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#1 Re: This is Cool » Find all the prime numbers in a given range.... » 2014-10-07 03:15:41

Yes. But should be able to work out a rule to find out the number of primes up to a limit.

#3 Re: This is Cool » Find all the prime numbers in a given range.... » 2014-08-27 06:05:20

Say x=29,

sqrt(29)=5 rd. dwn. to nearest prime,

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 Delete All Even No.'s,
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 Times remaining by 3= 3,9,15,21,27,33 etc.......Remove these:
1 5 7 11 13 17 19 23 25 29     Remaining No.'s Timesed by 5 = 5,25,35 etc. Remove theses again = 1 7 11 13 17 19 23 29.

Take all primes < or = to sqrt.(x).

#4 Re: This is Cool » Find all the prime numbers in a given range.... » 2014-08-27 05:06:56

Multiplied rather than Times by...? Here is an example:

x=17 sqrt(17)=3 rounded down to the nearest prime.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 (Delete ALL Even No.'s)

  1 3 5 7 9 11 13 15 17 (Multiply remaining no.'s by 3 = 3,9,15,21 etc.) Remove these No.'s Remaining = Primes bar 1

   1 5 7 11 13 17

#5 This is Cool » Find all the prime numbers in a given range.... » 2014-08-27 03:25:44

Primenumbers
Replies: 6

Take your Range x.
Delete All even No.'s then take all remaining no.'s and times by 3. Delete these no.'s Repeat for ALL PRIMES less than or = to sqrt.(x).

#6 Re: This is Cool » Did you know this about any Non-Prime? » 2014-08-18 22:47:48

Yes, but if we input the smallest values for x and y, and it results in a no.>z isn't z proven to be prime or 1...?

#8 Re: This is Cool » Did you know this about any Non-Prime? » 2014-08-18 07:31:24

bobbym wrote:

How about 8^2 - 7^2 = 15? x - y is not greater than 1 and 15 is a composite number.

Yes, but

   and   

Where 15=3*5 therefore x=(5+3)/2 and y=(5-3)/2.

#9 Re: This is Cool » Did you know this about any Non-Prime? » 2014-08-18 07:05:08

Agnishom wrote:

Are you saying that if x - y = 1, then x^2 - y^2 is a prime?

No, what I am saying is only Primes and 1 can not have (x - y)>1, when

.

#10 Re: Help Me ! » I have proved Goldbach's theorem can someone check it for me? » 2014-08-17 16:25:30

If Goldbach's conjecture is false and in the occurrence of e= a+b, (e) being any even no. (a) is always prime and (b) is always non-prime. The below statement is always true:

"For any area of (y)squared or greater there are y no. of primes separated by <y sized gaps."

This is not true.

For the high No.'s >(y) squared these bigger gaps are eradicated by the smaller one's at the beginning. (e)/2 will never = <(y)squared where y is the biggest prime square <e.

#11 Re: Help Me ! » I have proved Goldbach's theorem can someone check it for me? » 2014-08-17 07:26:02

If when (a) is prime, (b) always isn't, that would mean that (b) would have a prime within 12 numbers of b and not more. If it did have a bigger gap than 12 then there would not be a prime occurring every 13 numbers.
If primes are occurring within 12 numbers of b and a is occurring 13 times in 169, then non - primes (b) = (h,i,j,k,l,m,n,p,r,t,u,v,w,) fill in each gap with a prime and you get 14 primes i.e.don't forget h minus up to 12. Therefore one gap must =0 for there to be 13 primes and not 14. I.e. there must be at least one occurrence of (b) being prime when (a) is.

#12 Re: Help Me ! » I have proved Goldbach's theorem can someone check it for me? » 2014-08-17 04:31:13

But if there is a certain no. of primes in a certain range doesn't that mean that the proportion of primes is high enough to create a and b?
PROOF:
(step 1) Take 169 write it down on a piece of paper (1,2,3,4,5,6,7.........169) not literally....
(step 2) Make at least 13 holes with the biggest gap being 12 between them. These are the occurences of (a) which are unknown.
(step 3) Fold the piece of paper 13 times these are the occurences of (b) which are unknown. Then un-ravel it and look at the creases...will any of the columns not have at least one hole in it.........no. Neither will the creases.

#13 Re: Help Me ! » I have proved Goldbach's theorem can someone check it for me? » 2014-08-17 02:19:07

The Algorithm will always work because 13 primes with the biggest gap between them being 13-1=12 means that 13 squared (169) is the biggest range you would need to work with to find one occurrence of a and b.

#14 Re: Help Me ! » I have proved Goldbach's theorem can someone check it for me? » 2014-08-17 01:46:49

INPUT e
(STEP 1) Take e, sqrt it AND THEN Rd. dwn. to nearest prime:
(STEP 2) Select a prime <e at random and minus it from e.
            IF NON-PRIME THEN REPEAT UP TO A MAXIMUM OF, (sqrt(e) rd. dwn. to nearest prime) TIMES AND IF PRIME THEN STOP AND OUTPUT.

INPUT 268
(STEP 1)16.37 (2d.p.) AND 13
(STEP 2)1st go: 268 - 23 = 245 (a non-prime) REPEAT
             2nd go: 268 -167 = 101 (prime) STOP OUTPUT (268 -167=101)

Note: The smallest probability will occur when two primes selected from <289 and >169 (Probability =1/169)
Probability from lower ranges will give a higher probability than 1/169 because primes occur more often in lower ranges.

#15 Re: Help Me ! » I have proved Goldbach's theorem can someone check it for me? » 2014-08-16 23:58:41

I think I might have made a mistake which makes things simpler:
For No.'s<(z)squared the greatest gap between two primes will be (z - 1).
This is because the greatest prime we are concerned with is not z, it is the prime below z. Therefore you do not need to be concerned with remainders that are as big as z.

e.g. for primes <49 need only have remainders; when you try to factor primes 2,3 and 5.
You need not be concerned with a remainder >6 because; that would incorporate 7. We are not concerned with the prime 7 having a remainder at all.

Two numbers that add up to e can be prime because there is 1/z occurences of primes in (z)squared therefore the occurences of two random numbers being prime is 1 in (z)squared.

#16 Re: Help Me ! » I have proved Goldbach's theorem can someone check it for me? » 2014-08-16 22:52:55

Let e = a + b where e is any even no.
                    where a is any prime no.
                    where b is any prime no.

((sqrt(e)) rd. dwn. to nearest prime) = z

In (z)squared there must be 1 occurrence of two primes adding up to e.
In (z) squared the biggest gap will be (z - 1) apart from around (z)squared where the gap will be bigger. Because for all gaps to contain factors of primes up to but not including z must be (z-1) at the biggest. (See earlier post).

With primes occurring every z numbers, if a is prime there is a 1/z chance and for b to be prime as well there is a (1/z)*(1/z) chance which = 1/((z)squared). So this will occur once in (z)squared. So e=a+b is true.

We can use probability in this case because it is definite that primes occur every z no.'s not that they probably will, we know they do.

#17 Re: Help Me ! » I have proved Goldbach's theorem can someone check it for me? » 2014-08-16 21:07:22

(1) All I mean is; say e = a + b and there was only 1 prime that could = a and one prime that could = b. Then there would be only one occurrence of this happening in e and that's all we are looking for...at least one.


(2) Secondly (1/sqrt(e))*(1/sqrt(e)) would equal this.
If you round down......then you will equal a No. greater than sqrt(e)...........which means you can then say that the percentage must equal this. I can explain it as follows; if there were one prime occurring every 5 no.'s in the whole of 100. Then when I half 100..I get 50 and if there's one occurring every 5 No.'s on one side and one prime occurring every 5 No.'s on the other. Then +/- c = a or b, and a and b must occur because the probability of c causing a and b to equal primes is the above probability. There's a physics formula where if the No. of trees in a forest reaches a certain percentage, then if one of the trees catches fire they all do.


(3) Thirdly the last question is really complicated to explain but is all true:

To get a No. not factorable by a bunch of primes, you have to get those numbers multiplied together. Then minus....or add, a no. NOT factorable by them. Once you've grasped that, it's simple, to get these formula's such as 2m +/- 1 or 6m+/- 1 or 5 you just repeat the formula (x) no. of times and then remove all no.'s factorable by x....which will be exactly the PREVIOUS set multiplied by x. All primes will be found in < ((the next prime above (x))squared) otherwise they are at risk of being factorable by primes>x.

2m+/-1
6m+/-1 or 5
30m+/- 1 or 7 or 11 or 13 or 17 or 19 or 23 or 29
210+/- 1 or 11......all no.'s not factorable by 2,3,5 or 7....(therefore prime if<11 squared)


(4) Also the gap that I talked about alters the percentage possibly, but everywhere around that gap will have the correct percentage. e +/- c = a or b, how big does c have to be? sqrt(e) rd. dwn. I think because then you will have enough occurences for probability to take effect. This is just right because the big gap will occur exactly at (x) squared. So maybe round e down to the nearest PRIME and it's true.

Thanks.

#18 Help Me ! » I have proved Goldbach's theorem can someone check it for me? » 2014-08-16 15:45:10

Primenumbers
Replies: 19

For Goldbach's Theorem it states that any even no. can be made up of two primes.
e =the even no. then if there are more than 1 prime occurring in every square root(e) then the probability of this being true is 1/square root(e) timesed by 1/square root(e)
so if the gap between each prime is always smaller than sqrt(e) minus 1, then goldbach's conjecture will be true.
This is the case because;

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.

As you can see the biggest gap occurs around the beginning which could possibly be bigger than e but this is not the case because those gaps are filled with primes normally. The second time this gap will occur is around the previous prime squared. This is because the sets are interlinked where if I wanted the next set after 2+1 all I have to do is repeat it 3 times and remove any no.'s in the previous set x3 i.e. 1 x 3 = 3.... to get 1 and 5. So as there could be a gap bigger than e - 1 that is not a problem if you ignore no.'s around that gap and think that the rest of the primes will have a density needed to create e then goldbach's conjecture is true. Especially if you think there is only one gap and not two so they can't affect two primes.

#19 Re: This is Cool » Prime Numbers!! » 2014-08-16 02:50:25

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.

#20 This is Cool » Did you know this about any Non-Prime? » 2014-08-15 08:12:51

Primenumbers
Replies: 9

Did you know that for any Non-Prime that has a factor of >1, it can be produced with the below formula;

only when (x - y) can be> 1

I can prove it

x = (f + L)/2
y = (f - L)/2 therefore

   and

minus the two and it = fL which are the two factors....
And x - y = ((f + L)/2) - ((f - L)/2) = L which must be >1 to Not be, Prime or 1.

#21 Re: This is Cool » How to find whether a given number is prime or not? » 2014-08-14 03:38:30

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.

#22 Re: Help Me ! » I need help with primes! » 2014-06-29 20:52:43

Example 1:

= <11
109 - 2 = 107 (prime)
109 - 6 = 103 (prime)
109 - 20 = 89 (prime)
109 - 42 = 67 (prime)
therefore 109 is prime
Example 2:
= <11
119 - 6 = 113 (prime)
119 - 10 = 109 (prime)
119 - 112 = 7 (not prime>7)
therefore 119 is not prime
Example 3:
= <13
163 - 6 = 157 (prime)
163 - 50 = 113 (prime)
163 - 14 = 149 (prime)
163 - 66 = 97 (prime)
therefore 163 is prime

#23 Re: Help Me ! » I need help with primes! » 2014-06-29 07:01:42

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


this will prove the primality of

#24 Re: Help Me ! » I need help with primes! » 2014-06-28 21:43:46

So if I minus multiples of primes that are

  from
until reaching other prime numbers, it proves
is prime........

#25 Re: Help Me ! » I need help with primes! » 2014-06-28 07:07:03

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.

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