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#76 Re: Dark Discussions at Cafe Infinity » Purpose of life???? » 2016-03-17 02:50:57

I am quite lucky I am on sickness benefit so I don't have to work. I get quite a lot of enjoyment out of life really.

#77 Re: This is Cool » Mersenne Prime Theorem » 2016-03-16 21:37:31

I don't get it. Why would.................

Nehushtan wrote:

prove that all possible prime factors will =mn+1..........................?

#78 Re: Dark Discussions at Cafe Infinity » Purpose of life???? » 2016-03-16 21:18:14

Neither did I. I think enjoying life is No. 1 priority, otherwise there's no point having more of a bad thing. So maybe, the purpose of life is to enjoy it so that it's better to be alive than not.

#79 Re: Dark Discussions at Cafe Infinity » Purpose of life???? » 2016-03-16 05:26:03

I think the meaning of life is to survive and have children. Taking my theory from Darwin- "survival of the fittest". It's good to be alive right? Right. So the only reason you are alive is because your ancestors survived and had children. (If they hadn't done that you wouldn't be here). Therefore for your future offsprings life you must survive and have children. Therefore the purpose of life in general is to survive and have children.

#80 This is Cool » Mersenne Prime Theorem » 2016-03-15 15:06:37

Primenumbers
Replies: 3

For a mersenne prime............

, all possible prime factors will =
where n= prime and m=any multiple..........

This will definitely work because of Fermat's little theorem

is always factorable by p, if p is prime and the fact that the pattern of remainders repeats itself so remainder 1 =
and n=(p-1)/m, and p=mn+1!!!

wink

#81 Re: Help Me ! » Mersenne Primes. » 2016-03-15 04:23:58

Thanks so much!

Now I can continue with my mersenne prime theorem.

smile

#82 Help Me ! » Mersenne Primes. » 2016-03-15 01:54:55

Primenumbers
Replies: 3

Hello

I am doing some work on some Mersenne primes and can't figure out the following;

Why is this always divisible by p when p is prime?

#83 Re: This is Cool » Fast way to find primes. » 2016-03-13 23:53:11

 

A= all the primes below

    multiplied together
If (A-p) and p do not share a common factor then p = prime.

#84 Re: This is Cool » Fast way to find primes. » 2016-03-13 04:53:58

I'm not sure but I'm guessing this would be a faster way to compute finding primes. I.e. using Euclid's algorithm.
See............ 

      https://en.wikipedia.org/wiki/Euclidean_algorithm

#85 This is Cool » Fast way to find primes. » 2016-03-13 02:58:19

Primenumbers
Replies: 8

Take your number, p, which you want to find out if it's prime or not.
Square root it. Multiply all the primes below the square root together.
Minus p from this number and see if there are any common factors between the two. If there is p is not prime, if there isn't p is prime.

Example:

p=51
square root p=7ish
2*3*5*7=210
210-51=159 common factor=3
51 is not prime

p=53
square root p=7ish
2*3*5*7=210
210-53=157 no common factors
53 is prime.

Note: It is impossible if the two numbers add up to a no. factorable by everything, for one to be factorable by a number and the other not. They must either both be factorable or both not.

#86 Re: This is Cool » Prime number Theorem! » 2016-02-06 07:13:26

Don't know what happened there then. It worked for all the ones I tried.

#87 This is Cool » Prime number Theorem! » 2016-02-06 03:42:42

Primenumbers
Replies: 5

Using a prime number, multiply it by 2.
Separate it into 2 numbers separated by 10x.
Multiply one of them by 10 and add it to the other no.
You should now have another prime but you may need to divide it by 2 until odd.

Example:
13 X 2 = 26
Split into 23 and 3
3 X 10 = 30 plus 23 =53
23 X 10 = 230 plus 3 = 233

#88 This is Cool » Infinite No. of twin primes! (attempt 3) » 2015-11-08 06:29:12

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!

#89 Re: This is Cool » Infinite No. of twin primes! (attempt 2). SIMPLE... » 2015-11-04 04:48:09

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

another p
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.


smile

#90 This is Cool » Infinite No. of twin primes! (attempt 2). SIMPLE... » 2015-10-31 03:43:16

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

another p
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.


smile

#91 Re: This is Cool » The Pattern Of The Primes. » 2015-10-31 01:21:53

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

#92 This is Cool » The Pattern Of The Primes. » 2015-10-26 14:04:05

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

and so on............................................................................2310 up to
etc.

#93 This is Cool » A New Way Of Finding If "p" Is Composite. » 2015-10-19 12:04:07

Primenumbers
Replies: 13

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

#94 Re: This is Cool » Proof that there are an infinite number of twin primes attempt 2. » 2015-07-22 02:38:43

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=

otherwise A - 2p and A- 2p +/-2= twin primes.
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.

#95 This is Cool » Proof that there are an infinite number of twin primes attempt 2. » 2015-07-19 02:32:32

Primenumbers
Replies: 1

There are no more twin primes after prime, z.
A=3*5*7*11*13*17*19*23*29*31....................................*z

+/-
  where p and m are primes>z or =1


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.

#97 Re: This is Cool » What's everyone working on? » 2015-06-23 10:21:21

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...?

#98 Re: This is Cool » Goldbach's conjecture second attempt. » 2015-06-23 04:45:54

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.
sad

#99 Re: This is Cool » Goldbach's conjecture second attempt. » 2015-06-23 02:19:03

Hi Bob,
          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.

#100 Re: This is Cool » Goldbach's conjecture second attempt. » 2015-06-22 08:00:01

bob bundy wrote:

Sorry.  sad  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.

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