- 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.

I will start with the breakthrough realization that was like a

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 - --- In primenumbers@yahoogroups.com, "Bryan Bartlett" <valareos@...> wrote:

> Given: Pasquels Triangle where each row x is sequentionally numbered

I remember way back in high school coming across that result when we were studying the

> 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?

> --

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> 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!

CC