A strange idea......

[Maybeso]

Wow!
I am very surprised this is the first time I've ever heard about this curious pattern. Cheesehead has covered most of the explanation, so I will pick up where he left off.

Originally Posted by Cheesehead

Fourth: Why do the digits in the cycle add to 9 pairwise first-half-to-second-half?

Here is the hand-wavy proof that cheesehead almost came up with for when k is even.
(Since I have hands, and even an extra prime hand over there <--- ;) )

For those interested in references, I did a google on "digit patterns repeating decimal", and one of the many links was
http://www.mersenneforum.org/attachments/pdfs/422.pdf
Which contains a specific example of this general statement.[code:1]If 1/b has a decimal pattern of length k, then b must divide 10^k - 1 = 99999. If, in addition, b is a prime number, then k must divide b - 1.[/code:1]
Now, if 1/b has a decimal pattern of length 2k, then b must divide 10^2k - 1 = (10^k - 1)(10^k + 1).
When b = 7, 1/7 = 0.142857... a pattern of length 6 ==> 7 divides 10^6 - 1 = (10^3 - 1)(10^3 + 1) = (999)(1001)

1000/7 = 142.857 142 857 142 857 142
+ 1/7 = 000.142 857 142 857 142 857
-------------------------------------
1001/7 = 142.999 999 999 999 999 999
143 = 143

And this will hold for any 1/b with a repeating decimal pattern of even length 2k, indicating b | 10^k+1 rather than 10^k-1. (If b | 10^k-1, then the pattern would be length k.)
That is what your pattern indicates -- not that b is prime, but that b divides 10^k+1.

For example,
1/7 = 0.142 857... --> 7 | 1001
1/13 = 0.076 923.. --> 13 | 1001
1/91 = 0.010 989.. --> 91 | 1001
91 = 7*13 is not prime.

And in the case of 49,
1/7 = 0.142 857... --> 7 | 1001
1/49 = 0.020408163265306122448 979591836734693877551... --> 49 | 10^21+1.
So 7 is a factor of 1001, but 49 = 7*7 is not.

Maybeso

[epatka]

Thank you, Maybeso,

I guess this means I will have to find another way to become famous :)

Out of luck today.....

Let's get back to my Mersenne Prime testing. It says, my completion date will be June 5 23:21 2005. Is this a bug or is it going to take 2 years? This is my second testing, so I don't know.

[eepiccolo]

Originally Posted by epatka

Let's get back to my Mersenne Prime testing. It says, my completion date will be June 5 23:21 2005. Is this a bug or is it going to take 2 years? This is my second testing, so I don't know.

Either your computer is only running for a very small part of the day, or you are testing an exponent around 33,000,000 with a processor like a Pentium II. If you have a slower processor, you may just want to do Double Checks. You can select what kind of work you want in the Primenet menu under the Test menu in Prime95.

[epatka]

You are right!
I am testing a 33,000,000 on a Pentium II :)

The first number I tested took about a month and then it said it is a composite. This one took one month so far, 2 days ago it said "no factors". I thought it was done and the number is a prime. But I've never been lucky in anything, so I wasn't too concerned about it.
I knew there was a catch :D

I was right. It is going to take another 2 years......

:(

[eepiccolo]

Originally Posted by epatka

The first number I tested took about a month and then it said it is a composite. This one took one month so far, 2 days ago it said "no factors".

Yeah, that means that the first number you tested had a factor that the trial factoring found. Because a factor was found, the program knows the number can't be prime, so a Lucas-Lehmer (LL) test was never even preformed on that number. But on your current number, trial factoring didn't find a factor, so it has to preform the LL test to see if the munber is prime or not. Oh well, it sounds like you're a patient person, anyway. ;)

[epatka]

What makes you think I am patient? :)

Is there a way of speeding it up? I updated yesterday to the latest version. That means it will take 25% less time....... I might have it by next summer......

[jeff8765]

The only way you can really speed it up is to use a faster computer or get a smaller number. It will take a very long time to test a 33M number on a Pentium II no matter what you do.
I hope your number is prime :D

[jocelynl]

If you want it to go really fast then you might want to join the Lone Mersenne Hunter`s. That way you`ll find a hole bunch of factors in no time. You won`t find prime numbers that way but you`ll play a major part of the search.

Joss

[cheesehead]

Originally Posted by epatka

I was worried I said someting very dumb or broke some unwritten rules, because there was no reaction to my post.

It was just unfortunate timing. There were squabbles in other forum threads that drew attention. There was Fathers Day. And after I first read your thread, I wasn't ready to post a response until a day later.

Of course it doesn't change that you are right in this case.

Just trying to put you at ease. :)

a division is nothing but substructions. Of course, in this case it woul mean 10 million substractions of 2 10 million digits long numbers, probably 10 million times

Mathematicians have come up with ways to multiply and divide large numbers that are much, much faster than the methods we all learned in school.

But it's always possible to find a number large enough to make arithmetic as big a task as you want.

I noticed that when I do the division by hand, there is always a point, where the remainder is 1. And then the whole cycle starts all over.

That's the point where you've found the k for this p !! When the remainder is 1, you've found the k such that 10^k-1 is divisible by p.

Question is, if this whole thing is any help in understanding or recognizing prime numbers???

Well, since division is the inverse of multiplication, it seems likely that any statement or theorem about prime integers has a corresponding statement or theorem about the inverses of primes.

[code:1]ps. sorry, I coudn't figure out how to use the "quote" function.[/quote][/code:1]
Next time you're about to post something, when you're at the form for entering your post, look at the lower left for "BBCode is ON". Click on "BBCode", and you'll get a help page about all the text formatting commands.

[cheesehead]

Originally Posted by Maybeso

Now, if 1/b has a decimal pattern of length 2k, then b must divide 10^2k - 1 = (10^k - 1)(10^k + 1).

Similarly, if 1/b has a decimal pattern of length 3k, then b divides 10^3k - 1 = (10^k - 1)(10^2k + 10^k + 1).

In general, if 1/b has a decimal pattern of length jk, then b divides 10^jk - 1 = (10^k - 1)(10^(j-1)k + 10^(j-2)k + ... + 10^k + 1).

[epatka]

[quote]

Originally Posted by jocelynl

If you want it to go really fast then you might want to join the Lone Mersenne Hunter`s. That way you`ll find a hole bunch of factors in no time. You won`t find prime numbers that way but you`ll play a major part of the search.

Thanks Joss, but my computer is so slow that I wouldn't be much help.... Besides, I like the suspense that someday.... sometime..... in 2005 I might win some money. It makes it more interesting. Although by the time I am finished with my number, someone else will have discovered a 10 million digit prime.

I haven't chosen this number. The program started it automatically. What if I increase the memory available for the program? I have 256 MB and there is lots of room.......

Eva