July 2013 - mathfun@mailman.xmission.com

[math-fun] polar/cartesian
by Cris Moore 14 Jul '13

14 Jul '13

This took me a moment to get, but it was worth it: http://xkcd.com/1230/ Cris

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Re: [math-fun] FindSequenceFunction
by Bill Gosper 13 Jul '13

13 Jul '13

<< Was anybody surprised by Fibonacci[6*n]*Fibonacci[n]/Fibonacci[3*n]/Fibonacci[2*n] - Fibonacci[3*n]/Fibonacci[n] ? NeilB found a short Mma proof. >> Fred> I suppose it is a little unexpected that the answer is not exactly zero ... We still have the nice little puzzle of how NeilB got Mma to say -2*(-1)^n. He then quickly found In[149]:= FunctionExpand[ Fibonacci[10*n]*Fibonacci[n]/Fibonacci[2*n]/Fibonacci[5*n] - Fibonacci[8*n]/Fibonacci[4*n] + (-1)^n* Fibonacci[4*n]/Fibonacci[2*n], n \[Element] Integers] Out[149]= 1 FullSimplify instead of FunctionExpand also works, but slower. Oddly, discharging both barrels at once, In[146]:= FullSimplify[ FunctionExpand[ Fibonacci[10*n]*Fibonacci[n]/Fibonacci[2*n]/Fibonacci[5*n] - Fibonacci[8*n]/Fibonacci[4*n] + (-1)^n* Fibonacci[4*n]/Fibonacci[2*n]], n \[Element] Integers] Out[146]= ((-(-1)^n)*4^(5*n) + (1 + Sqrt[5])^(10*n))/ (-(-4)^(5*n) + (1 + Sqrt[5])^(10*n)) A point from which Mma seems unable to reach 1. ? --rwg

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Re: [math-fun] FindSequenceFunction
by Wouter Meeussen 13 Jul '13

13 Jul '13

Oops, replace "Now, adding that 2 Cos[Pi n] to it and evaluating at the integers .." by "Now, adding that 2 Cos[Pi n] to it and evaluating at the half-integers .." I just skipped over the fact their sum is exactly zero *at* the integers ... Wouter. -----Original Message----- From: Wouter Meeussen Sent: Saturday, July 13, 2013 4:09 PM To: math-fun Subject: Re: [math-fun] FindSequenceFunction If you plot Fibonacci[6*n]*Fibonacci[n]/Fibonacci[3*n]/Fibonacci[2*n] - Fibonacci[3*n]/Fibonacci[n] then it jumps out that it swiftly approaches -2*Cos[Pi n]. Now, adding that 2 Cos[Pi n] to it and evaluating at the integers gives Table[ c[n] f[n - 1/2]/f[3 n - 3/2] - f[3 n - 3/2]/f[n - 1/2], {n, 5}] with c[n] a set of integers. Which seq is hiding there? {2, 17, 122, 842, 5777, 39602, 271442, 1860497, 12752042, 87403802, 599074577, 4106118242} use FindSequenceFunction: c[n_Integer]:= (-2 + ((7 - 3*Sqrt[5])/2)^n*(3 + Sqrt[5]) - (-3 + Sqrt[5])*((7 + 3*Sqrt[5])/2)^n)/2 does the job. The Sqrt[5]-ish stuff hints at a cool Fibo-format. How to get there from here? Wouter. -----Original Message----- From: Fred lunnon Sent: Saturday, July 13, 2013 9:42 AM To: math-fun Subject: Re: [math-fun] FindSequenceFunction << Was anybody surprised by Fibonacci[6*n]*Fibonacci[n]/Fibonacci[3*n]/Fibonacci[2*n] - Fibonacci[3*n]/Fibonacci[n] ? NeilB found a short Mma proof. >> I suppose it is a little unexpected that the answer is not exactly zero ... << Here's van der Poorten's paper which Rumely just refers to as "to be published": http://maths.mq.edu.au/~alf/www-centre/alfpapers/a077.pdf Victor >> Coming back to detecting quotients, it seems from the handy 2-page summary for unenlightened monoglots in the above that (for example) the quotient of integer LFSR sequences _is_ another LFSR sequence if it comprises only integers --- or more generally (I think) if the denominators have finitely many distinct prime factors. But that doesn't help detect a quotient sequence where that is not the case. Back to square one, wiser but sadder. Fred Lunnon On 7/11/13, Bill Gosper <billgosper(a)gmail.com> wrote: > Fred> The term-by-term sum and product of such LFSR sequences are > also LFSR sequences, but not in general the quotient. BIll Gosper > recently asked for an algorithm to detect the quotient of LFSR sequences. > I have been waiting for sombody like Neil Sloane or Simon Plouffe to > wade in, but in vain ... maybe there's a trade secret to protect? > > Or maybe they just can't think of anything intelligent to say about it. > Like me. > > WFL > > I actually have an application for this problem--determining if and how > two > continued fraction formulÃ¦ are related. > > Fooling around with special cases, two heuristic methods seem to work. > Looking at arithmetic subsequences of numerators and denominators > separately > should eventually score, although there might be multivalued ambiguity in > filling in the missing values. Alternatively, taking repeated ratios of a > long > sample seems to smooth out sporadic cancellations, at the expense of large > numbers of large numbers. Neither method suggests when to give up. > > Was anybody surprised by > Fibonacci[6*n]*Fibonacci[n]/Fibonacci[3*n]/Fibonacci[2*n] - > Fibonacci[3*n]/Fibonacci[n] ? NeilB found a short Mma proof. > --rwg > On Fri, Jul 5, 2013 at 6:13 AM, Bill Gosper <billgosper(a)gmail.com> wrote: >> is sporadically impressive: >> In[127]:= FindSequenceFunction[Table[2^n - 1/n, {n, 11}], n] >> >> Out[127]= (-1 + 2^n n)/n >> >> In[133]:= FindSequenceFunction[Table[3^n/n - 2^n, {n, 11}], n] >> >> Out[133]= FindSequenceFunction[{1, 1/2, 1, 17/4, 83/5, 115/2, 1291/7, >> 4513/8, 1675, 48809/10, 154619/11}, n] >> >> In[130]:= FindSequenceFunction[{1, 2, 4, 8, 15}, n] >> >> Out[130]= 1/6 (8 n - 3 n^2 + n^3) >> >> In[131]:= Table[2^n/%, {n, 11}] >> >> Out[131]= {2, 2, 2, 2, 32/15, 32/13, 64/21, 4, 512/93, 512/65, 128/11} >> >> In[132]:= FindSequenceFunction[%, n] >> >> Out[132]= (3 2^(1 + n))/(n (8 - 3 n + n^2)) >> >> In[156]:= FindSequenceFunction[Table[1/(3^n - n), {n, 12}], n] >> >> Out[156]= FindSequenceFunction[{1/2, 1/7, 1/24, 1/77, 1/238, 1/723, 1/ >> 2180, 1/6553, 1/19674, 1/59039, 1/177136, 1/531429}, n] >> >> Is there known a generalization of reciprocal differences, say, that can >> detect rational_function(n,2^n,(-1)^n,Fib(n),...)? >> --rwg >> I have one that can detect >> Fibonacci[6*n]*Fibonacci[n]/Fibonacci[3*n]/Fibonacci[2*n] - >> Fibonacci[3*n]/Fibonacci[n] ;-) > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] FindSequenceFunction
by Bill Gosper 13 Jul '13

13 Jul '13

is sporadically impressive: In[127]:= FindSequenceFunction[Table[2^n - 1/n, {n, 11}], n] Out[127]= (-1 + 2^n n)/n In[133]:= FindSequenceFunction[Table[3^n/n - 2^n, {n, 11}], n] Out[133]= FindSequenceFunction[{1, 1/2, 1, 17/4, 83/5, 115/2, 1291/7, 4513/8, 1675, 48809/10, 154619/11}, n] In[130]:= FindSequenceFunction[{1, 2, 4, 8, 15}, n] Out[130]= 1/6 (8 n - 3 n^2 + n^3) In[131]:= Table[2^n/%, {n, 11}] Out[131]= {2, 2, 2, 2, 32/15, 32/13, 64/21, 4, 512/93, 512/65, 128/11} In[132]:= FindSequenceFunction[%, n] Out[132]= (3 2^(1 + n))/(n (8 - 3 n + n^2)) In[156]:= FindSequenceFunction[Table[1/(3^n - n), {n, 12}], n] Out[156]= FindSequenceFunction[{1/2, 1/7, 1/24, 1/77, 1/238, 1/723, 1/ 2180, 1/6553, 1/19674, 1/59039, 1/177136, 1/531429}, n] Is there known a generalization of reciprocal differences, say, that can detect rational_function(n,2^n,(-1)^n,Fib(n),...)? --rwg I have one that can detect Fibonacci[6*n]*Fibonacci[n]/Fibonacci[3*n]/Fibonacci[2*n] - Fibonacci[3*n]/Fibonacci[n] ;-)

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[math-fun] Killing Fibonacci's Rabbits
by Stuart Anderson 12 Jul '13

12 Jul '13

If Fibonacci had used more realistic assumptions (ie his rabbits dying) he would have produced the Padovan (or Perrin?) sequence. http://matheminutes.blogspot.com.au/2012/02/killing-fibonaccis-rabbits.html Another example of this sort of thing; summing A000079 (powers of 2), and subtracting previous terms 2 generations back gives A003945 Is there a general form for such sequences?

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[math-fun] STNATSNOC LACITAMEHTAM by RETTOP and NOSNIBOR
by Simon Plouffe 12 Jul '13

12 Jul '13

Hello, This is the famous print of Robinson and Potter, that someone graciouly gave me (neil sloane) so I could not resist scanning with OCR , I used 600 dpi to b sure not missing anything. The result is here : http://www.plouffe.fr/simon/math/ the file is STNATSNOC LACITAMEHTAM RETTOP and NOSNIBOR.pdf it will appear in a short time. 25 min. The strange title is easy to read but you have to think a litlle before. I really don't know if the content is copyright, there is a floating copy of the document somewhere but the scanning is poor. also we can part of it on google books, it appears on amazon but no price or availability of any kind, I presume that the 2 authors would not object to the diffusion of that document. If someone has an objection to this action I would remove the document. Best regards, Simon PLOUFFE

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[math-fun] it Shor helps to have a good compiler
by Veit Elser 12 Jul '13

12 Jul '13

20000-bit number factored by 2008 US Oklahoma commemorative quarter: http://www.nature.com/nature/journal/v499/n7457/full/nature12290.html?WT.ec…

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[math-fun] Mathematical proofs set to music
by Warren D Smith 11 Jul '13

11 Jul '13

http://us.metamath.org/mpegif/mmmusic.html

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Re: [math-fun] triangular tables [Was: Atrocious Mma ...]
by Bill Gosper 10 Jul '13

10 Jul '13

Fred> They've been lazy, that's all: they have assumed that symmetry applies across the board, overlooking the fact that the basic recursion which they have broken at the origin is implied by a global property of the Gamma function; while symmetry is not. rwg>No, it used to work! 9.0.1 Doc: As a bivariate function, Binomial is not continuous in both variables at negative integers: In[1]:= Binomial[-3, -5] Out[1]= 0 In[2]:= Binomial[-3.0000001, -5.0000001] Out[2]= 6. They up and broke it! (And forgot to break the Doc.) Probably in response to a misguided complaint. That's not lazy--that's industrious. Re misguided complaints, I've probably related here Petr Beckmann's unfortunate elision from his History of π, after a reader numerically determined that Product 1/(1-p^-2), p=2,3,5,7,9,..., was not π^2/6. The book never said 9. --rwg Fred> What intrigues me is that their identity reduction mechanism seems to give results correct also for integer arguments, despite the fact that they presumably must invoke properties of Gamma and hypergeometric functions which fail at the singularities. I conjecture that any identity supported in this fashion must hold for all flavours of pseudo-binomial coefficients defined by approaching the limit from a consistent direction. An example would be Rich's one plus two halves-wedge variant, where the limit is approached along x = 2 y rather than y = 0 . Which just goes to show that you can have half your cake and eat it! Fred If true, this should allow such identities to be rewritten generalized. On 7/10/13, Bill Gosper <billgosper(a)gmail.com> wrote: rwg>Here is an almost correct technique for printing number triangles, of which Julian must repeatedly remind me: pt[n_Integer, from_Integer: 0] := TableForm[Table[If[EvenQ[i + j], "", Binomial[-1 + i, -Floor[n/4] + (1 + i + j)/2] /. 0 -> ""], {i, from, n}, {j, -Ceiling[3*n/4], n + 1}]] [...] Duh, for actual symmetrical triangles, In[516]:= Column[Table[Row[Table[Binomial[n,k],{k,0,n}]," "],{n,0,5}],Center] Out[516]= 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 --rwg I somehow missed the distressing answer that Maple has been bodysnatched by the same brain virus (or is it Toxoplasma?) that's corrupted Mma: http://isc.carma.newcastle.edu.au/standardCalc accepts Maple input. Standard lookup results for *Pi^binomial(-2,-6)* Best guess: Pi^(5) Both leading CASs! Maybe they think it's just a matter of personal preference, like whether toilet paper spools off the front or the back?

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[math-fun] triangular tables [Was: Atrocious Mma ...]
by Bill Gosper 10 Jul '13

10 Jul '13

rwg>Here is an almost correct technique for printing number triangles, of which Julian must repeatedly remind me: pt[n_Integer, from_Integer: 0] := TableForm[Table[If[EvenQ[i + j], "", Binomial[-1 + i, -Floor[n/4] + (1 + i + j)/2] /. 0 -> ""], {i, from, n}, {j, -Ceiling[3*n/4], n + 1}]] [...] Duh, for actual symmetrical triangles, In[516]:= Column[Table[Row[Table[Binomial[n,k],{k,0,n}]," "],{n,0,5}],Center] Out[516]= 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 --rwg I somehow missed the distressing answer that Maple has been bodysnatched by the same brain virus (or is it Toxoplasma?) that's corrupted Mma: http://isc.carma.newcastle.edu.au/standardCalc accepts Maple input. Standard lookup results for *Pi^binomial(-2,-6)* Best guess: Pi^(5) Both leading CASs! Maybe they think it's just a matter of personal preference, like whether toilet paper spools off the front or the back?

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