Re: Prime Factor Testing- A Method

On Feb 28, 11:08 pm, Douglas Eagleson <eaglesondoug...@xxxxxxxxx>
wrote:
z a factorization method was to be understood as allowed.

given p as p = prime 1* prime 2;

Denote a floor as the integer divisor.  Allow a loop to exit as the
solved division fails.  So real answer simply begs how to prime test
any integer.

Test number z = z+1, divide and if integer solves apply test for a
prime? NO only one test was to be devised.  A valid integer as given p
and z.  Fast divisor means no divison as a real.. SO all integer as
solvable was to be:

single character naming convention will mean analogy.

char[z]  an array of z size  a division meant the place. So sort
char[p] as a relative list.

place char[z]  a place was to be searched to find z.  given a set p
find a z.  ANy z.  So take the set p and sort such that z exists.

for a= 1 to A loop
for z= 1 to Z loop
if z a=a+1;    /*the inner loop increment the outer answer loop
/*a prime test is conducted each incremeent of a*/
if a*z <p then exit inner loop
/*if z increments and is not a prime,exit the inner loop immediately.
z=z+1;
end loop
end loop

Integer multiplication as a rule is fast and loop outer or a appears
the smaller prime to be found.

The advantage was a simple single fast multiplication.

How does this idea look?

Douglas Eagleson
Gaithersburg, MD USA

A lousey method hihgly nonlinear, I will modifry it to decrement a
factor of ten, When A has a failing z then start over with A then
increment A by ones and decrement z. A power of ten NEVER has a prime
answer allowing all primes to be a set invertable now.

S a multiplication as ten was removable. All ten decrements at not
answerable So If z = p/10 and a*z>p,, then S. transfrom a=z z=a and
count down.. I get confused I will program it up and post it.
.

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