >> 9 Sqrt[5] + 2 Sqrt[10], >> 5 + 3 Sqrt[2] - 8 Sqrt[5] + 5 Sqrt[10]}, {2 + 4 Sqrt[2] + >> 6 Sqrt[5] + 9 Sqrt[10], 1 - 5 Sqrt[2] + 8 Sqrt[5] - Sqrt[10], >> 1 + Sqrt[2] - 6 Sqrt[5] + 9 Sqrt[10], -5 + 8 Sqrt[2] + 5 Sqrt[5] + >> 3 Sqrt[10]}, {-2 + 3 Sqrt[2] - 2 Sqrt[5] + 7 Sqrt[10], >> 5 - 6 Sqrt[2] - 6 Sqrt[5] - 9 Sqrt[10], -3 + 4 Sqrt[2] - >> 4 Sqrt[5] + 5 Sqrt[10], -1 + 7 Sqrt[2] - Sqrt[5] - 6 Sqrt[10]}, {1, >> 5 + 4 Sqrt[2] + 9 Sqrt[5] + 9 Sqrt[10], >> 4 - 5 Sqrt[2] + 9 Sqrt[5] - 9 Sqrt[10], >> 3 + 4 Sqrt[2] + 6 Sqrt[5] + 9 Sqrt[10]}} >> >> In[708]:= Times @@ # & /@ % >> >> Out[708]= {(6 + 7 Sqrt[2] + 8 Sqrt[5] - 4 Sqrt[10]) (-8 + 5 Sqrt[2] - >> 9 Sqrt[5] + 2 Sqrt[10]) (5 + 3 Sqrt[2] - 8 Sqrt[5] + >> 5 Sqrt[10]) (6 + 5 Sqrt[2] + 8 Sqrt[10]), (1 - 5 Sqrt[2] + >> 8 Sqrt[5] - Sqrt[10]) (-5 + 8 Sqrt[2] + 5 Sqrt[5] + >> 3 Sqrt[10]) (1 + Sqrt[2] - 6 Sqrt[5] + 9 Sqrt[10]) (2 + >> 4 Sqrt[2] + 6 Sqrt[5] + 9 Sqrt[10]), (5 - 6 Sqrt[2] - 6 Sqrt[5] - >> 9 Sqrt[10]) (-1 + 7 Sqrt[2] - Sqrt[5] - 6 Sqrt[10]) (-3 + >> 4 Sqrt[2] - 4 Sqrt[5] + 5 Sqrt[10]) (-2 + 3 Sqrt[2] - 2 Sqrt[5] + >> 7 Sqrt[10]), (4 - 5 Sqrt[2] + 9 Sqrt[5] - 9 Sqrt[10]) (3 + >> 4 Sqrt[2] + 6 Sqrt[5] + 9 Sqrt[10]) (5 + 4 Sqrt[2] + 9 Sqrt[5] + >> 9 Sqrt[10])} >> >> made four quadruples with single digit coefficients. Reciprocating, >> In[709]:= RationalizeDenominator /@ (1/%) >> >> Out[709]= {(-8740961902019984 - 5440587070445184 Sqrt[2] + >> 2455653039709912 Sqrt[5] + 1573532047536568 Sqrt[10])/ >> 510888132844109317136, (4549175633470544 + >> 1131438893999820 Sqrt[2] - 1605347641058392 Sqrt[5] + >> 706649915148436 Sqrt[10])/ >> 968178139395832169104, (508007526470048 - 194043325345184 Sqrt[2] - >> 158757589376544 Sqrt[5] + 99676189957856 Sqrt[10])/ >> 20517561741702420224, (-102601486842 + 402200749082 Sqrt[2] + >> 176375285826 Sqrt[5] - 343223324388 Sqrt[10])/7676200432663996}

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[math-fun] Looks fishy; Smells OK.
by Bill Gosper 22 Jul '18

22 Jul '18

FindIntegerNullVector constructed 47 products of linear combinations of √2, √5, and √10, all = 1. The very first thing it found was (3 + 3 √2 - 2 √5 - √10) (3 - 3 √2 - 2 √5 + √10). Is this even obviously rational? It's 1 iff (3 - 3 √2 + 2 √5 - √10) (3 + 3 √2 + 2 √5 + √10) is. Things got rapidly worse: (√2 - 1) (3 + √2 - √10) (43 + 26 √2 + 10 √5 + 11 √10)/71, until bewilderment: replaced doubt: (51 + 38 √2 - 15 √5 - 16 √10) (1 + 3 √2 - √10) (-1 + 2 √5 - √10) (1 - √2 + 2 √5 - √10) (-1 + 3 √5 - √10)^3 (4 + 3 √2 + √10) (3 + √2 - 3 √5 + √10) (3 + 7√2/3 + 4√5/3 + √10) (17 + 6 √2 + 8 √5 + √10)^3 (-29 - 20 √2 + 18 √5 + 11 √10)/43267971744 In:= RootReduce@% Out= 1 Such things ought to be trivial to construct: Just multiply a random pile of surds by their reciprocal, with denominator rationalized. But I'd expect these reciprocals to have coefficients a lot bigger than 51: In[707]:= RandomInteger[{-9, 9}, {4, 4, 4}].Sqrt[{1, 2, 5, 10}] Out[707]= {{6 + 5 Sqrt[2] + 8 Sqrt[10], 6 + 7 Sqrt[2] + 8 Sqrt[5] - 4 Sqrt[10], -8 + 5 Sqrt[2] - 9 Sqrt[5] + 2 Sqrt[10], 5 + 3 Sqrt[2] - 8 Sqrt[5] + 5 Sqrt[10]}, {2 + 4 Sqrt[2] + 6 Sqrt[5] + 9 Sqrt[10], 1 - 5 Sqrt[2] + 8 Sqrt[5] - Sqrt[10], 1 + Sqrt[2] - 6 Sqrt[5] + 9 Sqrt[10], -5 + 8 Sqrt[2] + 5 Sqrt[5] + 3 Sqrt[10]}, {-2 + 3 Sqrt[2] - 2 Sqrt[5] + 7 Sqrt[10], 5 - 6 Sqrt[2] - 6 Sqrt[5] - 9 Sqrt[10], -3 + 4 Sqrt[2] - 4 Sqrt[5] + 5 Sqrt[10], -1 + 7 Sqrt[2] - Sqrt[5] - 6 Sqrt[10]}, {1, 5 + 4 Sqrt[2] + 9 Sqrt[5] + 9 Sqrt[10], 4 - 5 Sqrt[2] + 9 Sqrt[5] - 9 Sqrt[10], 3 + 4 Sqrt[2] + 6 Sqrt[5] + 9 Sqrt[10]}} In[708]:= Times @@ # & /@ % Out[708]= {(6 + 7 Sqrt[2] + 8 Sqrt[5] - 4 Sqrt[10]) (-8 + 5 Sqrt[2] - 9 Sqrt[5] + 2 Sqrt[10]) (5 + 3 Sqrt[2] - 8 Sqrt[5] + 5 Sqrt[10]) (6 + 5 Sqrt[2] + 8 Sqrt[10]), (1 - 5 Sqrt[2] + 8 Sqrt[5] - Sqrt[10]) (-5 + 8 Sqrt[2] + 5 Sqrt[5] + 3 Sqrt[10]) (1 + Sqrt[2] - 6 Sqrt[5] + 9 Sqrt[10]) (2 + 4 Sqrt[2] + 6 Sqrt[5] + 9 Sqrt[10]), (5 - 6 Sqrt[2] - 6 Sqrt[5] - 9 Sqrt[10]) (-1 + 7 Sqrt[2] - Sqrt[5] - 6 Sqrt[10]) (-3 + 4 Sqrt[2] - 4 Sqrt[5] + 5 Sqrt[10]) (-2 + 3 Sqrt[2] - 2 Sqrt[5] + 7 Sqrt[10]), (4 - 5 Sqrt[2] + 9 Sqrt[5] - 9 Sqrt[10]) (3 + 4 Sqrt[2] + 6 Sqrt[5] + 9 Sqrt[10]) (5 + 4 Sqrt[2] + 9 Sqrt[5] + 9 Sqrt[10])} made four quadruples with single digit coefficients. Reciprocating, In[709]:= RationalizeDenominator /@ (1/%) Out[709]= {(-8740961902019984 - 5440587070445184 Sqrt[2] + 2455653039709912 Sqrt[5] + 1573532047536568 Sqrt[10])/ 510888132844109317136, (4549175633470544 + 1131438893999820 Sqrt[2] - 1605347641058392 Sqrt[5] + 706649915148436 Sqrt[10])/ 968178139395832169104, (508007526470048 - 194043325345184 Sqrt[2] - 158757589376544 Sqrt[5] + 99676189957856 Sqrt[10])/ 20517561741702420224, (-102601486842 + 402200749082 Sqrt[2] + 176375285826 Sqrt[5] - 343223324388 Sqrt[10])/7676200432663996} --rwg

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[math-fun] Everyone's a winner! (was Re: What is the simplex best at being best at?)
by Keith F. Lynch 21 Jul '18

21 Jul '18

James Propp <jamespropp(a)gmail.com> wrote: > In no sense did I intend a real math question. But I suspect that > the meta-game of asking ?Are there ten things I can do that nobody > else can do all of? How about nine? How about eight?? (or a > variant) has a name in pop (nerd) culture, and I was hoping to > elicit it from one of you. That would be a very difficult question to answer. Not the question of whether your first question has a name (though if it does, I don't know it), but your first question itself. (Speaking of questions, the digest option on this list turns so-called "smart" quotation marks and all other non-ASCII characters into question marks. So I wish people would stick to ASCII. Thanks.) A few weeks ago, I was in a discussion with friends, and they started speculating on which historical person had the highest-ever IQ. For instance did Archimedes have a higher IQ than Newton? I think that just might depend on what language the test was in. More generally, I think that's a profoundly misguided way of looking at the world, not much more meaningful than asking who would do best at playing your favorite video game. I made the perhaps over-bold claim that since there are more possible tasks than there are people, everyone can be the best in the world at *something* if they work at it diligently. I'll moderate it to say that almost anyone without any major disability can be in the top ten in something if they're willing to put in the effort. I also mentioned United Flight 232, which lost all its hydraulics, making it unflyable. The pilot managed to land it anyway. It wasn't a particularly good landing, but the majority of passengers did survive. In all other cases of a passenger jet losing all its hydraulics, everyone died. And there have been countless computer simulations of that situation, all of which ended in a crash. The Flight 232 passengers should be thankful that their pilot was someone very good at piloting rather than someone very good at taking IQ tests. How many things are there that it would be possible to be the best at the world in? You could start with the Olympic records and with the Guinness Book. Or with one of those lists of astonishing human achievements, such as Lynne Cox, who swam from Alaska to Russia through miles of freezing-cold water, wearing just an ordinary bathing suit. Or just look at some relatively narrow field such as ham radio, in which there are records for lowest-power trans-Atlantic contact (less than one microwatt), most countries communicated with in one day/week/month/year, shortest time to communicate with all countries, highest data rate, lowest data rate (I think the current record is one bit per day), lowest frequency, highest frequency, most contacts in one day, lowest power for bouncing a signal off the moon, etc. Then multiply the total by the number of hobbies, professions, sports, games, etc., that exist. Also, of course, for everything that's ever been done, someone holds the record for having done it first. Like most people here, there are a few math problems I was the first or only person to solve, and a few others for which I think I have the best algorithm for searching for solutions. This is of course doesn't mean I'm the person best able to solve such problems, but that they didn't happen to catch the attention of anyone better suited to solve them who had the free time available. I may hold the record for the geekiest ever Wikipedia update. I was casually browsing some of its math-related articles when I happened to notice that the twin primes constant was wrong in the 7th decimal place, so I fixed it. There are records that most of us would like to set, such as being the world's oldest person ever, beating Jeanne Calment, who lived to be 122. There are records that none of us would like to set, such as being the American wrongfully imprisoned for the longest. Richard Dan Phillips of Michigan holds the record for being exonerated and freed after 45 years (1972-2017), but the wrongfully-imprisoned record is almost certainly held by someone who was never exonerated.

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Re: [math-fun] 4-5-6
by Bill Gosper 21 Jul '18

21 Jul '18

Man, you guys work fast. On 2018-07-19 09:17, Neil Sloane wrote: > Steve, Thank you, that was very helpful, and I updated A316833. And ...834 and ...835. But Oy, that Maple code! In[886]:= ev@q_:=Evaluate@Echo@Sum[q^(2n)^2,{n,0,\[Infinity]}] 1/2 (1+EllipticTheta[3,0,q^4]) In[888]:= od@q_:=Evaluate@Echo@PowerExpand@Sum[q^(2n-1)^2,{n,\[Infinity]}] 1/2 EllipticTheta[2,0,q^4] In[893]:= #@od\[Subset]#@ev&[(#[q]^4-6#[q^2]#[q]^2+8#[q^3]#@q+3#[q^2]^2-6#[q^4])/24&] See 1998 mail below Out[893]= 1/24 (1/16 EllipticTheta[2,0,q^4]^4-3/4 EllipticTheta[2,0,q^4]^2 EllipticTheta[2,0,q^8]+3/4 EllipticTheta[2,0,q^8]^2+2 EllipticTheta[2,0,q^4] EllipticTheta[2,0,q^12]-3 EllipticTheta[2,0,q^16]) ⊂ 1/24 (1/16 (1+EllipticTheta[3,0,q^4])^4-3/4 (1+EllipticTheta[3,0,q^4])^2 (1+EllipticTheta[3,0,q^8])+3/4 (1+EllipticTheta[3,0,q^8])^2+2 (1+EllipticTheta[3,0,q^4]) (1+EllipticTheta[3,0,q^12])-3 (1+EllipticTheta[3,0,q^16])) In[897]:= PowerExpand[Series[#,{q,0,420}]&/@%893] Out[897]= q^84+q^116+q^140+2 q^156+q^164+q^180+q^196+2 q^204+q^212+2 q^228+q^236+q^244+2 q^252+3 q^260+2 q^276+2 q^284+3 q^300+2 q^308+2 q^316+3 q^324+q^332+2 q^340+2 q^348+2 q^356 +3 q^364+3 q^372+4 q^380+4 q^396+3 q^404+6 q^420+O[q]^421 ⊂ q^56+q^84+q^104+q^116+2 q^120+q^140+q^152+q^156+q^164+q^168+q^180+2 q^184+q^196+2 q^200+q^204+q^212+2 q^216+q^224+q^228+q^236+q^244+3 q^248+q^252+2 q^260+2 q^264+2 q^276 +2 q^280+q^284+3 q^296+2 q^300+2 q^308+4 q^312+q^316+2 q^324+q^332+2 q^336+q^340+3 q^344+q^348+2 q^356+4 q^360+2 q^364+2 q^372+4 q^376+2 q^380+3 q^392+2 q^396+3 q^404+3 q^408+q^416 +4 q^420+O[q]^421 E.g., 420 is the sum of four distinct odd squares six different ways and the sum of four distinct even squares four different ways --rwg > > Michael Hirschhorn's book looks very interesting and judging by the > excerpts that your link showed > (look at page 295!) could lead to many other new sequences. If anyone would > like to help entering them into the OEIS, please do! > > > 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 > > > On Thu, Jul 19, 2018 at 11:26 AM, Lucas, Stephen K - lucassk < > lucassk(a)jmu.edu> wrote: > >> I don’t know about formally published, but the proof is discussed on the >> Mathworld site on square numbers, about halfway down, and references >> Michael Hirschhorn. It gives the identity >> (4a+1)^2+(4b+1)^2+(4c+1)^2+(4d+1)^2 = 4[ (a+b+c+d+1)^2 + (a-b-c+d)^2 + >> (a-b+c-d)^2 + (a+b-c-d)^2 ]. >> >> You can find this mentioned in his own book at >> https://books.google.com/books?id=60QwDwAAQBAJ&pg= >> PA295&lpg=PA295&dq=hirschhorn+sum+four+distinct+odd+squares& >> source=bl&ots=P0bLaMkf63&sig=B1tCdmh6jDWeYBfza_arOJwvtgE&hl=en&sa=X&ved= >> 2ahUKEwiUrtmOuKvcAhWhct8KHTzfCe8Q6AEwB3oECAEQAQ#v=onepage&q= >> hirschhorn%20sum%20four%20distinct%20odd%20squares&f=false >> >> Steve >> >> On Jul 19, 2018, at 10:57 AM, Neil Sloane <njasloane(a)gmail.com<mailto:nj >> asloane(a)gmail.com>> wrote: >> >> Bill, "Sums of four distinct odd squares" seemed to be missing from OEIS >> so I created A316833 - could you check? Was Mike H.'s proof ever >> published? >> >> 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: https://urldefense.proofpoint. >> com/v2/url?u=http-3A__NeilSloane.com&d=DwIGaQ&c= >> eLbWYnpnzycBCgmb7vCI4uqNEB9RSjOdn_5nBEmmeq0&r=vge6KOo90zMf7Wx14WFtiQ&m= >> KlvEAtFe4zKCr3Aql2E0mi4T73AzYJ8kLGVSN9IC010&s=zft- >> IZ1e4jFVTGQ3CXQOdJ4AHPEdQ1E2Vr39GfbCVSc&e= >> Email: njasloane(a)gmail.com<mailto:njasloane@gmail.com> >> >From rcs Wed Oct 14 11:50:42 1998 Return-Path: <owner-math-fun(a)baskerville.CS.Arizona.EDU> Received: from optima.cs.arizona.edu (optima.CS.Arizona.EDU [192.12.69.5]) by baskerville.CS.Arizona.EDU (8.9.1a/8.9.1) with ESMTP id LAA04966 for <math-fun(a)baskerville.cs.arizona.edu>; Wed, 14 Oct 1998 11:50:42 -0700 (MST) Received: from NEWTON.Macsyma.COM ([192.156.175.1]) by optima.cs.arizona.edu (8.9.1a/8.9.1) with SMTP id LAA24261 for <math-fun(a)OPTIMA.CS.ARIZONA.EDU>; Wed, 14 Oct 1998 11:50:22 -0700 (MST) Received: from SWEATHOUSE.Macsyma.COM by NEWTON.Macsyma.COM via INTERNET with SMTP id 419503; 14 Oct 1998 14:49:02-0400 Date: Wed, 14 Oct 1998 11:50-0700 From: Bill Gosper <rwg(a)[192.156.175.1]> Reply-To: rwg(a)NEWTON.Macsyma.COM Subject: integer sequences and partitions, [arcsines -> Mt. Ararat] To: math-fun(a)optima.CS.Arizona.EDU cc: njas(a)research.att.com, wilf(a)math.upenn.edu In-Reply-To: <19980828061701.1.RWG(a)[192.233.166.105]> Message-ID: <19981014185051.6.RWG(a)[192.233.166.105]> Yesterfortnight, I muttered Suppose h(t) has a powerseries with coefficients in {0,1}, and is thus the generating function of the characteristic function of an integer sequence. E.g., to characterize the positive squares, h(t) := (theta_3(t)-1)/2 = sum(t^n^2,n,1,inf). Then the coefficient of s^i t^j in inf ==== i i i \ (- 1) s h(t ) - > --------------- / i ==== i = 1 (d1100) %e =: f(s) is the number of ways to partition j into i distinct elements of the integer sequence. To relieve the distinctness constraint, take all signs positive. Expanding both versions at s=0 gives the generating functions for partitions into 0 members, 1 member, 2 members, ... of whatever set h characterizes (e.g., positive squares.) (c3232) taylor(1/subst(-s,s,d3226),s,0,4); /* allowing repetitions */ Time= 68 msec. 2 2 2 (h(t ) + h (t)) s (d3232)/T/ 1 + h(t) s + ------------------ 2 3 2 3 3 (2 h(t ) + 3 h(t) h(t ) + h (t)) s + ----------------------------------- 6 4 3 2 2 2 2 4 4 (6 h(t ) + 8 h(t) h(t ) + 3 h (t ) + 6 h (t) h(t ) + h (t)) s + -------------------------------------------------------------- 24 + . . . (c3226) taylor(d1100,s,0,4); /* distinct */ 2 2 2 (h (t) - h(t )) s (d3226)/T/ 1 + h(t) s + ------------------ 2 3 2 3 3 (h (t) - 3 h(t ) h(t) + 2 h(t )) s + ----------------------------------- 6 4 2 2 3 2 2 4 4 (h (t) - 6 h(t ) h (t) + 8 h(t ) h(t) + 3 h (t ) - 6 h(t )) s + -------------------------------------------------------------- 24 + . . .

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Re: [math-fun] What is the simplex best at being best at?
by Dan Asimov 20 Jul '18

20 Jul '18

I've had a few days to think about this. At this point, I'm pretty sure *what* I think is the number one distinction between the extremes of the regular simplex and the sphere, though I haven't yet convinced myself I know exactly *why* I believe this. Meanwhile, if anyone else would like to express their opinion, I'd be curious to learn what others think about this. —Dan I wrote: ----- Let C_n be the space of convex bodies in R^n. (I.e., closed and bounded convex subsets of R^n that contain interior points.) (Topologized with the Hausdorff metric, C_n is compact. Hence for any continuous function F : C_n —> R there exists a global maximum and a global minimum on C_n. For many geometrically defined such F : C_n —> R, all spheres represent precisely the set of global maxima and all regular n-simplices represent precisely the set of global minima. (Or vice versa — same difference.) But there is an embarrassment of options. Which F : C_n —> R best characterizes the gradient between the sphere and the simplex? ... ... -----

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Re: [math-fun] Optimal embeddings of tangles
by Dan Asimov 19 Jul '18

19 Jul '18

I'm not sure I understand the part about mandating the optima in advance. If the tangles are well-tangled in space, I'm not sure how an optimum could be able to be a circle arc. (Further, aren't tangles supposed to happen *inside* something, like a cube?) —Dan Jim Propp wrote: ----- Consider the points A=(1,0,0), B=(-1,0,0), C=(0,1,0), and D=(0,-1,0). There are lots of ways to draw two disjoint 3D arcs joining A with B and C with D. What if we want them to be short without getting too close to each other? We might set up some “short-but-not-too-close” objective functional and optimize it using calculus of variations. There are probably many ways to do this, some solvable and some not. Does anyone know of work along these lines? In the spirit of calculus-of-variations-of-variations, we might mandate the optima in advance (each arc is a half-circle on the sphere x^2+y^2+z^2=1, one in the upper hemisphere and one in the lower hemisphere) and then ask, For what objective functional is this pair of arcs locally optimal? ... -----

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Re: [math-fun] What is the simplex best at being best at?
by Dan Asimov 19 Jul '18

19 Jul '18

Another thing I noticed that surprised me re the sphere and the simplex, is that if you put a one inside the other and then try to fill the space between them with surfaces, it seems a *lot* more natural to begin with a sphere inside a simplex than to begin with a simplex inside a sphere. But maybe this is just an illusion? —Dan

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Re: [math-fun] What is the simplex best at being best at?
by Dan Asimov 19 Jul '18

19 Jul '18

Okay. What I've decided is the most relevant distinction between the simplex and the sphere in R^ is that the simplex is the convex hull of only n+1 points, whereas the sphere is the convex hull of a set of points that must be dense in the sphere. I'm sure I could have expressed the question better. And I still hope to someday. —Dan ----- I've had a few days to think about this. At this point, I'm pretty sure *what* I think is the number one distinction between the extremes of the regular simplex and the sphere, though I haven't yet convinced myself I know exactly *why* I believe this. Meanwhile, if anyone else would like to express their opinion, I'd be curious to learn what others think about this. —Dan I wrote: ----- Let C_n be the space of convex bodies in R^n. (I.e., closed and bounded convex subsets of R^n that contain interior points.) (Topologized with the Hausdorff metric, C_n is compact. Hence for any continuous function F : C_n —> R there exists a global maximum and a global minimum on C_n. For many geometrically defined such F : C_n —> R, all spheres represent precisely the set of global maxima and all regular n-simplices represent precisely the set of global minima. (Or vice versa — same difference.) But there is an embarrassment of options. Which F : C_n —> R best characterizes the gradient between the sphere and the simplex? ... ... ----- -----

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Re: [math-fun] 4-5-6
by Bill Gosper 19 Jul '18

19 Jul '18

Man, that Hirschhorn is dangerous! billgosper$ python rattri.py 5 3 | head 2098817 2016840 220597 In[524]:= ratang[823543, 1024303, 220597] Out[524]= {(\[Pi] - ArcCos[1461083/1647086])/(\[Pi] - ArcCos[-(11472481/11529602)]), (\[Pi] - ArcCos[-(11472481/11529602)])/(\[Pi] - ArcCos[-(13/14)]), (\[Pi] - ArcCos[-(13/14)])/(\[Pi] - ArcCos[1461083/1647086])} In[525]:= N@% Out[525]= {26.7290445719883, 0.261887400469822, 0.142857142857143} In[526]:= FI /@ {823543, 1024303, 220597} Out[526]= {7^7,7 41 43 83,13 71 239} Anecdote: The only time I spoke with Erdös was at the Ramanujan Centenary at UIUC. I told him my conjecture that any sum of four distinct odd squares was the sum of four distinct even squares. After about 30 sec, he said "This is not my problem." Later I repeated the conjecture to Hirschhorn, who handed me a handwritten algebraic proof the next day. --rwg On 2018-07-16 10:03, Tomas Rokicki wrote: > Here's a python script that generates integer triangles with angles in any > rational > ratio (pass the numerator and denominator on the command line). > > I'm sure someone skilled in Python can write this much more simply. > > Given this, the only thing we need to show is that these are the only such > integer triangles. > > Sample usage: > > abacus:inttri rokicki$ python rattri.py 1 6 | head > > 1771561 2271060 566839 > > 46656 72930 30421 > > 4826809 8761896 4632263 > > 117649 240240 146329 > > 11390625 25591020 17151679 > > 262144 637560 458081 > > > # > # Given two integers h and k, find integer triangles with angles in the > # ratio h:k. Based on the generic parameterization given here: > # > # > https://en.wikipedia.org/wiki/Integer_triangle#Integer_triangles_with_one_a… > # > import sys > from fractions import gcd > import math > # > h = int(sys.argv[1]) ; > k = int(sys.argv[2]) ; > q = 1 > coscrit = math.cos(math.pi/(h+k)) > def choose(a, b): > if b < 0 or b > a: > return 0 > if b + b > a: > b = a - b > r = 1 > for i in range(1, b+1): > r = r * (a - i + 1) / i > return r ; > while True: > for p in range(1, q): > if q*coscrit >= p: > continue > if gcd(p, q) > 1: > continue > a = 0 > b = 0 > c = 0 > for i in range(0, (h+1)/2): > a += ((-1)**i)*choose(h, 2*i+1)*(p**(h-2*i-1))*((q*q-p*p)**i) > for i in range(0, (k+1)/2): > b += ((-1)**i)*choose(k, 2*i+1)*(p**(k-2*i-1))*((q*q-p*p)**i) > for i in range(0, (h+k+1)/2): > c += ((-1)**i)*choose(h+k, 2*i+1)*(p**(h+k-2*i-1))*((q*q-p*p)**i) > a *= q**k > b *= q**h > g = gcd(gcd(a, b), c) > a /= g > b /= g > c /= g > print a, b, c > q = q + 1 > > > > On Sun, Jul 15, 2018 at 2:55 PM Bill Gosper <billgosper(a)gmail.com> wrote: > >> Wow. I haven't even digested WDS's thing. --Bill >> Maybe it's time for MathOverflow, e.g.? >> >> On Sun, Jul 15, 2018 at 1:36 PM Tomas Rokicki <rokicki(a)gmail.com> wrote: >> >>> Any more progress on this? >>> >>> I have a large number of additional examples if you are interested, >>> such as this triple for 6: >>> >>> 117649 146329 240240 >>> >>> and these for 5: >>> >>> 231 1024 1220 >>> >>> 3024 3125 5555 >>> >>> 7776 12155 17214 >>> >>> 16807 34320 42833 >>> >>> 36400 59049 89001 >>> >>> 32768 79695 92168 >>> >>> 59049 162656 178695 >>> >>> 100000 302499 320210 >>> >>> 161051 206460 323851 >>> >>> I have not yet tried to find recurrences for these . . . >>> >>> It's abundantly clear empirically that your hypothesis works both >>> ways (with suitable caveats). There should be a way to prove >>> this. >>> >>> -tom >>> >>> On Sat, Jul 14, 2018 at 4:14 PM Bill Gosper <billgosper(a)gmail.com> wrote: >>> >>>> OMG! Nick Baxter somehow found >>>> 1024 | 1220 | 231 -> {23.09066960680706, 0.2165376788608164, 1/5}, >>>> 46656 | 72930 | 30421 -> {9.357241092137794, 0.6412146423203051, 1/6} >>>> >>>> 23.1--a toothpick! Even more toothpicular is >>>> 357 | 343 | 20 -> {18.99091913441869, 0.01755224857596323, 3}, >>>> because it's 1/19 of a smaller angle. >>>> [Massive chop!]

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[math-fun] Thanks for the replies about sums of squares ...
by Colin Wright 19 Jul '18

19 Jul '18

As is so often the case, once someone points out the answer you feel stupid for not having seen it yourself. My thanks to Tomas Rokicki and Gareth McCaughan for making me feel stupid <grin> Cheers! Colin -- Now available: "Bridges of Dee - Colour and Composition" http://www.williamsonfineart.co.uk/pages/bridges.html

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