Quetzalcoatl Posted May 24, 2011 Posted May 24, 2011 OK, This is what I did: 1. Built a vector with ones in its prime indices, and zeros for the rest. 2. Applied the Discrete Fourier Transform to the vector with a sliding (rectangular) window of 200 samples. 3. Plotted the results (see attached). The x-axis represents the numbers axis. The y-axis is the frequency components axis. The z-axis (color) is the intensity of the specific frequency components in the vicinity of the x number. You can definitely see the frequencies 0.16666Hz and 0.3333Hz going across the entire graph. Now it seems the real avg frequency is 1/0.1666Hz = every 6 numbers, and the 0.3333Hz is just a harmonic of that base frequency. My question is this: Do the prime numbers have an average interval of 6 between them? I've checked for higher primes too and got the exact same results...
DrRocket Posted May 24, 2011 Posted May 24, 2011 My question is this: Do the prime numbers have an average interval of 6 between them? I've checked for higher primes too and got the exact same results... No. http://en.wikipedia.org/wiki/Prime_number_theorem If you have a healthy skepticism for Wiki you can find this theorem proved in any good book on number theory, say Hardy and Wright's An Introduction to the Theory of Numbers. Rudin's Functional Analysis also contains a proof.
Hemi Boso Posted May 25, 2011 Posted May 25, 2011 My "wiriting through the curriculum" answer is that the average distance between prime numbers is ~3.75. All prime numbers >5 are distributed evenly (quantifiably speaking) along 8 diagonals in a 30-sectioned spiral (or as arrayed in 8-columns in a rectangular matrix) populated by all natural numbers not divisible by 2, 3 and 5: Modulo 30 for all numbers in this array, and therefore all primes >5 must be 1, 7, 11, 13, 17, 19, 23 or 29. It follows, therefore, that the distances between distributed primes are 6, 4, 2, 4, 2, 4, 6 then 2 to the next rotation of the spiral (or row of a matrix). The sum of these intervals = 30; the average distance between them = 30/8 = 3.75 (excluding a micro-adjustment for the 1st 3 primes, 2, 3 and 5) The full story, including why I can state with authority that all primes >5 are distributed evenly along the 8 diagonals described above, is here: http://www.primesdemystified.com -3
DJBruce Posted May 25, 2011 Posted May 25, 2011 My "wiriting through the curriculum" answer is that the average distance between prime numbers is ~3.75. No. See DrRocket's link above. All prime numbers >5 are distributed evenly (quantifiably speaking) along 8 diagonals in a 30-sectioned spiral (or as arrayed in 8-columns in a rectangular matrix) populated by all natural numbers not divisible by 2, 3 and 5: Modulo 30 for all numbers in this array, and therefore all primes >5 must be 1, 7, 11, 13, 17, 19, 23 or 29. This literally makes no sense. It follows, therefore, that the distances between distributed primes are 6, 4, 2, 4, 2, 4, 6 then 2 to the next rotation of the spiral (or row of a matrix). The sum of these intervals = 30; the average distance between them = 30/8 = 3.75 (excluding a micro-adjustment for the 1st 3 primes, 2, 3 and 5) Again this is non-sensical. Why do you ignore 2, 3, 5, and all primes greater than 29? See Dr.Rocket's link. The full story, including why I can state with authority that all primes >5 are distributed evenly along the 8 diagonals described above, is here: http://www.primesdemystified.com Yeah, I saw no formal proofs or even any well stated conjectures or theorems.
baxtrom Posted May 25, 2011 Posted May 25, 2011 You can definitely see the frequencies 0.16666Hz and 0.3333Hz going across the entire graph. I did a quick FFT of a similar vector from the first 1,000 primes and it looks pritty darn random to me Difference is I didn't do the sliding rectangular window thing. Perhaps your peaks are related to the windowing.
Quetzalcoatl Posted May 26, 2011 Author Posted May 26, 2011 (edited) I did a quick FFT of a similar vector from the first 1,000 primes From what you're saying, I think u tried something somewhat differrent than I had. The vector I put into the fft was this (1's at prime indices): 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0... not this: 2, 3, 5, 7, 9, 11, 13... Try that in the FFT. Using a regular fft (no windowing) I got this second plot (attached). Seems that the stronger frequencies are 1/n. The magnitude of each frequency is related to the amount of numbers divisible by the n. Edited May 26, 2011 by Quetzalcoatl
DrRocket Posted May 26, 2011 Posted May 26, 2011 (edited) I think this might explain the picture: All primes are of the form 6k ± 1 (with 2 and 3 being the only exceptions). So prime numbers tend to be on a ~6k slope, hence the 1/6 frequency. This is not even close to correct. If primes >3 were of thev form 6k ± 1 the minimun separation between those primes would be 4. Twin primes are primes of the form N and N+2. Here is a list of 100,00 twin primes http://primes.utm.ed...l/100ktwins.txt The twin prime conjecture -- that there are infinitely many twin primes -- is an open problem. Note in edit: This is incorrect as shown by uncool below. Edited May 27, 2011 by DrRocket
imatfaal Posted May 26, 2011 Posted May 26, 2011 Q(the winged serpent) - if you refuse to read the articles that prove mathematically that you are incorrect; could you explain why something that has a frequency of 1 in 6 in your theory doesn't even come to approximating that for the first 10000 primes. If you are even close to being right then the 10,000th prime should be around 60,000; well it is the work of two minutes to find that the 10,000th prime is 104,729. the prime number theorem linked to above is much closer than you and it gets better as the numbers get bigger. As another bigger example the 50millionth prime (again a minute on the internet) is at 982,451,653 - your theory would predict is around 300,000,000, the prime number theorem which is getting into its stride would estimate that you would get 50,000,000 million primes around 1,038,029,502. Regardless of the charts and your DFT and banding, in simple terms frequency can be looked at by how many in a given large spread - and your ideas do not hold water.
Quetzalcoatl Posted May 26, 2011 Author Posted May 26, 2011 Thanks for the pointers. I have read the article, and changed my post accordingly. Now, as for primes being of the form '6k±1', it's like saying every prime is of the form '2k+1' (odd numbers) except for 2 (which is the only even prime, of course). so, '6k±1' actually is true. Doesn't say much, but still true. Read about it in wiki's "Primality Test" article.
DrRocket Posted May 26, 2011 Posted May 26, 2011 (edited) Thanks for the pointers. I have read the article, and changed my post accordingly. Now, as for primes being of the form '6k±1', it's like saying every prime is of the form '2k+1' (odd numbers) except for 2 (which is the only even prime, of course). so, '6k±1' actually is true. Doesn't say much, but still true. Read about it in wiki's "Primality Test" article. It doesn't say much, and it is false. Read what I said. I don't care what wiki said. The difference between two numbers of the form '6k±1' is at least 4. If all primes (>3) were of the form '6k±1' there could be no twin primes other than 2 and 3. There are lots of them. Note in edit: This is still incorrect. Edited May 27, 2011 by DrRocket
imatfaal Posted May 26, 2011 Posted May 26, 2011 (edited) 6k +/-1 is just no back up at all to your theory. It is easily rationalised and doesnt tell us anything; 6k must be even, you cannot say much about 6k -1, 6k -2 again even, 6k-3 must be divisible by 3, 6k -4 must be even, 6k -5 = 6(k-1) +1. That accounts for all numbers the only ones not divisible by 2 or 3 are 6k+1 and 6k-1 Clearly the number between twin primes must be divisible by 6, ( ie one of the primes is 6k+1 and the other is 6k-1)- i it must be an even between two odds - divisible by two ii any three consecutive numbers one must be divisible by three, obviously neither of the primes can be divisible by 3 therefore the middle one must be iii any number divisible by 3 and 2 is divisible by 6 But this still doesn't help you. DocRock has provided a link to the Wikipedia page which briefly outlines prime number theorem, and a reference for further reading. And a simple numerical test shows that your prediction is miles out Edited May 26, 2011 by imatfaal
uncool Posted May 27, 2011 Posted May 27, 2011 It doesn't say much, and it is false. Read what I said. I don't care what wiki said. The difference between two numbers of the form '6k±1' is at least 4. If all primes (>3) were of the form '6k±1' there could be no twin primes other than 2 and 3. There are lots of them. . 5 = 6*1 - 1 7 = 6*1 + 1 7 - 5 = 2. The twin primes are of the form (6k - 1, 6k + 1) for some k. =Uncool- My "wiriting through the curriculum" answer is that the average distance between prime numbers is ~3.75. All prime numbers >5 are distributed evenly (quantifiably speaking) along 8 diagonals in a 30-sectioned spiral (or as arrayed in 8-columns in a rectangular matrix) populated by all natural numbers not divisible by 2, 3 and 5: Modulo 30 for all numbers in this array, and therefore all primes >5 must be 1, 7, 11, 13, 17, 19, 23 or 29. It follows, therefore, that the distances between distributed primes are 6, 4, 2, 4, 2, 4, 6 then 2 to the next rotation of the spiral (or row of a matrix). The sum of these intervals = 30; the average distance between them = 30/8 = 3.75 (excluding a micro-adjustment for the 1st 3 primes, 2, 3 and 5) The full story, including why I can state with authority that all primes >5 are distributed evenly along the 8 diagonals described above, is here: http://www.primesdemystified.com First, you have to define what you mean by the "average distance between prime numbers". Do you mean: [math]\lim_{n \rightarrow \infty} \frac{p_n}{n}[/math]? If so, then you are wrong, as that limit is infinite. To see why you are wrong, consider a 210-sectioned spiral. Then there are a total of 48 (mod 210) possibilities, meaning that according to your argument, the average must be 210/48 = 35/8 = 4.375, which is substantially larger than your number. Not only that, but it is always possible to keep going up (next, try 2310, which gets you 4.8125). =Uncool- 1
baric Posted May 27, 2011 Posted May 27, 2011 The full story, including why I can state with authority that all primes >5 are distributed evenly along the 8 diagonals described above, is here: http://www.primesdemystified.com Why the emphasis on numerology on your page? You do understand that all of your focus on the final digits of numbers and also numbers like 11,111,111 have absolutely no numerical significance but are simply artifacts of our use of the base-10 system. In addition, your sieve automatically incorporates primes up to 29 as best as I can tell. That means it will produce a correct list of primes until 961.
DrRocket Posted May 27, 2011 Posted May 27, 2011 (edited) 5 = 6*1 - 1 7 = 6*1 + 1 7 - 5 = 2. The twin primes are of the form (6k - 1, 6k + 1) for some k. =Uncool- You are correct. I overlooked the obvious. Edited May 27, 2011 by DrRocket
baric Posted May 27, 2011 Posted May 27, 2011 My "wiriting through the curriculum" answer is that the average distance between prime numbers is ~3.75. Since the average distance between primes increases in an unbounded manner as you approach infinity, this can only be true for a specified range of integers.
DrRocket Posted May 27, 2011 Posted May 27, 2011 Since the average distance between primes increases in an unbounded manner as you approach infinity, this can only be true for a specified range of integers. Right. An awful lot of what is being said here violates the prime number theorem.
baxtrom Posted May 27, 2011 Posted May 27, 2011 The vector I put into the fft was this (1's at prime indices) I was unclear, that was exactly what I did. The plot looks similar too. Scilab?
Quetzalcoatl Posted June 2, 2011 Author Posted June 2, 2011 (edited) I was unclear, that was exactly what I did. The plot looks similar too. Scilab? Matlab... Right. An awful lot of what is being said here violates the prime number theorem. I'm not trying to prove anything, man. Just interested in those spikes on the graph. Do you think it's just random noise? Doesn't looks so to me. That's why I posted this, wanted some help with interpretation of the plot from the math community... No reason to get angry. Just interested in some spikes on a Matlab plot. I am however thankful for your quick answer to my original question, and you're quite correct - the average interval certainly isn't 3 or 6 or whatever number. I'm still curious, though, with regard to the fft plot. Edited June 2, 2011 by Quetzalcoatl
DrRocket Posted June 4, 2011 Posted June 4, 2011 (edited) I'm not trying to prove anything, man. Just interested in those spikes on the graph. Do you think it's just random noise? Doesn't looks so to me. That's why I posted this, wanted some help with interpretation of the plot from the math community... No reason to get angry. Just interested in some spikes on a Matlab plot. I am however thankful for your quick answer to my original question, and you're quite correct - the average interval certainly isn't 3 or 6 or whatever number. I'm still curious, though, with regard to the fft plot. Angry ? Nobody is angry. You have the right to be as silly and obtuse and you please. It certainly is no skin off my nose. While most engineers read "FFT" as the "Fast Fourier Transform" mathematicians also know it as the "Finite Fourier Transform" because that is what it really is -- the Fourier transform on a finite abelian group. The exact nature depends on the arbitrary choice of the number of points involved, the order of the cyclic abelian group. Now why would you expect the Fourier transform of some sampled data on [math] \mathbb Z/ n \mathbb Z [/math] to tell you anyrhing at all about the distribution of prime numbers in [math]\mathbb R [/math]? Edited June 4, 2011 by DrRocket
baxtrom Posted June 17, 2011 Posted June 17, 2011 While most engineers read "FFT" as the "Fast Fourier Transform" mathematicians also know it as the "Finite Fourier Transform" because that is what it really is -- the Fourier transform on a finite abelian group. Personally I see the FFT as an optimized computer algorithm for obtaining the DFT, or Discrete Fourier Transform.
Quetzalcoatl Posted June 23, 2011 Author Posted June 23, 2011 (edited) Now why would you expect the Fourier transform of some sampled data on [math] \mathbb Z/ n \mathbb Z [/math] to tell you anyrhing at all about the distribution of prime numbers in [math]\mathbb R [/math]? Because usually we look at prime numbers in the context of the natural numbers set, Mr Rocket. And yes, I am an engineer. Edited June 23, 2011 by Quetzalcoatl
baric Posted June 23, 2011 Posted June 23, 2011 Because usually we look at prime numbers in the context of the natural numbers set, Mr Rocket. And yes, I am an engineer. If you are seeing a significant pattern in a set of data where none is expected, my advice would be to apply the same techniques to different sets of data (i.e. different intervals of integers) and see if the pattern is consistent for all intervals or is just an artifact of your initial sample.
DrRocket Posted June 23, 2011 Posted June 23, 2011 (edited) Because usually we look at prime numbers in the context of the natural numbers set, Mr Rocket. And yes, I am an engineer. The first sentence is true but irrelevant. The second is no excuse. Edited June 23, 2011 by DrRocket
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