Re: Mersenne: re: Mersenne prime exponent binary

1999-07-13 Thread Brian J. Beesley
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

Re: Mersenne: Benford's law (was exp. representations)

1999-07-13 Thread Henrik Olsen
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

Re: Mersenne: Benford's law (was exp. representations)

1999-07-13 Thread Jud McCranie
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

Re: Mersenne: re: Mersenne prime exponent binary

1999-07-13 Thread Chip Lynch
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

Mersenne: Hyperbola

1999-07-13 Thread Kris Garrett
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.

Re: Mersenne: re: Mersenne prime exponent binary

1999-07-13 Thread Jud McCranie
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.

Re: Mersenne: re: Mersenne prime exponent binary

1999-07-13 Thread Todd Sauke
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

Re: Mersenne: Benford's law (was exp. representations)

1999-07-13 Thread Chris Nash
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