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**Primenumbers**- Replies: 0

**Assumptions:**

1. There are regular gaps of **4** between primes

2. These gaps of **4** have varying remainders for 4, (either 1 or 3)

A= 3x5x7x11x13x17............(x next prime in series)

x = the next prime above highest prime in A used

p and p+4 are prime

(A - p)/2 = twin prime

(A-(p+4))/2 = twin prime

**Example:**

A= 3x5x7= 105

x=11

A has remainder=1 for 4

2 Primes = 67 and 71

67 and 71 have remainder=3 for 4, which is different from A

(105-67)/2 = 19

(105-71)/2 = 17

Twin primes!

**I realise my first post is incomplete so I have made some amendments...............**

A= 3, 15, 105, 1155, 15015 x next prime in the series................

p= prime or not factorable by any primes in A

and a gap of 2, therefore a p+1 can NOT be a prime because p+2 can not be prime

and a gap of 4, therefore a gap of 4 can not occur because p+2 can not be prime

OR and a gap of 4, therefore a gap of 4 can not occur because p+1 can not be prime

With a gap of 4, one number minused from A will be divisible by 4 or 2. If one is divisible by a higher than 4, the other can't be because the gap would have to be higher than 4 for the remainders to match. Therefore we can express a gap of 4 using the above equations, where if one number is divisible by a greater than 4 this will be p+1 an even no. and p is what was divisible by 4 but no higher.

and a gap of 8, therefore a gap of 8 can not occur because p+4 can not be prime

and a gap of 8, therefore a gap of 8 can not occur because p+2 can not be prime

and a gap of 8, therefore a gap of 8 can not occur because p+1 can not be prime

With a gap of 8, one number minused from A will be divisible by 2,4 or 8. If one is divisible by a higher than 8, the other can't be because the gap would have to be higher than 8 for the remainders to match. Therefore we can express a gap of 8 using the above equations, where if one number is divisible by a greater than 8 this will be p+1 an even no. and p is what is divisible by 8 but no higher.

Divide A by 3 and multiply by 2

another p

and a gap of 6, therefore a gap of 6 can not occur because p+2 can not be prime

With a gap of 6, one number minused from A will be divisible by 3. If one is divisible by a higher than 3, the other can't be because the gap would have to be higher than 6 for the remainders to match. Therefore we can express a gap of 6 using the above equation, where if one number is divisible by a greater than 3 this will be p+2 and p is what is divisible by 3 but no higher.

Divide A by 5 and multiply by 2

another p

and a gap of 10, therefore a gap of 10 can not occur because p+2 can not be prime

With a gap of 10, one number minused from A will be divisible by 5. If one is divisible by a higher than 5, the other can't be because the gap would have to be higher than 10 for the remainders to match. Therefore we can express a gap of 10 using the above equations, where if one number is divisible by a greater than 5 this will be p+2 and p is what is divisible by 5 but no higher.

**..........And so on until a gap that must be >2 becomes a gap that must be >infinity, there therefore must be an infinite No. of twin primes.**

**Primenumbers**- Replies: 1

A= 3, 15, 105, 1155, 15015 x next prime in the series................

p= prime or not factorable by any primes in A

and a gap of 4, therefore a gap of 4 can not occur because p+2 can not be prime

and a gap of 8, therefore a gap of 8 can not occur because p+2 can not be prime

Divide A by 3 and multiply by 2

another p

and a gap of 6, therefore a gap of 6 can not occur because p+2 can not be prime

Divide A by 5 and multiply by 2

another p

and a gap of 10, therefore a gap of 10 can not occur because p+2 can not be prime

**..........And so on until a gap that must be >2 becomes a gap that must be >infinity, there therefore must be an infinite No. of twin primes.**

Thanks for the post. I have checked out the website. Very interesting!

**Primenumbers**- Replies: 2

I have posted this on other topics, but I wanted to make sure everybody has got it!

My theory is simple, Primes have a pattern and I shall tell you what it is.

The pattern is based on primes multiplied together. 2, 6, 30, 210, 2310, 30030.

Then just multiply by the next prime in the series to get, 30030*17=510510 the next number in the series.

From now on you will regard 1 as a prime number as it behaves in much the same way as the primes, but from now on if you see a 1 just remember that it is indeed not a prime.

Okay, let's start with 2.

Write down 2 cross it out, then put a 1 before it. Now just add 2 up to 6........1 3 5. Now times 3 by all no.'s below 2 = 3. Cross it out.

Now let's start with 6. Now add 6 up to 30.............1 5 7 11 13 17 19 23 25 29. Now times 5 by all no.'s below 6 = 5,25. Cross it out.

Now 210. Now add 30 up to 210...........1 7 11 13 17 19 23 29 31 37 41 43 47 49 53 59 61 67 71 73 77 79 83 89 91 97 101 103 107 109 113 119 121 127 131 133 137 139 143 149 151 157 161 163 167 169 173 179 181 187 191 193 197 199 203 209. Now times 7 by all no.'s below 30 = 7,49,77,91,119,133,161,203. Cross them out.

Keep doing this and you will generate all the primes.

Note: In using 210 I have used the primes 2,3,5 and 7. The next prime is 11, so I have only generated the primes up to

**Primenumbers**- Replies: 0

p=any number ending in 3

check p is not a square

where a+b is a factor and a-b is another factor

if p is prime you will not find any factors

if p is composite you will find factors.

Example:

p=33

27*33=891

46+35=81=27*3 46-35=11

factors for 33 are 3 and 11

** Simpler version**

There are no more primes after prime, z.

A=3*5*7*11*13*17*19*23*29*31............................................*z.

p= No. not factorable by any primes in A or 2.

p+/-1=

Some gaps must be >4 as p+1 and p+4-1=.

No room for .

These gaps must be >6 as p+1 and p+3 and p+6-1=.

Again no room for .

And so on until gaps that must be >2 become gaps that must be >infinity.

**Primenumbers**- Replies: 1

There are no more twin primes after prime, z.

A=3*5*7*11*13*17*19*23*29*31....................................*z

so long as or whichever is greater rd. dwn. to nearest prime = z or less+/- must exist in the correct range otherwise m+/-1=*composite then in a gap of 4 there are no more even no.'s left to as m+1=*composite and m+4-1=*composite

Therefore some gaps must be >4. In these gaps we now know m+/-3=*composite therefore these gaps must be >6 as m+1=*composite m+3=*composite and m+6-1=*composite.

Still no even no.'s left for .

And so on until gaps that must be >2 become gaps that must be >infinity.

Ok, cool.

zetafunc wrote:

But is that really any better than, say, the Sieve of Eratosthenes?

You would have thought knowing what a and b end in you could determine that a number is composite and therefore not prime, turns out it doesn't seem to work that way.

It wouldn't -- using your notation, x is prime iff (a,b) = (1, x) or (x,1). Even if a or b is 1 modulo 10, that doesn't guarantee that a or b is 1, and for large x, will give you a very large number of possibilities.

I still don't get this post..................? Is 1 modulo 10 a mathematical way of saying, ends in 1.....? I was trying to prove x is composite not prime......? a or b ending in 1 wouldn't guarantee x is prime...?

**You were right............it doesn't work.**

I made a big mistake:

Primes minus No.'s with remainder e would = (not >1) using my method..........which is not true.

Thanks for looking at my posts. I will try to recreate it into simpler style and put it in an algorithm. Might take me a little while. Get back to you later................Primenumbers.

bob bundy wrote:

Sorry. I got totally lost at "but we don't want it to have the same remainders as e".

We have to delete remainders for e otherwise if a had r.=x, then e-a=b: e(with r.x)- a(with r.x)=r.0 for b, so b would not be prime.

Does that help Bob?

Think of it this way;

Way of finding primes=

2m+1

6m+1 or 5

30m+1 or 7 or 11 or 13 or 17 or 19 or 23 or 29

210m + 1 or 11 or 13.....(no.'s not factorable by 2,3,5 or 7)...........209

e might have a remainder if you try to factor 3, 5 or 7.

e will never have a remainder for 2.

1) Let's pretend e does have a remainder for 3. I.e. remainder for 3=x

There will only be 1 no. with r.x in 6.

2) Let's pretend e does have a remainder for 5 also. I.e. remainder for 5=y

There will only be 2 no.'s with r.y in 30.

3) Let's pretend e does have a remainder for 7 also. I.e. remainder for 7=z

There will only be 8 no.'s with r.z in 210.

You can now see that minusing these no.'s from no. of primes there will still be enough left to make a and b.

bob bundy wrote:

So you want two odd primes for c and d.

c and d are not prime. Don't worry, I don't think they work. But take this new idea:

e=any even no.

e= a + b

We want a to be prime but we don't want it to have the same remainders as e, otherwise e-a= non-prime.

Finding the numbers that have a different set of remainders than e will be the same as the number of primes.

Because we are just trying to delete a different set of remainders. (instead of r.=0, r.=x). But hang on a minute r. for 2 will be the same in both sets as e is always even.

Therefore there will always be < no. of primes for the no. of numbers that don't have the same r. for e.

Therefore there will always be an instance for (a) and therefore an instance for (b).

Example: e=84

Finding the no. of primes in e:

2m

6m+3

30m+5 or 25

210m+7 or 7x7 or 7x11 or 7x13 or 7x17 or 7x19 or 7x23 or 7x29

For 84 count the No. of numbers it passes and minus it from 84 then -1 for 1 and + 4 for 2, 3, 5, and 7, I got 23 which is the No. of primes below 84.

Finding the instances of e remainders:

84 has remainder 0 for 2,3, and 7 but remainder 4 for 5. So lets delete the no. of primes that =(5y + 4);

There are 2 no.'s in 30 which have a remainder 4; 19 and 29

30m +19 or 29

84 passes through 5 of them but one of these is 49 so 23-4 but one of these is 2 also so 23-5=18 but don't forget to minus 3 and 7 so 18-2=16......

(5,79)(11,73)(84,71)(17,67)(23,61)(31,53)(37,43)(41,43).

**I am working on a simpler theory;**

e=any even no.

e=(c-a)+(d-b)

1) primes <e excluding 2

2) 2

c and d factorable by 1) not factorable by 2)

a and b factorable by 2) not factorable by 1)

Therefore (c-a) and (d-b) are prime.

I am not however sure whether I need to prove that e will = a certain size....?

**Rule for primes;**

x! is factorable by x only once when x is prime and more than once when x is not prime. This only happens when x>4.

If x=ab, ab occurs more than once, i.e. x!=axbxabx?

If x=

, occurs more than once i.e. x!=ax2ax. Unless =2 or less and x= 4 or1. a>2 is fine.I think this is simpler than x/x! factors down when x is not prime and doesn't when x is prime!

**Primenumbers**- Replies: 10

Goldbach's conjecture states; Every even integer greater than 2 can be expressed as the sum of two primes.

How about this;

e= any even no.

c+d= e! factored down with factors of e, but not 2.

i.e. 12! factored down with factors of 12 but not 2 = 12!/3 as many times as possible = 12!/243=1971200.

a+b=(c+d)-e

(1) No.'s <e excluding factors of e, except 2

(2) e factors

f.= factorable by

e is f.(2) not f.(1)

c+d is f.(1) not f.(2)

a+b is f.- not f.(1)(2)

a is f.(2) not f.(1)

b is f.- not f.(1)(2)

c is f.(1) not f.(2)

d is f.(1)(2) not f.-

e= (c-a)+(d-b) = two primes added together. (They are prime because they are not f.(1) or f.(2).)

The above is true because of the following rule:

A=B+C If B is factorable by x but C isn't then A won't be.

i.e. A=B+C

59=21+38 21/7=3 38/7=5 remainder(3)

59/7 = 8 remainder(3). The remainder carries.

zetafunc wrote:

Even if a or b is 1 modulo 10, that doesn't guarantee that a or b is 1, and for large x, will give you a very large number of possibilities.

If x = (10c + d)(10e + f) where d and f = 1,3,7 or 9 I just have to prove c and e > 0 granted that x is not factorable by 2,5,3,7. i.e. a and b > 10.

I know how to prove x is not factorable by 2,3,5 or 7, I don't know how to prove that c and e are both >0.

But say I knew what d and f were, you would have thought I'd be able to work out what c and e were................ Don't you think?

zetafunc wrote:

Like this?

More like this;

x is not factorable by 2 or 5.

x = ab

x ends in 1; a and b end in (1,1)(3,7) or (9,9).

x ends in 3; a and b end in (1,3) or (7,9).

x ends in 7; a and b end in (1,7) or (3,9).

x ends in 9; a and b end in (1,9)(3,3) or (7,7).

i.e. 3x9=27.......ends in 7.

by knowing what x ends in we can determine what a and b might end in.

You would have thought knowing what a and b end in you could determine that a number is composite and therefore not prime, turns out it doesn't seem to work that way.

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

It's fine if you work with odd numbers only.

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: 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).