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The primes there are in a certain range can be estimated because there are;
1 No. not factorable by 2 in (2)
There are 2 No.'s not factorable by 2 or 3 in (6)
There are 8 No.'s not factorable by 2 or 3 or 5 in (30)
There are 48 No.'s not factorable by 2 or 3 or 5 or 7 or in (210)
times 48 by (prime -1) to get the next number of no.'s. i.e. =480 no.'s in (2310) not factorable by 2,3,5,7, or 11.....and so on.
Did you know that for any Non-Prime that has a factor of >1, it can be produced with the below formula;
I can prove it
x = (f + L)/2
y = (f - L)/2 therefore
x! - 2^y = Primes when divided by 2 until odd, and is < x squared.
because a no. divisible by 7 minus a no. that isn't, always = a no. NOT divisible by 7 this will work for any no. because the remainder remains when you minus from the no. which has a remainder of zero.
Example 1:
Let (n) = any number...
(n) - 2 - 2 -2 -2 -2 -2 -2 -2 ..............= any prime > 2
(n) - 3 -3 -3 -3 -3 -3 -3 -3 ...............= any prime > 3
(n) - 5 -5 -5 -5 -5 -5 -5 -5 ...............= any prime > 5
if continue to prime
So if I minus multiples of primes that are
from until reaching other prime numbers, it proves is prime........So 7 for example is not factorable by 2.
7/2=3 + remainder 1
because 7 is <9 I need only to see if it is divisible by 2 for it to not be prime.
<25, 3 <49,5 etc.
I need help with these prime numbers...
I can't seem to find them, what do you think of these ideas;
1. You multiply a prime by a certain number. That will somehow keep all remainders of p, a prime number, prime. Like 1.25 x a No. <25 etc.
2. You get groups of multiples of primes that do not align with each other. i.e. say you have 3, 7, 11 primes multiply each of them by separate multiples of 10(to the power of x) and they won't divide any of them by that No.
Please help I've been working on primes.
By basis for my theory is that a prime no. is not divisible by primes < sqrtp only. And primes if you work with the factors of primes you can generate them with various methods!
2+/-1..
6+/-1 5..
30+/- 1 7..
210 +/- 1 11..
2310 +/- 1 13..
equals primes......to a certain point above highest prime squared. e.g.>9,25,49,121 or 169 accordingly. And so on but the numbers get very big but you can do this up to infinity.
Prime Number TEST!: (where p is Prime)
Repeat process until y<p, exchanging y for p. Then when y<p, y will be a recognizable prime (though not found in A), if this is not the case then p is NOT prime.
2m
6m+3
30m+5 or 25
210m+7 or 7x7 or 7x11 or 7x13 or 7x17 or 7x19 or 7x23 or 7x29
For 121 count the No. of numbers it passes and minus it from 121 then -2 for 1 and itself and + 4 for 2, 3, 5, and 7, I got 30 which is the No. of primes below 121
Sorry, that's prime when >1 and <p(x+1) squared.
Okay. Let p1=2, p2=3, p4=5 p5=7 p6=11 and so on up to infinity.
Let x and y both =any positive integer, and z = any No. that does not have p1, p2, p3, or p4...up to px, as it's factor.
Then the
Rule= [p2*p3*p4*px] +/- [(p1^y)*z] = prime when >1 and <p(x+1).
Theory: This No. will not be factorable by p up to px because one half will be factorable by it. And the other half will always be a remainder. So when you add or minus the numbers together, the remainder will remain.
It is prime when < p(x+1) because: if P=a*b a will be >square root P and b will be< square root P when a and b are whole numbers. And either a or b can be prime because all P's can be reduced to prime factors so the smallest factor will always be prime and we can try to find the numbers which don't have any there. No.'s that don't have any there are Prime in the theory.
Another way of saying it is;
[All the primes in the series up to a point, multiplied together, multiplied by any odd No.], minus, [2^y multiplied by any No. that is not divisible by all the primes in the series you used]. Will equal a prime No. when >1 and <the next No. in the series after the point multiplied by itself by definition. Oh and y must be at least 1.
So 3x5x7 - 2x11 = 83 which is prime as is >1 and <11 squared=121.
Or 3x5x7 -2^3 = 97 is prime as is >1 and <121 etc.
This is slightly different from what I said earlier.
That is not under 13 squared which is the prime No. just above 11. The idea is you have two groups; one divisible by 2 only the other by all possible odd factors. You fiddle the values to get the number in range but is very difficult for big numbers. I want to try and link it to Mersenne Primes because they are in the form 2^any integer - 1. For instance 2^4 minus 3x5 = 1 so 2^5 -1 is like 2^5 - (2^4 - 15)= 2^4 + 15 =31 (Prime) which is < 49.
It has to be in the series, apart from 2, so starting with 3 then 3x5, then 3x5x7, then 3x5x7x11 etc.
Here is an example:
3x5=15, 15 plus or minus 2, 4, 8, 16, 32 in the range of >1 and <49 = 7,11,13,17,19,23,31 and 47, which are all prime.
A way of finding any prime number:
2(to the power of any integer) + or - [3x5x7x11x13...etc. x any odd No.] = A Prime No. when>1 and <the prime No. just above the highest prime used in the series, squared.
For prime numbers (p), under 3 squared; p = 2a + 1 (where a = any whole number)
For prime numbers (r), under 5 squared; r = 3p +/- 2 or 4
For prime numbers (i), under 7 squared; i = 5r +/- 6 or 12 or 18 or 24
For prime numbers (m), under 11 squared; m = 7i +/- 30 or 60 or 90 or 120 or 150 or 180
For prime numbers (e), under 13 squared; e = 11m +/- 210 or 420 or 630 or 840 or 1050 or 1260 or 1470 or 1680 or 1890 or 2100
(A no. divisible by a, but not divisible by a group b +/- A no. divisible by group b, but not a = A no. not divisible by a or b.)
* Some rules give 1 but this is not prime.