1. I've just been reading about Pascal's Triangle over at mathsisfun and I'm amazed at how many "easter eggs" there are in it. It's almost like a Rosetta Stone of numerical sequences.

Are there any other patterns or things of numbers with as many cool secrets to discover hidden in plain sight like Pascal's Triangle?

2.

3. I find that the numbers in every other row used as coefficients make great smoothing functions.

For example, I typically use xn' = (1∙xn–1 + 2∙xn + 1∙xn+1)/4

4. Hi Daecon,
I don't know if there are other things to be discovered...

I think that the Pascal's Triangle allows cells to be Turing machines. That's because the Triangle has the Fibonacci sequence within it. Thus this function can be computed.

5. A Turing machine must accept a program of some kind. (It's like the difference between a program and a programming language.)

6. Hello
Originally Posted by MagiMaster
(It's like the difference between a program and a programming language.)
I didn't quite understand what you said.

7. Pascal's triangle is more like a single program, while a Turing machine is more like a programming language. The two aren't really comparable.

8. I created a similar triangle, but it's like this —

.....................25
................16
............9.......21
........4......13
....1......7.......19
......3......12
....1......7.......19
........4......13
............9.......21
................16
.....................25

9. The Fibonacci sequence and the geometry of Phi would be a couple of others with a large set of relationships involved in them.

Fibonacci number - Wikipedia, the free encyclopedia

Golden ratio - Wikipedia, the free encyclopedia

10. Hi.
How did you make this triangle?
1+1+•=3
1+3=4
4+3=7
I understand that the numbers at the edge of the triangle are perfect squares.
The ....1......7.......19 may be +6, +12
Can you tell me what the next column is?

11. I understand that now... Can a Turing machine compute a Pascal Triangle?

12. Yes. It's a fairly simple program since each row is just the sum of the previous row with itself shifted over by 1. [1 4 6 4 1] = [0 1 3 3 1] + [1 3 3 1 0]. (Practical implementations need to deal with the limited number of symbols available to a Turing machine. The easiest way of just reserving space for wide numbers works, but only until you get a number wider than that.)

Actually though, a Turing machine can compute anything computable. It's just as powerful as any modern programming language (if not quite as easy to actually use).

13. Originally Posted by nellythedog
How did you make this triangle?
I understand that the numbers at the edge of the triangle are perfect squares.
Can you tell me what the next column is?
Nellythedog, I'm sorry, I didn't keep up with this thread.

The next column is

36
31
28
27
28
31
36

and the one after that is

49
43
39
37
37
39
43
49

followed by

64
57
52
49
48
49
52
57
64

This triangle is actually a template that shows the inverse density of a hexagonal field for a given distribution pattern of occupied cells. By density, I mean of course that

density = # of occupied cells divided by # of total cells,
and it follows that,
inverse density = # of total cells divided by # of occupied cells

I just found it easier to work with inverse densities because they were whole numbers.

I actually generated the template by working backwards from repeated patterns of occupied cells to inverse densities for those patterns. From there, I found how to generate the inverse densities without going through my original process, which was getting tedious.

This is how the inverse densities run: You recognized the squares running along the edges of the triangle. So, for a cell with an inverse density of , the next cell in the column is n–1 less than the square, the next cell is n–3 less than that number, and so on. Calculate these numbers in from both ends of the column and, of course, stopping when you get to the middle of the column. You'll notice that the inverse density of the center cell(s) of a column is ¾ of the square at the end of the column (where n is even), or slightly more than ¾ (where n is odd).

If you want me to show you a hexagonal field for a given distribution pattern of occupied cells, then that's another matter entirely, as I need to show you visible examples (really hard to explain by words).

This triangle is actually a template that shows the inverse density of a hexagonal field for a given distribution pattern of occupied cells. By density, I mean of course that

So this triangle corresponds to a beehive with occupied cells?
I would like to know where you found this shape. Is it some sort of fractal?
I was working on the Fibonacci sequence in cell division. I found out that cells compute a variety of functions.
Thanks again.

15. 1. Sure, it could be. Actually, it corresponds to a repeating pattern of occupied cells.

2a. I set up various repeating patterns, and then measured their densities. Later I noticed a pattern in the inverted densities that I could more easily calculate without the patterns and the measuring.

2b. No, it's not a fractal because there's nothing to zoom in on. It's just a repeating pattern.

I began with a field of hexagons, and I called one cell the "origin", which I designated as occupied. You'll notice that the field around the origin (or any cell) can be divided into six triangles — this is where my template comes in. Place the dot of my template on the origin, and select one of the cells in the triangle. Designate the cell in the field under that selected cell as occupied. Do this for all six triangles around the origin. Next repeat this procedure for all the newly occupied cells you just created. Continue repeating this procedure, which will create a repeating pattern. The inverse density of that pattern is the number in the cell of my triangle that you had selected.

I'm having trouble posting artwork, but if you post a pattern that you generate, then we can discuss it. A trivial pattern is a solid pattern, every cell occupied. Of course, this happens when you use either cell with an inverse density of 1 on my template.

Here's two found on the Internet.

Designate one of the colors as occupied and the others as empty. Its pattern has a density of 1/3.

The density below is 1/4.

16. Thanks jrmonroe.
I didn't know anything about patterns for a long time, until I came across some interesting problems (in biology). I'll try to draw the hexagon fields. Thanks for replying.

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