An introduction.

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• Greetings to all. First, allow me to introduce myself. I am Bryan, currently living near Brisbane, AUstralia. I had been for the last 5 or so years working
Message 1 of 3 , Dec 4, 2008
Greetings to all.

First, allow me to introduce myself. I am Bryan, currently living
near Brisbane, AUstralia.

I had been for the last 5 or so years working independantly with prime
numbers, for no other reason than it gives a mental challenge for me
to do so.

I have, after this time, came up with a solution to a problem I
started with, mainly a better way of testing and finding prime
numbers.

Now, my methods may have already been discovered and published,
however, i feel a bit of pride about coming to this solution using my
own intuition.

catalyst for my research.

For the following, I will be dealing with Pasqual's triangle, with the
following conditions

Given: Pasquels Triangle where each row x is sequentionally numbered
from 0 from top to bottom, and each column y is sequentionally
numbered in each row from 0 from left to right, each cell can be
identified by [x,y]. #

With this triangle, the following can also be assumed
[x,0]=1, [x,x]=1, and y<=x

The breakthrough for me was the realization that prime numbered rows
have a UNIQUE feature that sets it apart from all other rows...
While 1 < y < x, [x,y]/x is an integer. Simply put, a prime row has
all cells in that row (other than the first and last cell, which are
1) that are multiples OF that row number.

This allowed an ability to test for prime numbers using pasquals
triangle. of coarse, when dealing with large numbers, you may not
have the values of that row handy.

Thus the formula can be adjusted for calculating the row as you test
the prime. This is done by replacing [x,y] with the calculation for
finding that [x,y], which is (x-y+1)/y*[x,y-1], which simply is a
formula that is dependant on the knowladge of the previous number.
since we know that [x,0] is always 1, we can calculate for every value
of y in row x.

To summarize where we are at, while 1 < y < x, x is a prime number if
(x-y+1)/y*[x,y-1]/x is an integer for all integer values of y

Stopping here would be an acomplisment, but the number of tests
required for each prime number gets larger. The second step was to
try to cut down the number of calculatons.

There are two ways of doing this. 1, Pasquals triangle only needs
HALF of the columns calculated, because it is symmetrical. 2, the
largest number needing to test is the square root of x, which is a
common method of testing smaller numbers for primeness.

While exploring the Triangle, another pattern developed. Rows would
fail the prime test at columns that were prime numbers, OR multiples
of primes. In fact, the prime numbers they failed at were in fact one
of its factors!

This means that the most that a row needed to be tested was row
numbers less than the square root of x, and only the prime number
rows.

This took most of the time to reach a conlusion. One these truths
became evident, the formula built itself. For the most minimum of
calculations, the order of primes up to square root of x needs to be
known, but a good computer can begin calculating those from 1. a php
program i build could do it up to the mathematical limitations of php
(row 16,000 tends to get up to the 10^300 level) in about a miniute,
calculating the first 2000 or so primes and verifying htem.

here it is, completed, but i dont have the math skills to do a proper
proof on it.

Given: Pasquels Triangle where each row x is sequentionally numbered
from 0 from top to bottom, and each column y is sequentionally
numbered in each row from 0 from left to right, each cell can be
identified by [x,y]. [x,0] = 1, [x,x] = 1, y <= x

Postulate: P(1) = 2, P(2) = 3 (first two primes are 2 and 3)

Theorem:
While P(n)^2 < x < P(n+1)^2,
If order of primes is known, and row x of Pasquels Triangle is known,
Then y is all prime integers 1 to Pn
x is a prime number if [x,y]/x is an integer for all values of y

If order of primes is known, and row x of Pasquels Triange is unknown,
Then y is all prime integers 1 to Pn
x is a prime number if (x-y+1)/y*[x,y-1]/x is an integer for all values of y

If order of primes is unknown
Then y is all integers 1 to Sqrt(x)
x is a prime or if (x-y+1)/y*[x,y-1]/x is an integer for all values of y

This has proven true findint all primes up to number 16132.
I can not use the program to auto verify numbers above that due to the
large numbers row 16133 and higher reach at about the 118th column

THoughts, comments? has someone beaten me to the punch?
--
Bryan Bartlett
• ... I remember way back in high school coming across that result when we were studying the binomial expansion and Pascal s triangle. So yes you have been
Message 2 of 3 , Dec 4, 2008
--- In primenumbers@yahoogroups.com, "Bryan Bartlett" <valareos@...> wrote:

> Given: Pasquels Triangle where each row x is sequentionally numbered
> from 0 from top to bottom, and each column y is sequentionally
> numbered in each row from 0 from left to right, each cell can be
> identified by [x,y]. [x,0] = 1, [x,x] = 1, y <= x
>
> Postulate: P(1) = 2, P(2) = 3 (first two primes are 2 and 3)
>
> Theorem:
> While P(n)^2 < x < P(n+1)^2,
> If order of primes is known, and row x of Pasquels Triangle is known,
> Then y is all prime integers 1 to Pn
> x is a prime number if [x,y]/x is an integer for all values of y
>
> If order of primes is known, and row x of Pasquels Triange is unknown,
> Then y is all prime integers 1 to Pn
> x is a prime number if (x-y+1)/y*[x,y-1]/x is an integer for all values of y
>
> If order of primes is unknown
> Then y is all integers 1 to Sqrt(x)
> x is a prime or if (x-y+1)/y*[x,y-1]/x is an integer for all values of y
>
> This has proven true findint all primes up to number 16132.
> I can not use the program to auto verify numbers above that due to the
> large numbers row 16133 and higher reach at about the 118th column
>
> THoughts, comments? has someone beaten me to the punch?
> --

I remember way back in high school coming across that result when we were studying the
binomial expansion and Pascal's triangle. So yes you have been beaten to the punch
regarding the fact that all interior numbers of Pascal's triangle on a prime row p are
multiples of the prime p. Regarding whether the converse is true, I would suspect so.
That is, given a row whose interior numbers are all multiples of that row number, is the
row number necessarily prime? The formula is n!/(k!*(n-k)!) , (k from 0 to n) , and if
n=p*v, (p prime, v>1) I'm pretty sure it can be shown that when k is a multiple of p then
n!/(k!*(n-k)!) will not have a factor of p, and hence will not be a multiple of n.

Bryan, welcome to the group. Like you I like the joy of discovering things for myself, but
at the same time realize there is about a 100 percent chance that any results of
significance have been known for perhaps hundreds of years.

We use the tools that we are familiar with and build from there. It is almost always the
people who have acquired the most tools - who have actually studied the subject in depth
from books and such - who are conceptually equipped to make the truly new discoveries
and not simply rediscover old ones. But here's to the fun of rediscovery! :)

Mark
• From: primenumbers@yahoogroups.com [mailto:primenumbers@yahoogroups.com] On Behalf Of Mark Underwood ... would suspect so. ... row number, is the ... from 0 to
Message 3 of 3 , Dec 4, 2008
On Behalf Of Mark Underwood
> multiples of the prime p. Regarding whether the converse is true, I
would suspect so.
> That is, given a row whose interior numbers are all multiples of that
row number, is the
> row number necessarily prime? The formula is n!/(k!*(n-k)!) , (k
from 0 to n) , and if
> n=p*v, (p prime, v>1) I'm pretty sure it can be shown that when k is a
multiple of p then
> n!/(k!*(n-k)!) will not have a factor of p, and hence will not be a
multiple of n.

Suppose n is not prime and let p be a prime divisor of it (as above).
Now let n = (p^m)*v
where p does not divide v (e.g., is n=12, p=2, then m=2 and v=3.).

Now consider n choose p, this is

[ (p^m*v) * (p^m*v-1) *...* (p^m*v-p+1) ] / [ p * (p-1) * ... * 1 ]

Note p only divides the first term of the top and bottom, so the power
of
p which divides this term is p^(m-1). This shows n does not divide this

coefficient.

Indeed and old gem, but as Mark says, nice to rediscover!

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