2014.04.28 I have solved the twin prime conjecture.
- Thread starter RD-BH
- Start date
RD-BH
Human
True. Note however that it is not divisible by your set of primes.
This is why I take it to the set of all primes.
Its period is parallel to the sequence of primes up to that point (not divergent, not convergent).
What that means is that for I*P, it is prime at 1P (ie a series can be referenced from any point).
RD-BH
Human
Holy gravy. How'd you do that? Figure out this example, I mean.
Rashiminos
Fool Prophet
My first guess would be writing a program that tests the divisibility of numbers generated as described by RD-BH from n, x, and I as inputs.
Part of the problem here is that by using the infinite product of all primes, you're trying to make a statement about the divisibility of infinity that will not make sense like it would for the divisibility of a set of integers generated by the finite versions of your numerator functions (+-).
(prod_{n=0}^{INFINITY}{P[n]})*((2I+-1)/2)+-3+-1) does not map to an integer for any value of I.
There are no two integers x any y such that xy = INFINITY.
There are no three integers x, y, and z such that xy + z = INFINITY.
It does not make sense to discuss the divisibility of INFINITY as if it were an integer, nor conclude that INFINITY is prime.
Spoiler :
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We have found plenty of other applications, but originally that is exactly the kind of thing computers were invented for.
I see the argument you are trying to make: For any given x you can find an n such that the equation holds. But I can make the reverse argument, that for any given n, I can find an x which serves as a counterexample of your equation. I would only be convinced by the formula if you can show that it holds for any given n and for any given x. As it doesn't, I am not convinced and if a referee is not convinced in a peer review he is going to argue for a fail.
If possible, you should reformulate the formula in such a way that it only contains one infinity (for x or n) and make the other variable dependent on that. Or find some other means to put your argument into a formula.
RD-BH
Human
The equation shows the relationship between the set of primes and the set of integers at the 1/2 of the period. It does not matter how many primes are in the set, as no primes (2,3,>=5) in the set can divide parallel infinite series of odd primes.
Spoiler :
It is late and I have gotten wordy ...
RD-BH
Human
Hi 1,618033
Not sure what the red progression is, but it could form in a slight variation a nice right angle with the -x and x in the Phi-related points.
Re(GR1(x)+GR2(x)) = pi (the red line)
Of course, the joke is that you cannot say the limit of GR1 + the limit of GR2 = pi.
... but the limit of GR1+GR2 does equal pi.
^Don't know what role Pi is presented as playing there, but if you indeed have a practical equation which has Pi as its limit, then you might be interested in using this with it:
because:
http://en.wikipedia.org/wiki/Spiral_of_Theodorus
Besides, in that spiral you have only right-angled triangles to deal with, and all of the hypothenuses are square roots of integers. If i personally ever tried to examine phi and pi, along with a link between them, i would use such geometrical shapes since they appear more direct to work with.
because:
Wiki spiral of Theodoros said:
http://en.wikipedia.org/wiki/Spiral_of_Theodorus
Besides, in that spiral you have only right-angled triangles to deal with, and all of the hypothenuses are square roots of integers. If i personally ever tried to examine phi and pi, along with a link between them, i would use such geometrical shapes since they appear more direct to work with.
I'm with Rashiminos. I'm struggling to understand exactly what you're saying in the OP due to the strange notation, and I think the major issue is the way you treat infinity.
The bit above in the spoilers, this bit makes sense:
But the rest, not so much. You're manipulating a finite set, mostly using division, to show something, but I don't see any reason why it should apply to an infinite set, because infinity/x doesn't make sense.
I think the bolded bits highlight the problem. Given the primes 2, 3, 5, you get a pattern that repeats with period 30. Given the primes 5, 11, 23, 31, you'll get a pattern that repeats with period 39215. Given an infinite number of primes, you won't get a repeating pattern at all. Instead, you'll get an infinite, non-repeating string of remainders. So to talk about the first & subsequent periods when given an infinite number of primes does not make sense.
In fact, using the method you set out in the spoilers:
then at that last step, won't you get
012345678901234567890123456789... numbers
010000000000000000000000000000... non-zeroes?
Because every single number (except 1) will have a residue of zero for at least one prime, as you have included all primes. Therefore when you multiply the columns vertically, each column will have at least one zero, and the product will be zero. So that clearly won't leave any other indivisible integers, much less any pairs.
You've certainly proved there are an infinite number of primes. You've proved that for a finite set of primes, there exists a twin pair that has prime factors not included in the finite set (product of the set +-1 will always satisfy that). But I can't see how it's a proof of anything beyond that.
The bit above in the spoilers, this bit makes sense:
But the rest, not so much. You're manipulating a finite set, mostly using division, to show something, but I don't see any reason why it should apply to an infinite set, because infinity/x doesn't make sense.
I think the bolded bits highlight the problem. Given the primes 2, 3, 5, you get a pattern that repeats with period 30. Given the primes 5, 11, 23, 31, you'll get a pattern that repeats with period 39215. Given an infinite number of primes, you won't get a repeating pattern at all. Instead, you'll get an infinite, non-repeating string of remainders. So to talk about the first & subsequent periods when given an infinite number of primes does not make sense.
In fact, using the method you set out in the spoilers:
then at that last step, won't you get
012345678901234567890123456789... numbers
010000000000000000000000000000... non-zeroes?
Because every single number (except 1) will have a residue of zero for at least one prime, as you have included all primes. Therefore when you multiply the columns vertically, each column will have at least one zero, and the product will be zero. So that clearly won't leave any other indivisible integers, much less any pairs.
You've certainly proved there are an infinite number of primes. You've proved that for a finite set of primes, there exists a twin pair that has prime factors not included in the finite set (product of the set +-1 will always satisfy that). But I can't see how it's a proof of anything beyond that.
Is residue a specific mathematical term? Is it just the remainder after a modulus process?
Also, note that this forum employs the
Also, note that this forum employs the
Code:
tag if you need to format a table:
1 2 3 4 5[I] [inserting spaces doesn't do anything in the first two lines][/I]
6 7 8 9 10
11 12 13 14 15
[code]1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
Residue is typically used like this:
https://en.wikipedia.org/wiki/Residue_(mathematics)
This is probably the clearest paragraph on that page:
https://en.wikipedia.org/wiki/Residue_(mathematics)#Simple_poles
Rashiminos
Fool Prophet
I'm wondering that if given an arbitrary positive integer M, you could find a mapping g: Z+ -> Z+ and use g(M) to bound the index of your product so that you would only need to consider a finite set of primes, perhaps {Pn: Pn <= (g(M))^(1/2)}, and try to show that there is a twin prime pair a, a+2 such that M < a < a +2 < g(M) using your approach.
This is how the term residue is used in number theory.
http://mathworld.wolfram.com/Residue.html
RD-BH
Human
If you were to do that, then yes everything would be zero except for -1,1.
It is a good thing that I am not doing that.
When you find a prime you zero its odd line, the remaining pattern of residue is true for all 2*3*5*...*newP*I.
Example:
01 Period 0: You have 1 vertical line of residues after zeroing P=2.
01 Period 1: this residue represents 3 (2,3)
01
==
31 Now zero 3s
03
01 Period 2: this residue represents 5 (4,5)
==
010001 Period 0: You have 2 vertical lines of residue after zeroing P=2 and P=3
010001 Period 1: these residues represent 7 and 11 (6,7,8,9,10,11)
010001
010001
010001 Note the first residue represents 5*5 and would be zeroed along with Period 0s 2nd residue (representing 5)
This is incorrect. We are dealing with integers, and 39215/2 is not an integer.
Rather, consider that sequence to be 5*11*23*31*i^2,5*11*23*31*i^4,3*5*11*23*31*i^4.
The period is 2*39215 and 39215 is at the 1/2.
This means that no prime in the set {5,11,23,31} can divide 39215+-3+-1 (39211,39213,39217,39219).
RE: infinite primes, period one, and subsequent pattern
If it is true for all subsets it is true for the whole set.
If my use of infinity seems confusing to you, replace both infinitys with variable b and examine results of b from 0 to infinity. {P[0]=2,P[1]=3,P[2]=5,etc}.
RE: -1,+1 satisfying equation
All P exist as -P,+P (P*i^2,P*i^4)
This means that all P exist on the odd 1 line (-1,1,3,5,7,9,...) as opposed to the even 1 line (0,2,4,6,8,10...). 2 is an exception in that it exists on its own odd line (-1*2,1*2).
The odd line of each P can be zeroed against the odd 1 line to sieve the next prime.
This means that -1,1 are not on any prime's odd or even lines. and cannot occupy the 1P value in the first period for any prime. Also, due to 2s unique nature its powers are on its even line while all the periods of prime combos exist on its odd line. For odd primes all powers exist on the odd P line (-1*P,1*P,3*P,5*P,...).
RD-BH
Human
RD-BH
Human
Consider:
How many P and PnP between (P[x]^2) and (P[x+1]^2)-1?
... the -1 is due to 2, (-1^2)=1,(2^2)-1=3
So,
... from 1..3 = [2,3]
... from 4..8 = [5,7] (we could switch to (P[x+1]^2)-2 from here on, if we wanted)
... from 9..24 = [11,13,17,19,23]
... from 25..48 = [29,31,37,41,43,47]
... etc, etc
Can 2 divide {ODD-4,ODD-2,ODD+2,ODD+4)? No.
Can 3 divide {3*I-4,3*I-2,3*I+2,3*I+4}? No.
Can P>=5 divide {P*I-4,P*I-2,P*I+2,P*I+4}? No.
Can any combination of ODD Primes (basically 2^x combos) divide {product+-3+-1}? No.
If it is true for all the subsets, it is true for the whole set.
RD-BH
Human
PI =limit(x=1..infinity){((i^(4/x)) - 1)/(2*(i^(1+(2/x))))}
If I tried to use this equation for PI, I might melt my cpu trying to plot it 8)
Rashiminos
Fool Prophet
None of that proves or disproves the twin prime conjecture. It just shows that P0,..., Pn does not divide P0*...*Pn*(2*I+-1)/2 +-1+-3 for arbitrarily large n. In the infinite case, none of those expressions are integers , and we're probably looking at {INFINITY, INFINITY, INFINITY, INFINITY} @ . In a finite case, it's not proven that out of {prod -4, prod -2, prod +2, prod +4) there must be 2 consecutive odd integers which are both prime, and not just relatively prime to P0,...,Pn.
@ and i^INFINITY is undefined.
The problem with using a variable b in place of both infinities simultaneously is that there could be primes greater than Pb that are not elements of {[0]prod(Pn)*(2*I+-1)/2 -4, [0]prod(Pn)*(2*I+-1)/2 - 2, [0]prod(Pn)*(2*I+-1)/2+2, [0]prod(Pn)*(2*I+-1)/2+4) } and divide enough of those elements so that there isn't a twin prime within the set@. Again, the limit of the product of b primes as b -> INFINITY does not exist.
@That is to say, you haven't shown that there is a twin prime pair outside of each finite subset, only some of them.
Edit: 39213 = 3*3*4567
39219 = 3*71*193
Edit2: 2*3*5*7*11*13 = 30030
30030 * (2*7 - 1)/2 = 195915
195191 = 47*4153
195199 = 29*53*127
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