Re: Mersenne: re: Mersenne prime exponent binary
On 12 Jul 99, at 17:45, Lucas Wiman wrote: That's the point of Benford's law, it is supposed to be relatively independent of the set of numbers. ... within reason ? If I take the (decimal) powers of 0.999 and get bored after 100 trials, I find they _all_ start with a 9 ;-) Note that in the set of mersenne prime exponents (so far), the leading digit 1 (in decimal), turns up 10 times as opposed to the 4.2 times expected by equal leading digit distribution... Actually we should expect an excess of smaller leading digits over that predicted by "Benford's Law" in this case. A smaller exponent is more likely to be prime than a larger exponent, and a smaller prime exponent is more likely to give rise a Mersenne prime than a larger prime exponent. "Benford's Law" would follow if _every_ exponent (prime or composite) was equally likely to give rise to a Mersenne prime. [Different message, same author] Yes. Though they were talking about the exponents... Weird, I would have thought that it wouldn't affect powers of two... Why not? Looks like a perfect model to me! Regards Brian Beesley Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: Benford's law (was exp. representations)
On Tue, 13 Jul 1999, Lucas Wiman wrote: So for numbers 2^n (in Base 10), [or is it 2^p?] there are a lot more leading ones than one would "expect" naievely (you would expect 1/9 to start with "1", I imagine). Yes. Though they were talking about the exponents... Here are the percentages for the first 3000 powers of 2. The first collumn is the percentage, the second is the difference from the predicted Benford percentage. Weird, I would have thought that it wouldn't affect powers of two... .30110036678892964321 .7037112494844799 .17639213071023674558 .0003008716540349 .12470823607869289763 -.00023050052960705550 .09703234411470490163 .00012233110664848727 .07935978659553184394 .00017854054790701622 .06702234078026008669 .7555114964688849 .05768589529843281093 -.00030605167925394399 .05168389463154384794 .00053137218416255899 .04534844948316105368 -.00040904107751407172 Benfords Law is a direct consequence of the fact that most sets of numbers obtained by actual measurements are either resulting from exponentially distributed data (eg. river lengths), or come from uniformly distributed numbers, gathered from multiple ranges of exponentially distributed length (eg. house numbers). Since powers of numbers are just about the cleanest exponentially distributed set of numbers you can get, it shouldn't really come as a surprise that they fix the law:) -- Henrik Olsen, Dawn Solutions I/S URL=http://www.iaeste.dk/~henrik/ Leonardo DiCaprio: Your social class is stuffy. Let's dance with the ship's rats and have fun. Kate Winslet: You have captured my heart. Let's run around the ship and giggle. (The ship SINKS.) Leonardo DiCaprio: Never let go. Kate Winslet: I promise. (lets go) Titanic, the Movie-A-Minute version Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: Benford's law (was exp. representations)
At 12:38 AM 7/13/99 -0400, Lucas Wiman wrote: Here are the percentages for the first 3000 powers of 2. The first collumn is the percentage, the second is the difference from the predicted Benford percentage. Weird, I would have thought that it wouldn't affect powers of two... That's the type of thing that follows the law precisely in the long run. (the repeated multiplications). A way to look at this is to think of a slide rule (if you remember them). Anything that has a uniform distribution on the slide rule follows Benford's law. The distance from 1.0 to 2.0 on the slide rule is 0.301... of the length of the scale, etc. A repeated multiplication by a number other than 0 or 1 gives a uniform distribution along the scale of the slide rule. +--+ | Jud "program first and think later" McCranie | +--+ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: re: Mersenne prime exponent binary
Note that in the set of mersenne prime exponents (so far), the leading digit 1 (in decimal), turns up 10 times as opposed to the 4.2 times expected by equal leading digit distribution... Actually we should expect an excess of smaller leading digits over that predicted by "Benford's Law" in this case. A smaller exponent is more likely to be prime than a larger exponent, and a smaller prime exponent is more likely to give rise a Mersenne prime than a larger prime exponent. "Benford's Law" would follow if _every_ exponent (prime or composite) was equally likely to give rise to a Mersenne prime. But that's part of the interesting thing... the size of the exponent is only vaguely associated with the LEADING digit. I disagree that we'd expect smaller leading digits, at least noticeably many, since it's the order of magnitude, not just the leading digit that makes the nth mersenne prime larger or smaller. I mean M20 is some 50 digits longer than M19... at this distance, I don't see how the leading digit is affected by the larger likelyhood that smaller exponents would make more likely primes. But I'm probably wrong. This is what makes this a really interesting fact, tho, I guess. Yes. Though they were talking about the exponents... Weird, I would have thought that it wouldn't affect powers of two... Why not? Looks like a perfect model to me! In some vague attempt to not take the Benford issue off topic, it's interesting that numbers 2^n (for all Natural numbers n) follows the pattern VERY closely, but it's also interesting (perhaps moreso), that 2^p follows the pattern, as do (apparently) the Mersenne primes themselves, as do (from quick examination) the exponents for the mersenne primes. Thanks for adding this to the FAQ, btw. ---Chip \\ ^ // (o o) ---oOO--(_)--OOo | Chip Lynch| Computer Guru| | [EMAIL PROTECTED] || | (703) 465-4176 (w) | (202) 362-7978 (h) | Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Mersenne: Hyperbola
I've noticed that with any odd number you can make the formula x^2 - y^2 = n where n = the odd number and x - y and x + y are factors of n. I was just wondering if one could use the graph of a hyperbola to see only the possible integer values of x and y. _ Do You Yahoo!? Get your free @yahoo.com address at http://mail.yahoo.com Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: re: Mersenne prime exponent binary
At 09:05 AM 7/13/99 -0400, Chip Lynch wrote: In some vague attempt to not take the Benford issue off topic, it's interesting that numbers 2^n (for all Natural numbers n) follows the pattern VERY closely, In the limit as n - infinity, 2^n must follow the law exactly. Almost by definition. +--+ | Jud "program first and think later" McCranie | +--+ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: re: Mersenne prime exponent binary
Brian Beesley wrote: Actually we should expect an excess of smaller leading digits over that predicted by "Benford's Law" in this case. A smaller exponent is more likely to be prime than a larger exponent, and a smaller prime exponent is more likely to give rise a Mersenne prime than a larger prime exponent. "Benford's Law" would follow if _every_ exponent (prime or composite) was equally likely to give rise to a Mersenne prime. This is not true. Actually, it is the fact that smaller primes are more likely to give Mersennes that theoretically should result in a "Benford's Law" type behavior of the second leading bit. It is in some sense an "accident" that Mersenne exponents SHOULD follow Benford's Law (at least the second bit generalization of it), and an irony that, due to small number statistics, they actually DON'T! (68% zeroes instead of predicted 58% or whatever) Benford's Law comes about because of power law scaling of some numbers. Many of the referenced web links emphasized that Benford's law is NOT for "regular" numbers, but ONLY for numbers expressing amounts in some (human selected) units, and that it is some property of "power law scaling" and/or "logarithmic invariance" of arbitrary choice of units to express AMOUNTS (NOT numbers) of things that result in Benford's law. As has been dealt with in many many recent posts regarding the density of Mersenne exponents, there is an expected (and observed) uniform density of Mersennes in LOG space, ie. equal numbers of Mersennes per factor of two in candidate space of about 1.5 or 1.48 or whatever. It is this logarithmic scaling or invariance that theoretically SHOULD result in Benford Law type behavior. It is ONLY this decreasing likelihood of primes to generate Mersennes that should cause Benford law behavior, not a reason for deviation from it. In fact, if the likelihood of numbers generating Mersennes followed some other law, say decreasing faster than logarithmically, then Benford's law might not apply, or only approximately. This thread has been very interesting in it's own right mathematically, but it's only "accidental" in some sense that Mersenne exponents should follow the law, and as I said before, ironic that it doesn't! The fact that we KNOW (or believe) the logarithmic scaling of Mersennes allows us to bypass all of the heuristic arguments about logarithmic scaling of human selected units for expressing quantities, and to directly DERIVE the law for Mersennes, as opposed to heuristically generating it for the other cases. Todd Sauke Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: Benford's law (was exp. representations)
Steven Whitaker wrote: Maybe it's my imagination, but it seems to me that the factors of the prime exponent Mersenne numbers start with a 1 more often than a 2 or 3 etc. Are they obeying Benford's law too? For instance, for the 10 primes from 5003 to 5081, there are 20 known factors. 10 of them start with a 1. Something similar. The Benford's law distribution works because we 'expect' natural, boundless, data to have the decimal part of the logarithm "fairly uniformly" distributed, and a quick look at a slide rule (younger readers, ask your Dad!) has 30.103% of its length with initial digit 1. By Merten's theorem, the probability *any* large number N has no factor smaller than X is C/log X, C is exp(-gamma) if I remember rightly. (strictly, X has to be much bigger than 1, and much smaller than sqrt(N) for this to make any sense). By that sort of estimate, suppose N has a factor F between 10^k and 10^(k+1). Then the probability that F begins with a 1 is something like [1/k-1/(k+0.30103)]/[1/k-1/(k+1)] which tends to log10(2) as k tends to infinity - so the distribution does approach Benford's law. In fact, if you plot the distribution above, it's more generous to 1's for smaller factors. It seems a skewed distribution, but remember it's based on our insistence of observing the world base 10, and believing that 2-1 is "just as important" as (M38)-(M38-1). The same observation is repeatable in any base - of course, at the most ridiculous, ALL factors begin with a 1 when written in base 2. Chris Nash Lexington KY UNITED STATES = Still a co-discoverer of the largest known *non*-Mersenne prime Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
