10 is known to be a generator mod 7^m for any m. This is an infinite family of FPD generators mod 10.

It's true that 10 is a primitive root mod 7^m. But if I understand the definition of FPD correctly, we're looking for a number n such that 1/n has a period of n-1 digits. For this purpose, n must be prime. 1/49, for example, has a period of 42 digits.

You also write:

I think a similar family could be constructed for any base n.

Maybe so, but I think that's still an open conjecture.

Unfortunately, I can't now get a handle on the problem 'are there infinitely many FPD generators to base 10'
I seems like there ought to be, but I can't prove it.
Is it a standard result?

The same comments apply to a general base n (except the question of whether it is a standard result of course: We know it to be open here)

Thankyou

sfwc
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Another note I might add concerns what I said: "It can be proven that a periodical decimal with period n, if multiplied by an integer k, will result into another periodical decimal with period n as well, though not necessarily the same sequence of digits. At the case of 2/7 we have: 0,285714..."... well that is valid for any integer k other than a multiple of b, being b the denominator of the FPD, as it'seems clear enough.

Well, sfwc, I still don't get the idea of "generator modulo". Because I'm brazilian, I'm not used too much to the english specific terms of math... I guess "modulo" is the operation that gives the remainder of a division of two terms, but I don't know enough of "generator modulo". Please clear me out this if you don't mind.

Now, up to some other questions:

In fact, I can't prove what I mentioned above, it's just an intuitive view to me. How can you prove (1) that n and b must be coprime?

Also, I tried to prove (2):

Let a/b be a FPD. It can be written as: 0,(stuff)(stuff)...
Let d be the number of digits of "stuff"

a/b * n^d-1 = (stuff),(stuff)(stuff)... - 0,(stuff)(stuff) = (stuff)

No prob till now.

So, thats equivalent to: a/b * n^d-1 = (a * n^d-1)/b = stuff

But stuff is an integer, so (a * n^d-1)/b must be reductible to k/1.
In order to do this, b must be a factor of (a * n^d-1). If we let "a" be the cannon numerator which is only a multiple of itself (i.e. 1), so we have that b must be a factor of n^d-1. OK.

But accordingly to the definition of FPD, d = b-1, so we have:

b ~ n^d - 1
b ~ n^(b-1) - 1
b ~ (n^b / n) - 1

or

b ~ n^d - 1
d+1 ~ n^d - 1

And I´m stuck into that. How do you get at n^d ~ 1 mod b?

Finally, when I ask "Are there any number that is a FPD generator in any base?" I guess I miswrote, I mean, is there some integer k that is a FPDgen in every base? We know 2 isn't a FPDgen at base 10 (it's true that the number of decimal digits of an a/b rational, with b not ~ 2, is 2-1 = 1, but the decimal is not periodic).

Hope I didn't brought you too much work.

I would like also to suggest a program algorithm to test if a number is a FPD generator at a given base.

Thanks again

Zakatos

>How can you prove (1) that n and b must be coprime?
Imagine (as you did earlier) performing the long division. let b and n have common factor k. Then eventually at any stage you are subtracting a number divisible by k (because it is a multiple of b) from another number also divisible by k (since it's last digit is 0 in base n). So the remainder which you cary at any stagemust be divisible by k. This restircts the number of possible remainders, and so possible length of the period, by a factor of k. So if k>1 then the period cannot be as large as b-1. It can only reach b/k - 1.

>How do you get at n^d ~ 1 mod b?
Well, if b is prime, then this is true whenever n is not divisible by b (FLT). We also require that b - 1 be the LEAST value of h such that n^h ~ 1 mod b. By fermat-euler, n^f(b) ~ 1 mod b. But f(b) <= b - 1 with equality iff b is prime (this is why I felt so dumb). Finally, the remaining condition is precisely equivalent to 'n is a generator mod b'

>Finally, when I ask "Are there any number that is a FPD generator in any base?" I guess I >miswrote, I mean, is there some integer k that is a FPDgen in every base? We know 2 isn't a >FPDgen at base 10 (it's true that the number of decimal digits of an a/b rational, with b >not ~ 2, is 2-1 = 1, but the decimal is not periodic).
Well, evidently no b is an FPDgen to the base b, or to any base with a common factor with b.
So we restrict this to the more interesting question 'is there a b which is an FPDgen to any base coprime to b'. 2 has this property, but no other number does. For no b can be an FPDgen mod b+1 except 2.

>Hope I didn't brought you too much work.
It's ok. it was interesting.

>I would like also to suggest a program algorithm to test if a number is a FPD generator at >a given base.
ok. Outline structure: First test b for primality. In a loop calculate the powers of n mod b up to n^(b-1)/2. If none of them are 1, then b is an FPDgen mod n. Otherwise, it isn't. I guess there must be a quicker algorithm, but I don't know it.

Thankyou

sfwc
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