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16 June 2009

Slashdot: 47th Mersenne Prime Confirmed

The new prime, 2^42,643,801 - 1, is actually smaller than the one discovered previously.

18 May 2009

Primality of 1

When I was in school, I remember them teaching that a prime number was any number divisible by only 1 and itself. Thus, in school, they taught that 1 was prime. Since then, I have noticed that everyone insists that 1 is not prime. Since many of the patterns I find work better with 1 being prime, I decided to try to figure out why the discrepancy...

From Wikipedia:

Until the 19th century, most mathematicians considered the number 1 a prime, with the definition being just that a prime is divisible only by 1 and itself but not requiring a specific number of distinct divisors. There is still a large body of mathematical work that is valid despite labelling 1 a prime, such as the work of Stern and Zeisel. Derrick Norman Lehmer's list of primes up to 10,006,721, reprinted as late as 1956,[2] started with 1 as its first prime.[3] Henri Lebesgue is said to be the last professional mathematician to call 1 prime.[citation needed] The change in label occurred so that the fundamental theorem of arithmetic, as stated, is valid, i.e., “each number has a unique factorization into primes.”[4][5]
So there you have it. They decided in the last 43 years that 1 could no longer be prime so that one of their theorems could be correct. Why does that remind me of someone fudging statistics?

I know, I know... soapbox, right? First the divide-by-zero post, then this... How about this, as soon as we can explain all our math in simple fractals without saying things like "except 0 or 1" or "with n>1", then I'll change my position on it.

Thus Theorem 1 of Hardy & Wright (1979) takes the form, “Every positive integer, except 1, is a product of primes”
bah humbug ;)

17 May 2009

Prime digits

A coworker made a comment to me Friday about the distribution of digits in prime numbers. He was referring to a recent Slashdot article talking about the first digit in the prime number...

but I have never liked thinking of the 1st digit as the far left... I'd prefer to think of the far right digit as digit 0. Can't help it - it's the whole positional notation thing (10^3 10^2 10^1 10^0 . 10^-1 10^-2 etc)...

So I started looking at the distribution of the primes on the right side.

The last digit (digit 0, far right digit) is directly related to the number of primes...
Now 2 and 5 are both used once (for themselves).
0,4,6 and 8 are never used.
1,3,7,9 all follow this pattern:

Assume {x} = {1,3,7,9}
Assume {y} = {all primes < 10^n}
let z = ({y} ≡ x (mod 10^n))
z is also the number of times that x was the last digit in {y}

So let's look at a simple example...
{y} = {all primes < 10000}

0 was never the last digit
1 was the last digit 307 times (if you count 1 as prime, 306 otherwise)
2 was the last digit once ('2')
3 was the last digit 310 times
4 was never the last digit
5 was the last digit once ('5')
6 was never the last digit
7 was the last digit 308 times
8 was never the last digit
9 was the last digit 303 times

there were 307 primes ≡ 1 (mod 10000) [assuming 1 counts as prime]
there was 310 primes ≡ 3 (mod 10000)
there was 308 primes ≡ 7 (mod 10000)
there was 303 primes ≡ 9 (mod 10000)

For {y} < 100000, we have
0 2388 1 2402 0 1 0 2411 0 2390

For {y} < 1000000, we have
0 19618 1 19665 0 1 0 19621 0 19593


Checking OEIS, I found those sequences listed here:
A073505 for 1 (if 1 is *NOT* prime)
A073506 for 3
A073507 for 7
A073509 for 9

A006880 seems to be the SUM of each line
Example: 0+307+1+310+0+1+0+308+0+303=1230 [-1 for the sequence due to 1 being prime or not] < 10^4

So it doesn't really get us much... but it does tell us that the number of primes ending in X is equal to the number of primes congruent to X mod 10^n.

27 April 2009

Division by Zero

I've been on this kick lately where I have started question some of the basic assumptions I learned in math... things like "you can't easily define pi" or the concept of an "irrational number". Today's philosophical debate is on the idea of division by zero.

This came up because I was creating a ContinuedFraction interface for integers... The number 3, for instance, is 3 + 0/0 + 0/0 .... etc... instead of zeros I had to replace them with nulls in order to prevent the app from trying to compute the 0/0 bit... That got me thinking... maybe it should be allowed...

Let's take a simple concept. Let's say there are 10 pennies on the table, and it is decided that the pennies will be split evenly amongst everyone at the table. [Yes, pennies - I don't want anyone getting the bright idea to make change].

Now, if there are 5 people, everyone gets 2 pennies. Simple.

If there are 3 people, everyone gets 3 pennies and 1 is left on the table.

If there are no people, then there is still 10 pennies on the table.

What's so difficult about that?

So, in essence:
10/5 = 2r0
10/3 = 3r1
10/0 = 0r10
Seems like Occams Razor would say that the whole concept has been arbitrarily inflated to give us errors, exceptions and NaN.

What about the other way around?

If there are 0 pennies...
0/5 = no pennies for anyone
0/3 = same
0/0 = same

Seems like we really don't have an issue there.

Of course, this would mean that things such as limits get screwed up -- but so what. If we think we have irrational numbers, why not stir the pot a bit? Maybe they were not correctly defined in the first place.

Besides, that will open up the ContinuedFractions to more interesting options... like

26 April 2009


While working on my ContinuedFractions library today, it occured to me that our current representation of pi could be simplified by just specifying the numerators and denominators as series...

I went ahead and changed the implementation to use a new SeriesCF class instead of the default ContinuedFraction class and I think the new approach is much cleaner.

25 April 2009


While writing this package today, I came across the PresenTeX website. Very nice...

cf = d_0 + \cfrac{n_0}{d_1 + \cfrac{n_1}{d_2 + \cfrac{n_2}{d_3 + \cfrac{n_3}{\ddots\,}}}}
I got:

10 February 2009

Continued Fractions

I was working on some continued fractions, and came across the fundamental recurrence formula above. It worked pretty well for some simple continued fractions (like √2) but seemed to be a bit unfriendly towards calculating π.

I'm using this definition of π:

Now, using that definition with the above equation, it was always completely wrong...

So I tried changing the equation...

So you'll notice a few specific changes... 1) A1 has been redefined in terms of previous results. 2) A1 reduced the index on a1 to a0. 3) An+1 changed the index on an+1 to an. 4) Bn+1 changed the index on bn+1 to bn.

This may seem like a lot of changes, but really #1 didn't change the outcome at all and the rest were all about reducing the index of a/b.

Net result? All the previous stuff (√2, golden mean, etc) give the same results as before... but now π works.