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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 37

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

Hi Primenumbers;

I am not following your formula, please provide a clear example.

**In mathematics, you don't understand things. You just get used to them.**

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 37

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

5 * 7 - 2^1 = 33 composite.

5 * 7 - 2^3 = 27 composite.

**In mathematics, you don't understand things. You just get used to them.**

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 37

It has to be in the series, apart from 2, so starting with 3 then 3x5, then 3x5x7, then 3x5x7x11 etc.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

Okay, let me see if I am doing what you want.

3 x 5 x 7 x 11 - 2^3 = 1147 which is not prime.

**In mathematics, you don't understand things. You just get used to them.**

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 37

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

Okay you want to subtract a product of primes from 2^n. I half got it. Please let me see a few more examples.

**In mathematics, you don't understand things. You just get used to them.**

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bobbym wrote:

Okay you want to subtract a product of primes from 2^n. I half got it. Please let me see a few more examples.

I think Primenumbers is saying this: let

where *n* is a positive integer and *p[sub]i[/sub]* is the *i*th prime (so *p*[sub]2[/sub] = 3, *p*[sub]3[/sub] = 5, etc).

Then Primenumberss assertion is this:

*Last edited by Nehushtan (2014-02-03 22:37:57)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

If that is the assertion it is easily false.

So r = 3 and p(r+1) = 7, p(r+1)^2 = 49

but 34 is not a prime.

**In mathematics, you don't understand things. You just get used to them.**

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 37

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

Hi;

I think I understand what you are doing now. I will look and see.

**In mathematics, you don't understand things. You just get used to them.**

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bobbym wrote:

Hi;

I think I understand what you are doing now. I will look and see.

Please explain me his formula

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

Here is the code I have written for it:

```
y = 11;
r = 2;
{Prime[2]*Prime[3]*Prime[4] - 2 y,
PrimeQ[Prime[2]*Prime[3]*Prime[4] - 2 y]}
P = Prime[2]*Prime[3]*Prime[4] - 2^r;
{P, PrimeQ[P]}
1 < P < y^2
```

**In mathematics, you don't understand things. You just get used to them.**

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I do not understand how to use that code but did you understand what Primenumbers has been telling?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

To write the code I had to understand it. I would now use the code to test for counter examples.

**In mathematics, you don't understand things. You just get used to them.**

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Please explain whatever his formula is

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

First, did you run the code because my explanation will depend on it. Second, I might still be misunderstanding what he is doing.

**In mathematics, you don't understand things. You just get used to them.**

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 37

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.

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**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 37

Sorry, that's prime when >1 and <p(x+1) squared.

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Hi bobbym;

I ran your code.

y=11;

r=2;

*Last edited by Agnishom (2014-02-05 01:56:03)*

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

Something is wrong with your latex. I can not see any of it.

**In mathematics, you don't understand things. You just get used to them.**

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Agnishom:

Use two backslashes (\\) for a linebreak and four backslashes (\\\\) for a double linebreak.

Agnishom wrote:

Hi bobbym;

I ran your code.

y=11;r=2;

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,964

Hi Nehushtan;

Thanks for cleaning the answer up.

Hi Agnishom;

Hmmm, that is not what you are supposed to get.

The output is supposed to be:

{83, True}

{101, True}

True.

**In mathematics, you don't understand things. You just get used to them.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,979

Hi Bobby,

My output is exactly the same as yours.

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