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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).

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

**Primenumbers**- Replies: 5

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).

I have edited formula. Thanks.

bobbym wrote:

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

Yes, but

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

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

.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.

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.

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.

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.

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.

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.

(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.

**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.

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.

**Primenumbers**- Replies: 8

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

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.

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:

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

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

So if I minus multiples of primes that are

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

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.

**Primenumbers**- Replies: 8

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.