RE Being a little less guarded!
- I haven't had the time to think much about it (I have a lot of work from University
classes!) but I can say that we can restate your problem as follows:
Given a prime p, look if all the numbers from 0 to p$-1 can be originated by the formula f
= IntegerPart(p$*k / R), where p$ is the product of the odd prime numbers up to p, k is
any integer and R is so that GCD(R,p$)=1, R < p^2. Note that R can be any number with
factors of 2 and factors of every prime number between p and p^2 (caring that R < p^2).
We can reformulate the problem like that since:
a) Every "chain" will have its numbers separated by multiples of 3 (all of them have the
same modulo in base 3), since the worse case is that with two 1s in p=3 (110,101,011).
b) We take a combination of Ca = (p-2)*(q-2)*... numbers from the chain, and this
combination can be any of the C(p$, Ca) possible combinations.
c) All the chains and combinations are possible, so we need a R to match every number from
0 to p$-1.
I don't think it is a trivial problem anymore. I'm not sure if it is as hard as GC, or if
it is easier, though.
Another simple observation:
p$ / R_i = c + r / R_i with 0 < r < R
Since GCD(p$, R_i) = 1, then GCD(r, R_i) = 1 and GCD(r, p$) = 1 and so, ( GCD(r,p$1) > 1
or GCD(r,R_j) > 1 (i<>j) or r = 1).
I don't think it leads anywhere, but maybe you find the condition useful for something.
----- Original Message -----
From: "chrisdarroch" <chrisdarr2@...>
Sent: Sunday, April 24, 2005 12:14 PM
Subject: [PrimeNumbers] Being a little less guarded!
Since you guys seem to be looking at my question: as a measure of my
respect for that, I shall give a little more.
Each data set of 1's and 0's refers directly to a particular even
e.g. for m = 68 we have.
For m = 66 we have.
There can be more than one data set for each even number, e.g.
for m = 68 we have.
Each data set uses a specific RelativePrime term.
For the first and second datasets it is 31 and for the third it is 29.
The RelativePrime term has a maximum at around m/2 and a minimum at
around m/4 or may be somewhat lower.
I will find that throughout the dataset there are columns which
contain all 0's.
It is the columns immediately prior to those that I am interested in.
I have called these columns A or n in previous posts (apologies).
Let me call them A.
For Dataset 1
There are A's at columns 8, 17, 29, 32, 38, 47 etc as these are
columns immediately prior, to columns containing all 0's,
Can any of those A's fit the equation A = IntegerPart [105*n/31],
where n is some whole number divisor, which is unknown?
I have used RelativePrime = 31 in this case, as it is within the range
for RelativePrime, for even number m = 68
By observation, I can see that there is such an A for that specific
Dataset 1, at column 47.
At this column 47 = IntegerPart [105 * 14/31]
For Dataset 2, I will use 31 again and for data set 3, I would use
29 as the RelativePrime term.
I need to find the general case, that given a dataset with the
1. Each row has a repeating pattern of 0's and 1's with period equal
to the prime at the head of that row.
2. There are no more than two 1's in each of those periods.
And given the RelativePrime term is limited as above.
And given that the PrimeProduct can only include the odd primes (thus
it should be called the OddPrimeProduct); is calculated by taking the
squart root of m and finding only the odd primes lower than it.
Is it true that I will find an A in every dataset (given the dataset
limitations) which will fit the formula
A = IntegerPart[OddPrimeProduct/RelativePrime]
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