February 2012 - mathfun@mailman.xmission.com

[math-fun] about pdf readers
by Robert Munafo 05 Feb '12

05 Feb '12

The reason we lose more performance than Moore's law gives back is because there are so many programmers (undeserving of the title "software engineers") who would rather use some huge, unoptimal software package or library than spending days, weeks or even years learning how to do it efficiently. (Market forces play a role too :-) On Sun, Feb 5, 2012 at 09:09, Joerg Arndt <arndt(a)jjj.de> wrote: > * Henry Baker <hbaker1(a)pipeline.com> [Feb 05. 2012 14:48]: > > There are now some free pdf reader apps that are about 1/10th the size > of Adobe Reader. > > [In Adobe Reader] You can disable Javascript by renaming some directory > in the installation btw. (could not find out what the > name is in <=60 secs). > Wikipedia says: [To] disable JavaScript in Adobe Reader [...] [use] the General preferences dialog (Edit, Preferences, JavaScript, and un-check "Enable Acrobat JavaScript"). And there is xpdf: executable < 300kB (but loads a couple > of dynamic libs). Memory footprint about 13MB on my system: > Ghostscript is useful too. xpdf and ghostscript are both available to Mac users via Macports [...] acroread used to be better in "repairing" broken pdfs, > nowadays it has clearly lost it's lead in this respect. > > Still, some documents, notable scanned books from arxiv, > take bloody ages (>10secs) to render a single page. Somehow > we are losing more performance over time than Moore's law > gives us. > > -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

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[math-fun] last nite, Corey and Julian
by Bill Gosper 05 Feb '12

05 Feb '12

ported my path-invariant stuff to Mma, after a week of arguing over how to think about it. At bedtime, (actually, long past), Julian casually tossed off an identity of the form trigamma(n0) = three contiguous pFp-1[1/4] . So I did FullSimplify[%/.n0->1/2] and went home. It's still running. Mma turned just one of the pFp-1 into a screenful of radicals and trigs of logs and arccoths, and wishes to wallow in it for eternity. Macsyma says it's equivalent to sum((((12 * j^2 + 30 * j + 19) * (2 * j + 1)!)/((j + 1) * (2 * j + 3) * 4^j * (2 * j + 5/2)!)),j,0,inf) = 2 * %pi^(3/2) - 16/(sqrt(%pi)) but when asked for a closed form for the lhs, temporized with Is 4 - g1928 * g1941 * g1960 positive, negative, or zero? Is 4 - g12326 * g1928 * g1941 positive, negative, or zero? Is 4 - g153423 * g1928 * g1941 positive, negative, or zero? and then the trump card Is false zero or nonzero? (As Jeff Golden used to say: "It's positive. Everybody knows that.") Then ten more impudent questions. Now it's just sulking. --rwg

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[math-fun] Fwd: infinite prime pattern
by Henry Baker 05 Feb '12

05 Feb '12

This email showed up on the "Khan Academy" (aka home school academy!) email list. I have no idea how old this person is, but he seems to be interested in learning more. >From: Travis Rogers <cyanember(a)msn.com> >To: <khan-academy-comments(a)googlegroups.com> >Subject: infinite prime pattern >Date: Wed, 1 Feb 2012 22:02:43 -0800 > >Well, here is two way of many to look at the pattern I found. first (not much faster, if at all, than conventional methods) square root of N =low Z(rounded up), N/3 = high Z(rounded up). N/ each Z from low to high =X and the remainders = Y N/each Y if none = whole number then N=prime. e.g say N=37 sq root 6.something so 7. 37/3= 12.3 so 13. >37/7 = x5y2 >37/8= x4y5 >37/9= x4y1 >37/10= x3y7 >37/11= x3y4 >37/12= x3y1 >37/13= x2y9 > >now with this method you can ignore any y's that are not prime's greater than 5 or are repeats from x(5,4,3,2) or z(7,8,9,10,11,12,13) so in this case 37 can be confirmed without further calculation. > >the second way I looked at the pattern seems to work much faster and the pattern is found in what is left behind > - >4 >6,8,9 >6,8,10,12,14,15,16 >18,20,22,24,25 >20,21,24,26,27,28,30,32,33,35,36 >30,32,35,40,42,44,45,48,49 >34,36,40,42,44,45,48,50,51,52,54,56,58,60,63,64 >38,39,40,42,44,45,48,50,54,56,57,60,63,66,70,72,74,75,76,78,80,81 >42,44,45,48,50,52,54,55,60,62,64,65,66,70,72,75,77,80,82,84,85,88,90,92,93,95,96,99,100 > >note 46 hasn't been found yet so accurate only to less than 44 but as you can see the first line is the numbers >left out of the eliminated in the first 10 squares, and the second is the actual primes in the first 10 squares. > >1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,63,67,68,69,71,73,79,83,86,87,89,91,94,97,98, > 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 > >Okay now to try and explain how to see the pattern. One of my biggest problems is thinking in three dimensions, and that is exactly where the pattern lies. Each number squared will eliminate a finite number of non primes (with the exception of 1, which by the standards of this pattern would be prime) so lets look at the first few squares. >1+ > >1+2+ >3*4- > >1+2+3+ >4*5*6- >7*8-9- > >1+ 2+ 3+ 4- >5* 6- 7* 8- >9* 10- 11* 12- >13* 14- 15- 16- >+=confirmed prime - =eliminated in this square *=prime candidate according to this square > >Z always equals the top right corner. x1 is the row below the top row ( or what I usually call the prime row) y's are the numbers in the prime row. If a number makes it to the prime row without being eliminate then it is a prime. to tell if a number is eliminated you must divide it by x and y if either comes out whole then it is not prime and is eliminated(for all numbers below the prime row). it is important to figure out each square separately and not automatically eliminate the numbers you know aren't primes. the pattern really comes to life when you build a 3d model even to just 10^2 (I bet Legos would work awesome) but once you build the model you then look at each xy xz and yz planes the patterns are impossible to miss. there are some obvious things like (if n = top right corner) x1 to n-1 diagonal on the xy plane never has any prime candidates while x=any y1 is always a prime candidate and x=any y=n never is a candidate. the way I found the above elimination patte rn was taking the lowest even number that has not been eliminated yet and starting at y1 of whatever x that number fell on. then just listed each number eliminated in the square from that starting point up. it creates some repeats but never eliminates a prime and as far as I can tell doesn't let anything slip through the cracks. I didn't test this real far but looks real promising. > >but look at cross section of say x2 and z it will look like > >o x x >o x o x >o x o x x >o x x x o x >o x o x o x x >o x o x o x o x >o x x x o x o x x >o x o x x x o x o x > >o is candidate x is non-candidate the first two rows are obvious all o's and all x's. third row x 2 o's fourth row all x's fifth x 4 o's 6th all x's 7th x 6 o's etc.. > >I wish I had more time to work on stuff like this but last week I put in 70+ hours at my regular job and I just can't dedicate much time to it. matter of fact this e-mail is taking up about 90% of my awake time today not spent working. I just wanted to share this with people who appreciate knowledge, and if anyone in the world can poke holes in it it will be you guys :) I hope somebody there will find the time to play with this a little. > >Travis Rogers >cyanember(a)msn.com

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Re: [math-fun] More fun with sequences embedded in decimals
by Michael Reid 05 Feb '12

05 Feb '12

Mersenne primes, apart from 2^2 - 1 , have the form 2 a^2 - 1 , (as do all "Mersenne numbers", M_n = 2^n - 1 , when n is odd). The product of two numbers of the form 2 a^2 - 1 does not necessarily have the same form, although it will be of the form 2 x^2 - y^2 . Consider the Diophantine equation (2 a^2 - 1)(2 b^2 - 1)^2 = (2 c^2 - 1) . This has the "obvious" solutions where b = 0, 1, -1 and a = +/-c . It also has the not-so-obvious solutions where b = 2a , and c = a (8 a^2 - 3) . Take a = 4 in the not-so-obvious family, to get b = 8 and c = 500 , which gives David's factorization. > Yes, in fact you cracked it. > > What I thought was notable about this solution was: > > - It arises naturally from a common length conversion. > > - We know that > > 0.3937 = 0.3937 > 1/2.54 = 0.3937007874... > > Given that the numbers themselves are so close, it would stand to reason > that their geometric mean would be near > their arithmetic mean, so that > > sqrt(0.3937 * (1/2.54)) .= (0.3937+(1/2.54))/2 = 0.3937003937... > > Both accurate to the decimals shown. This gives us a square root with a > pretty repeated start block, which you observed and reproduced in > another example. > > - But then there is the nice common factor between the block values > > 3937 = 31*127 > 254 = 2*127 > > This arises from the unusual square factor in the prime factorization: > > 999998 = 2*31*127^2 > > This has the peculiar side benefit that the quotient of the repeating > blocks is a terminating decimal: > > 3937/254 = 15.5 > > Going back to our geometric mean identity, we get > > sqrt(0.3937 * (1/2.54)) = sqrt(0.155) = 0.3937003937... > > Which gives us a pretty square root of a terminating decimal. > Multiplying by 20 on both sides then gives > > sqrt(62) = 7.874 007874 ... > > I will guess that small integer square roots with pretty blocks like > this are rare in any base. > > - And an aside about the Mersenne primes > > You can't miss that 31 and 127 are Mersenne primes, which at first might > seem incidental. However, consider that > > 2^20 = 1048576 .= 10^6 > > is close to a power of 10. Obviously this is well known, since we call > 2^10 bytes a kilobyte, 2^20 bytes a megabyte, etc. > > We can break this up to > > 2 * 2^5 * (2^7)^2 = 1048576 .= 10^6 > > And if we tweak a couple of these factors slightly, we shouldn't affect > the product too much, so lets tweak 2^5 and > 2^7 down a notch: > > 2 * (2^5-1) * (2^7-1)^2 .= 999998 .= 10^6 > > So perhaps it is now less surprising that Mersenne primes might multiply > to a value near 10^6, since Mersenne > primes are slightly smaller than powers of 2 and 2^20 is slightly larger > than 10^6. > > But of course, this is a hindsight analysis, since I knew the > factorization to begin with. > > On 2/5/2012 3:09 AM, Robert Munafo wrote: > > I don't see how the Mersenne factors matter. > > > > The first two of David's properties, 1/2.54 = 00.3937 007874... and > > 1/39.37=00.0254 000508... are explained by their product being 99.9998 > > = (10^N-2) / 10^K, as I suggested before. > > > > The other of David's properties, regarding the geometric mean, also > > comes from (10^N-2) / 10^K, as you can see here: > > > > 1/sqrt(99.9998) > > = .100000100000150000250000437500787501443752681255027353246111 > > 1/sqrt(99.998) > > = .100001000015000250004375078751443776813002743871274179225808 > > 1/sqrt(99.98) > > = .100010001500250043757876444018175282935398352327750882087668 > > 1/sqrt(99.8) > > = .100100150250438288946436286857920839193514369563735147583946 > > > > These all have the same pattern, related to the series expansion of > > sqrt(1/(1-2x)), which you can get by "Series[sqrt(1/(1-2*x)),{x,0,8}]" > > in Mathematica or Wolfram Alpha. The coefficients are: 1, 1, 3/2, 5/2, > > 35/8, 63/8, 231/16, 429/16, 6435/128, ... The numerators are > > https://oeis.org/A001790, which are also the numerators of the > > sqrt(1/(1-x)) series expansion because only the denominators change. > > >

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Re: [math-fun] number theory puzzle about squarefree number density
by Henry Baker 05 Feb '12

05 Feb '12

There are now some free pdf reader apps that are about 1/10th the size of Adobe Reader. I stopped "upgrading" Adobe Reader a couple of years ago when the download exceeded 100Mbytes. I have since found out that Adobe "Reader" has all kinds of cr*p that you don't use & can hurt you -- including full Javascript (I think that Javascript is what blew it past 100Mbytes). You don't need or want Javascript for pdf files, because Javascript enables all kinds of bad behavior, including worms, viruses, etc. Javascript is primarily used for filling out forms, which you probably need to do at most 1x per year. The rest of the year, Javascript is a gaping security hole. Apple licensed pdf a very long time ago & uses it extensively, but the Apple "Preview" reader is much, much smaller than the Adobe Reader. At least one of Apple's word processing file formats is ".pages", which is really a .zip file in drag with a .pdf file hidden inside. At 10:02 AM 2/4/2012, Fred lunnon wrote: >On 2/4/12, Warren Smith <warren.wds(a)gmail.com> wrote: >> In fact I'm still not aware of it since my computer refuses to open >> the pdf file, says it is >> "damaged." > >I had trouble with this phenomenon until I twigged that it happens when >the PDF file has been generated with a newer version of the Adobe software, >but the reader is an older version. Pity they couldn't just tell you >that, isn't it? > >Adobe Acrobat Reader version 7.0.8 is compatible with the newer PDF files >and the older operating systems that people like us are (for various reasons) >stuck with. It can be downloaded from > > http://mac.oldapps.com/adobe_reader.php > >--- but it's a dreadful memory hog, compared with (say) version 5. > >Fred Lunnon

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Re: [math-fun] More fun with sequences embedded in decimals
by Michael Reid 05 Feb '12

05 Feb '12

I think there's more than this, if I get what David was hinting at. Rot13, so as not to give it away: Qnivq'f ahzore snpgbef nf n cebqhpg bs gjb fznyyre ahzoref, juvpu fgvyy funer n pbzzba snpgbe. Zberbire, nyy bs vgf cevzr snpgbef ner Zrefraar cevzrf! Truly remarkable, David! > > The 27, 37 example is the smallest pair that isn't just a trivial repeat of > 1/9=.1111... or 1/3 = .3333... > > But yes, you're right. I didn't want to give away too much of the answer of > David Wilson's exercise to the reader. . . but, spoilers and many examples > at [1] > > - Robert > > [1] http://mrob.com/pub/math/numbers-12.html#lb999 > > On Sun, Feb 5, 2012 at 00:44, Tom Duff <td(a)pixar.com> wrote: > > > Robert Munafo <mrob27(a)gmail.com> writes: > > > >> > >> Hey that's cool, and they're much like my favorite such pair, > >> 1/27=0.037037037... and 1/37=0.027027027... I'll add them to my numbers > >> pages! > >> > > > > I suspect these are pretty thick on the ground. For example we have > > 1/303 = 0.0033 0033 0033 ... > > and > > 1/33 = 0.0303 0303 0303 ... > > > > -- > Robert Munafo -- mrob.com > Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - > mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

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Re: [math-fun] More fun with sequences embedded in decimals
by Tom Duff 04 Feb '12

04 Feb '12

Robert Munafo <mrob27(a)gmail.com> writes: > > Hey that's cool, and they're much like my favorite such pair, > 1/27=0.037037037... and 1/37=0.027027027... I'll add them to my numbers > pages! I suspect these are pretty thick on the ground. For example we have 1/303 = 0.0033 0033 0033 ... and 1/33 = 0.0303 0303 0303 ... (Hint for finding more: 27x37=999, 303x33=9999) -- Tom Duff. "Gore puts the hammer on me," Mattson said. "I folded like a cheap suit."

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[math-fun] Wilson's theorem on primes (no, not that one, the one about decimals)
by Warren Smith 04 Feb '12

04 Feb '12

> From: David Wilson <davidwwilson(a)comcast.net> > Let p >= 10^d, and let 0 <= B < 10^d be any d-digit block of digits. > By the definition of the ceiling function > [1] pB/10^d <= ceil(pB/10^d) < pB/10^d + 1. > Dividing by p gives > [2] B/10^d <= ceil(pB/10^d)/p < B/10^d + 1/p > But p >= 10^d, so 1/p <= 1/10^d, and > [3] B/10^d + 1/p <= B/10^d + 1/10^d = (B+1)/10^d > [2] and [3] together give > [4] B/10^d <= ceil(pB/10^d)/p < (B+1)/10^d > Letting n = ceil(pB/10^d), we have integer n with > [5] B/10^d <= n/p < (B+1)/10^d > [5] says that the decimal fraction n/p starts with block B. > So if p >= 10^d, there is a fraction n/p (0 <= n < p) starting with any > given d-digit block B, namely n/p = ceil(pB/10^d)/p. > > If p > 10^d, then by the previous paragraph, there is a fraction n/p (0 > <= n < p) starting with any given d-digit block B. But because p > 10^d, > since 0/p and 1/p both start with block B = 0, so we can strengthen the > condition on n to 1 <= n < p. In other words: > > If p > 10^d, there is a fraction n/p (1 <= n < p) starting with any > given d-digit block B. > > Now suppose p > 10^d and p has primitive root 10. Because of the > primitive root 10, every n with 1 <= n < p is expressible as n = 10^k > mod p for some 0 <= k < p-1. By our previous paragraph, there is then a > fraction of the form (10^k mod p) / p starting with any desired d-digit > block B. From this we conclude that every d-digit block occurs somewhere > in the decimal expansion of 1/p. --this is a nice theorem, and it might be a new one. It is also of some importance because it proves a randomness property of the Marsaglia type of pseudorandom number generator. To determine if it is new, I suggest looking to see if it is in Hardy & Wright chapter 9 as step one. Frankly though, it looks too easy to have escaped previous notice. > In other words, any prime p > 10^d with primitive root 10 includes every > d-digit block in the decimal expansion of its reciprocal. This is why I > must choose the smallest such prime if I am looking for the first number > with that property. > > I can show that the smallest number n whose decimal reciprocal includes > every d-digit block must satisfy n >= 10^d. --well, that seems trivial since the period<n. > I can also show that the > reciprocal of the smallest prime p > 10^d with primitive root 10 > includes every d-digit block. I can show that if p < 2*10^d, --you do not need this "if" because of the known theorem that there is always a prime between X and 2X if X>1: http://en.wikipedia.org/wiki/Bertrand's_postulate > then there > is no other prime q with 10^d <= q < p whose reciprocal includes every > d-digit block. What I can't show is that there is not some composite q > with 10^d <= q < p whose reciprocal includes every d-digit block, though > I seriously doubt such a q could exist. --if q=a*b*...*c where a,b,...,c are primes with 10 primitive modulo each of them, then every n with 1<=n<q and gcd(n,q)=1 is expressible as n = 10^k mod q for some 0<=k<p-1. However, each n with gcd(n,q)>1 definitely is NOT thus-expressible. This tells me that the decimal expansion of 1/q with q composite, q>10, NEVER includes all k-digit blocks where k=floor(log10(q)), because those blocks are exactly the opening bars of 10^j/q. To prove that, there are two cases: case i: q>10 contains a prime p with p>10. Then the multiples of q/p are unrepresentable and include examples below 10^k case ii: q>10 contains a prime p<10, then the multiples of p do the job. QED.

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[math-fun] prime products - the next one
by Warren Smith 04 Feb '12

04 Feb '12

0.125486980905809 = prod[p prime] (1 - 3/p^2) is the probability that n, n+1, n+2 all are squarefree. This number does not seem to be in Plouffe invertor, google, etc.

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Re: [math-fun] computing products like (1 - 2/p^2)
by N. J. A. Sloane 04 Feb '12

04 Feb '12

Rich, I have fixed the broken link in http://oeis.org/A001692/a001692.html Also the constant in question is A065474. Neil

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