The Odd Only Spiral (A twist on the ulam spiral with an interesting property)

[trini]

Hi guys, I've been playing around with structure generated patterns and have come across one which has caught my attention. I am only just starting to learn about number theory and so I am sure someone might be able to provide an explanation for this. Let me describe what I did then show you the results:

The motion of the path is the same as a Ulam spiral. For those of you not familiar with this structure, you can read about it here:

http://mathworld.wolfram.com/PrimeSpiral.html

If you go to that link, you can observe the square spiral path the numbers follow. Notice that the path progresses by counting all the positive integers, starting with 1 (1,2,3,4...).

This is where I made my change. I figured that since half of the numbers on the path of a standard Ulam grid were even, and if all we wanted to do was color the prime numbers, then half of the space was being wasted. By only considering odd numbers on the path, blank spaces would have more 'meaning' from a visual point of view, since there was at least the possibility that the space could have been prime.

So I set the starting point to 0, and the from that point assigned successive ODD NUMBERS ONLY as the path progressed (0,1,3,5,7...). What I found was that there were two columns and two rows which contained absolutely NO primes (except for 7). I found this interesting, as I had verified this for up to around the first 250,000 primes.

I am attaching 3 pictures of the sequence, here are some things to keep in mind:

1) most examples of the Ulam spiral (like the one in the link provided) start at the center, move right, and then follow an anticlockwise pattern.

In my algorithm, I start at the center, move up, then follow a clockwise pattern. Note this just changes the paths orientation on the grid, it doesn't affect the pattern.

2) I refer to each full rotation around the center as a turn.

3) I highlighted the columns and rows I was talking about in green for ease of viewing them. the MATLAB file doesn't do that on its own.the three pics are at 10, 50, and 250 turns. I am also attaching the MATLAB file I used(see next post). I did it up to turn 690 but it froze before I saved the results >.< the property still held tho. Feel free to play around with it if you like. there may be more interesting patterns that a more trained eye can see.

This is probably trivial, but I thought I should share =) Hopefully someone could explain how this works to me.

 

[trini]

The MATLAB file:

 

[trini]

Oh my goodness, after reviewing my post, I noticed that there are actually 2 more columns and rows with blank spaces, meaning there is a double column underneath the start point, a double row to its right, a single above it, and a single to its left (from the frame of reference I started with)

Here are the pics showing the 4 quadrants:

 

[trini]

sigh, nobody? =(

 

[Mandlebra]

sigh, nobody? =(

Would you mind upping one with the squares labeled?

 

[trini]

no prob, I'm uploading one with 3 turns of the spiral labelled and another showing the path

 

[Mandlebra]

Well, I'm sure you you know that given k consecutive integers, one of them is a multiple of k. Why the multiple s of 3 and 5 are picked out idk. Maybe if you figure a closed form for where a number is placed you'll see that your windings place 3s and 5s only in those columns. I'm only guessing here. But you should readup on number sieves, I thinkthatll help you out.

Typing on a tablet is toodamnhard

 

[trini]

well that was my initial thought, for the first 6 turns or so all the numbers in the columns and rows I highlighted were multiples of 3,5, or 7. However there reaches a point when that trend stops (on the number 187) after that there's no real order to what the highlighted columns are multiples of.

 

[Hurkyl]

The vertical column going downwards from the blue square consists of the points
$$8n^2 + 2n - 1$$
which factors as
$$(2n+1)(4n-1)$$

The line going leftwards is
$$8n^2 + 6n + 1 = (2n+1)(4n+1)$$

The other downwards line is
$$(2n-1)(4n+1)$$

My expectation is that every product $$(2n+a)(2n+b)$$ with a,b odd gives you an eventually empty vertical or horizontal line, and conversely every such line is of this form.

 

[trini]

thanks hurkyl! ^_^

I'll try that equation out and use it to hunt for all the lines in the image. Is there any use to things like these?