barkerhugh.blogspot.com

Prime Numbers

Friday, 20 February 2015. Collatz - restated without the halving. Here's the way of reformulating Collatz chains that I have been using. This gives us a way to express the conditions for 1) the existence of a loop and 2) the conjecture being true. Start from n, any odd number. Multiply by 3 and add 1 = 3n 1. Multiply by 3 = 9n 3. Factor as (2 a)(an odd number). Add 2 a = 9n 3 2 a. Multiply by 3 = 27n 9 3(2 a). Factor as (2 b)(an odd number). Add 2 b = 27n 9 3(2 a) 2 b. Repeat until we reach 2 z(n) or 2 z.

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Prime Numbers | barkerhugh.blogspot.com Reviews

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Friday, 20 February 2015. Collatz - restated without the halving. Here's the way of reformulating Collatz chains that I have been using. This gives us a way to express the conditions for 1) the existence of a loop and 2) the conjecture being true. Start from n, any odd number. Multiply by 3 and add 1 = 3n 1. Multiply by 3 = 9n 3. Factor as (2 a)(an odd number). Add 2 a = 9n 3 2 a. Multiply by 3 = 27n 9 3(2 a). Factor as (2 b)(an odd number). Add 2 b = 27n 9 3(2 a) 2 b. Repeat until we reach 2 z(n) or 2 z.

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Prime Numbers | barkerhugh.blogspot.com Reviews

https://barkerhugh.blogspot.com

Friday, 20 February 2015. Collatz - restated without the halving. Here's the way of reformulating Collatz chains that I have been using. This gives us a way to express the conditions for 1) the existence of a loop and 2) the conjecture being true. Start from n, any odd number. Multiply by 3 and add 1 = 3n 1. Multiply by 3 = 9n 3. Factor as (2 a)(an odd number). Add 2 a = 9n 3 2 a. Multiply by 3 = 27n 9 3(2 a). Factor as (2 b)(an odd number). Add 2 b = 27n 9 3(2 a) 2 b. Repeat until we reach 2 z(n) or 2 z.

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barkerhugh.blogspot.com

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Prime Numbers: A binary representation of the Collatz conjecture part 2

http://barkerhugh.blogspot.com/2013/10/a-binary-representation-of-collatz_24.html

Thursday, 24 October 2013. A binary representation of the Collatz conjecture part 2. We can make several probabilistic observations from this method. And also there are a few things that we can note about the way the patterns develop (I’ve mentioned these in previous posts, but I think using binary gives more clarity). 1 If the active part of the number ends 111110 (any number of 1s) then these digits will end up as 000000. So 10 - 00, 110 - 000 etc. 7 So if this chain doesn’t reach 1 (eg is a coun...

2

Prime Numbers: May 2012

http://barkerhugh.blogspot.com/2012_05_01_archive.html

Monday, 28 May 2012. I've been working on some stuff about prime quadruplets. For reasons too dull to explain just now I started by looking for a formula that always generates composite numbers. At some point when I get the chance I'll explain this a bit more, but after a bit of thought I came up with this:. 2x -1 4y(x y) for all positive integer values of x and y. It does work, which is nice - it generates all the odd composites (which is all you need since all evens are composite except 2. If you conti...

3

Prime Numbers: Collatz - restated without the halving.

http://barkerhugh.blogspot.com/2015/02/collatz-restated-without-halving.html

Friday, 20 February 2015. Collatz - restated without the halving. Here's the way of reformulating Collatz chains that I have been using. This gives us a way to express the conditions for 1) the existence of a loop and 2) the conjecture being true. Start from n, any odd number. Multiply by 3 and add 1 = 3n 1. Multiply by 3 = 9n 3. Factor as (2 a)(an odd number). Add 2 a = 9n 3 2 a. Multiply by 3 = 27n 9 3(2 a). Factor as (2 b)(an odd number). Add 2 b = 27n 9 3(2 a) 2 b. Repeat until we reach 2 z(n) or 2 z.

4

Prime Numbers: A binary representation of the Collatz conjecture

http://barkerhugh.blogspot.com/2013/10/a-binary-representation-of-collatz.html

Thursday, 24 October 2013. A binary representation of the Collatz conjecture. Here is a restatement of the Collatz conjecture expressed in binary numbers. It is a compression and extension of recent posts on this subject which I think gives a bit more clarity. First I’ll explain the method, then I’ll explain why it works. Starting from any odd number N, write it down in binary, but turn the final 1 into a 0 (in other words write down the even number N – 1. I’ll run an example for N = 7. 2N = 1110 (=2x 7.

5

Prime Numbers: November 2012

http://barkerhugh.blogspot.com/2012_11_01_archive.html

Wednesday, 21 November 2012. Twin primes, fractals and frequencies. Following on from the previous post. Here are a few thoughts about twin primes and frequencies. I was picturing primes in terms of frequencies between 0 and 1. We can also reduce each set of twin primes to a single frequency, by looking at their product. So we find the frequency 1/35 that hits both 1/5 and 1/7 (badly drawn above). This frequency is symmetrical within the space, never intersects with the frequency 1/6 (or for 1/36). You c...

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Prime Number Symmetry | Prime Number Patterns

https://primepatterns.wordpress.com/2010/08/17/hello-world

Patterns in the Prime Numbers. The balance of the primes – conjecture. Most texts on Prime Numbers (and even a professor of Maths at Oxford University? Symmetry is not given much attention when considering prime number distribution. Hardly surprising as prime numbers don’t occur symmetrically! The next line of symmetry starts at 5#/2 = 15 AND the pattern it explains starts at a larger prime, 11. The primes occur at intervals of multiples of 5#=30 from 8 bases:. Iii) there are two lines of primes that nev...

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barkerhugh.blogspot.com

Prime Numbers

Friday, 20 February 2015. Collatz - restated without the halving. Here's the way of reformulating Collatz chains that I have been using. This gives us a way to express the conditions for 1) the existence of a loop and 2) the conjecture being true. Start from n, any odd number. Multiply by 3 and add 1 = 3n 1. Multiply by 3 = 9n 3. Factor as (2 a)(an odd number). Add 2 a = 9n 3 2 a. Multiply by 3 = 27n 9 3(2 a). Factor as (2 b)(an odd number). Add 2 b = 27n 9 3(2 a) 2 b. Repeat until we reach 2 z(n) or 2 z.

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