March 2013 - mathfun@mailman.xmission.com

Re: [math-fun] How often does every bit matter?
by Keith F. Lynch 02 Apr '13

02 Apr '13

Warren D Smith <warren.wds(a)gmail.com> wrote: > Each of the following is a prime number P, such that if any single > bit of P's binary representation is changed, then you no longer have > a prime. (And this is the full list of such P<10000.) I take it that leading zeros aren't allowed. If they are, I suspect the list would be empty, as any entry would be an odd prime such that if you add any power of 2 to it the result will always be composite. I doubt such a number exists, but I can't see how to prove it.

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[math-fun] How often does every bit matter?
by Warren D Smith 01 Apr '13

01 Apr '13

Each of the following is a prime number P, such that if any single bit of P's binary representation is changed, then you no longer have a prime. (And this is the full list of such P<10000.) 127, 173, 191, 223, 233, 239, 251, 257, 277, 337, 349, 373, 431, 443, 491, 509, 557, 653, 683, 701, 733, 761, 787, 853, 877, 1019, 1193, 1201, 1259, 1381, 1451, 1453, 1553, 1597, 1709, 1753, 1759, 1777, 1973, 2027, 2063, 2333, 2371, 2447, 2633, 2879, 2917, 2999, 3083, 3181, 3209, 3313, 3511, 3593, 3643, 3767, 3779, 3851, 3877, 3889, 3967, 4013, 4177, 4283, 4441, 4451, 4561, 4597, 4603, 4679, 4813, 4889, 4951, 5051, 5099, 5209, 5323, 5557, 5801, 5867, 6007, 6073, 6151, 6203, 6211, 6287, 6323, 6379, 6481, 6521, 6971, 6977, 6997, 7027, 7039, 7043, 7103, 7109, 7151, 7207, 7297, 7307, 7331, 7369, 7507, 7573, 7583, 7841, 7883, 8017, 8087, 8111, 8171, 8231, 8243, 8311, 8363, 8627, 8747, 8831, 8849, 8867, 8923, 9137, 9151, 9161, 9319, 9323, 9697, 9767 The Mersenne primes P=2^p-1 also have this "every bit matters" property when p = 7, 31, 127, 607, 1279, 4423 for the p<5000. I would presume that asymptotically, a constant fraction C of all primes are "every bit matters" primes. What is that constant? I would guess that C = exp( (-1/ln2) * (1-1/2)/(1-1/3) * (1-1/4)/(1-1/5) * (1-1/6)/(1-1/7) * ... ) = exp( (-1/ln2) * (1*3)/2^2 * (3*5)/4^2 * ... * (p-2)*p/(p-1)^2 * ... ) = exp( (-1/ln2) * 0.660162) = 0.38581 the product being over all odd primes p. However, my guess seems to be wrong in view of the actual counts 89486 / 664579 = 0.13465 from the primes below 10^7 794760 / 5761455 = 0.13794 from the primes below 10^8 I think I see my error now... getting this right should be tricky. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Weird leap-year rule
by Keith F. Lynch 31 Mar '13

31 Mar '13

"Adam P. Goucher" <apgoucher(a)gmx.com> wrote: > http://cp4space.wordpress.com/2012/09/12/lunisolar-calendars/ > I briefly mentioned the use of the Rabbit sequence to optimally > approximate a leap year of length 364 + phi. Interesting. Thanks. And very appropriate for Easter. :-) You may have noticed that all my rules involve divisibility. That's because that's the most obvious property integers have that approaches a limit, i.e. some proportion of the first N integers have this property (e.g. are odd), and that proportion approaches a limit other than 0 or 1 as N grows large. It's difficult to come up with other examples of such a property. (There's the divisibility of the number of 1-bits in the binary representation of the number, but that has nothing to recommend it over the divisibility of the number itself.) You've come up with such a property that I hadn't previously heard of Thanks. Can anyone think of any others? Responding to your discussion (at the above URL) of our calendar: Pope Gregory's intention when he established the current calendar wasn't to track the average length of the tropical year as you implicitly assume, but to track the average time between vernal equinoxes, since his goal was to make sure Easter is always celebrated on the right day. (To us moderns, the idea that there's a "right day," that the placement of holidays isn't completely arbitrary, seems odd, but not to scholars in those days, with the exception of Kepler.) Yes, these are the same thing over the very long term, but not during any given millennium, thanks to the eccentricity of Earth's orbit and the precession of the equinoxes. The average time between consecutive vernal equinixes over the past 2000 years is significantly different from the average time between consecutive autumnal equinixes over the same period. And the difference will be about the same over the next 2000 years too. Also, there's no point in making a calendar more accurate than one day in a thousand years, since the number of days in a year is dropping quickly and unpredictably enough due to tidal drag that we'll need a new calendar rule every two or three millennia anyway. J.R.R. Tolkein made this mistake. In appendix D of _The Lord of the Rings_, he said: The Calendar of the Shire differed in several features from ours. The year no doubt was of the same length,^1 for long ago as those times are now reckoned in years and lives of men, they are not very remote according to the memory of the Earth. ... ^1 365 days, 5 hours, 48 minutes, 46 seconds This was wrong. Well, strictly speaking, the length of the *year* may have been the same within a few seconds, but the length of the *day* certainly wasn't. So the number of days per year was different. I've even seen a few silly speculations as to whether the number of days in a year is transcendental. The "correct" answer is that it usually is, but there are infinitely many times during every nanosecond when it isn't. :-)

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Re: [math-fun] How often does every bit matter?
by Keith F. Lynch 31 Mar '13

31 Mar '13

"Fred W. Helenius" <fredh(a)ix.netcom.com> wrote: > 271129 is such a number; it is prime and 271129 + 2^k is always > divisible by (at least) one of 3, 5, 7, 13, 17 and 241. Interesting. Thanks. I think allowing leading zeros makes the problem better, since you otherwise have an annoying asymmetry where it's not always the case that if you can get A from B via a single bit flip you can get B from A the same way. Does this generalize to powers of numbers other than 2? To the Fibonacci sequence, i.e. is there a number which when added to any Fibonacci number is always composite? To any other interesting sequences? Thanks.

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Re: [math-fun] Weird leap-year rule
by Adam P. Goucher 31 Mar '13

31 Mar '13

> I'm still looking for good rules for phi, [...] http://cp4space.wordpress.com/2012/09/12/lunisolar-calendars/ I briefly mentioned the use of the Rabbit sequence to optimally approximate a leap year of length 364 + phi. Just choose your favourite rule for generating it: http://en.wikipedia.org/wiki/Rabbit_sequence I seem to recall that there was a Google Turing machine thingy that generated the Rabbit sequence. Sincerely, Adam P. Goucher

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Re: [math-fun] Weird leap-year rule
by Keith F. Lynch 30 Mar '13

30 Mar '13

Last month I wrote: > I also came up with simple rules for calendars which gave pi in the > average year length. Can anyone else find such a rule? Or rules > that give e, Euler's constant, or other interesting numbers in the > average year length? It looks like nobody else is interested in working on it. So I'll give my answers. By a leap-year rule, I mean a rule that adds some whole number of days (possibly zero or negative) to some years based on the year number. If a year has a negative length, the necessary number of days are borrowed from the following year. If that causes the following year to also have a negative length, the necessary number of days are borrowed from the year after that, etc. If this is impossible, i.e. all future years put together don't have enough days to pay off the accumulating debt, the calendar breaks down. The calendar rule I mentioned last month was that a day is added for each odd number (including 1) that the year number is divisible by, and subtracted for each even number that the year number is divisible by. The average year length converges to 365 plus the natural log of 2. Calendar rule pi: Add four days for each number congruent to 1 mod 4 that the year number is divisible by (e.g. 1, 5, 9, 13), and subtract four days for each number congruent to 3 mod 4 that the year number is divisible by (e.g. 3, 7, 11, 15). The average year length converges to 365 + pi. Calendar rule pi^2: Add six days for each square that the year number is divisible by (e.g. 1, 4, 9, 16). The average year length converges to 365 + pi^2. Calendar rule e: Add one day for each factorial number that the year number is divisible by (e.g. 1, 1, 2, 6, 24). Note that 1, which all years are divisible by, counts as two factorial numbers (0! and 1!). The average year length converges to 365 + e. Calendar rule e^2: Add N^2 days for each factorial number that the year number is divisible by, where N is which number's factorial it is. The average year length converges to 365 + e^2. (This generalizes to e to any natural-number power.) Calendar rule 1/e: Add or subtract one day for each factorial number that the year number is divisible by. Add if it's the factorial of an odd number, subtract if it's the factorial of an even number. The average year length converges to 365 + 1/e. I'm still looking for good rules for phi, Euler's constant, and the square root of 2. Can anyone find any? Or for any other "interesting" numbers? (It's always possible to find a solution using greedy Egyptian fractions, but that's ugly.) Some rules don't converge. For instance what if you add a day for each distinct prime number that the year number is divisible by? The average length grows without limit, but very slowly. Turn it around. What if you *subtract* a day for each distinct prime number that the year number is divisible by? This calendar will eventually break down into a cascade of day-borrowing that can never be caught up. But not for a very, very, very long time. Can anyone figure out when? I do know that the first negative-length year won't occur until after the year 10^1050. But that year will have no trouble borrowing from the following year; the breakdown won't be until much later.

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[math-fun] Question on real functions
by David Wilson 30 Mar '13

30 Mar '13

Let f be a continuous bijection between two real intervals, which remains a bijection when restricted to the rationals. The only examples of such f that I could construct are piecewise of the form y = (ax+b)/(cx+d), with a, b, c, d integer, cd != 0. Are there other examples?

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Re: [math-fun] How often does every bit matter?
by Henry Baker 29 Mar '13

29 Mar '13

So this means that most primes are 'dangerous', in the sense that if you get 1 bit wrong, it is not a prime, and in fact, may be quite composite. Soooo.... Are there primes which _incorporate_ their error-correcting code into their own bit-string ? I.e., these primes don't need a separate set of error-detecting/correcting bits. BTW, I was thinking of primes over the usual integers represented in base 2. But I guess this wouldn't be possible for prime polynomials over GF(2)... At 11:34 AM 3/29/2013, Warren D Smith wrote: >>Charles Greathouse points out paper by T.Tao... > >--oh, great, scooped by Terry Tao. >Terence Tao: >A remark on primality testing and decimal expansions, >J.Australian Math'l. Soc. 91,3 (2011) 405-413. >http://arxiv.org/abs/0802.3361

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Re: [math-fun] speed-of-light-may-not-be-constant?
by Keith F. Lynch 29 Mar '13

29 Mar '13

"Adam P. Goucher" <apgoucher(a)gmx.com> wrote: > The speed of light is actually constant. Well, that clears that up. Hans Havermann <gladhobo(a)teksavvy.com> wrote: > Theoretical. There are two papers in Springer's 'The European > Physical Journal D'. Here are the abstracts: Thanks. I don't see anything in the first abstract that Feynman didn't say decades ago. This tiny variation in the time light takes to travel a fixed path is more commonly expressed as Heisenberg uncertainty in the length of that path. Thomas Knight <tk(a)csail.mit.edu> wrote: > Perhaps the Casimir effect can be used to detect a shift, given its > change in the population of virtual particles in the gap. Maybe. Intuitively, this quantum effect should be much larger than the general relativistic effect due to the negative mass-energy in the gap, and both effects should cause light to go faster. However, since faster light could cause causality violations, there's probably some reason this won't work. Dan Asimov <dasimov(a)earthlink.net> > Naturally, since the speed of light is constant by definition, it > must be the length of a meter that potentially can vary. And what does *that* mean? It varies relative to what?

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Re: [math-fun] speed-of-light-may-not-be-constant?
by Keith F. Lynch 28 Mar '13

28 Mar '13

I'm not sure what this would even mean. Since 1983, SI has defined the length of the meter in terms of the speed of light and the length of the second. So the speed of light in a vacuum as a constant by definition. Before that, the meter was defined in terms of a wavelength emitted by krypton. The second was then, as it is now, defined in terms of a frequency emitted by cesium. So the constancy of the speed of light then depended on whether the relative behaviors of those two elements varied. Before that, the meter was defined as the length of a specific object in Paris. If the speed of light seemed to vary then, it could be that the size of the atoms in that object varied. Farfetched? Not really. The sizes of atoms depend on the values of various fundamental physical constants, including the speed of light. I think it's only meaningful to speak of changes in dimensionless numbers. For instance the fine structure constant, whose value is the same in all possible systems of units. It depends on the speed of light, Planck's constant, and the charge of the electron. It's meaningful to talk about the fine structure constant changing. If it does change, I think it's completely arbitrary whether that change is attributed to a change in the speed of light, Planck's constant, the charge of the electron, or some combination of these. (On the other hand, it seems to me that if Earth were to suddenly shrink, apparent gravity would drop as the shrinkage began, and would increase when it ended. Obviously, there would be no perceived change in gravity if Earth "shrank" only due to a redefinition of the meter. (The numerical value of the gravity might change due to redefinition of the meter. But readings on scales would remain unchanged until the scales were recalibrated or replaced.)) Can anyone find an online copy of the original study that makes this claim? From the news coverage, I can't even tell whether they claim to have measured an effect, or whether this is entirely theoretical. Warren D Smith <warren.wds(a)gmail.com> > Does this contradict the observed fact that light from halfway > across the universe, hitting the left side of your telescope mirror, > is still perfectly in phase with light hitting right side of your > telescope mirror, allowing interferometry? No, so long as the speed varies smoothly with time and location, rather than there being any abrupt transitions. Exactly as happens with curvature of space-time due to gravity. In fact, this alleged variability would be indistinguishable from such curvature. Indeed, it's arbitrary to say that space-time is curved near masses rather than to say that c is slower there and that objects there are shrunken and slowed by just enough that they don't notice the slowdown. One possible difference is that space-time curvature near masses is always equivalent to slower light, not faster. If this new effect allows light to go faster than c, it might allow causality violations, i.e. a way to send messages back through time. On second thought, space-time curvature near *negative* masses would also be equivalent to faster light. Maybe that's why there are no negative masses. Unless the Casimir effect counts. That's a very weak effect, and probably any speedup of light due to the negative mass-energy between the plates would be swamped by the positive mass-energy *of* the plates.

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