>>5562

For each (x+n) = 1, 3, 5, etc, there is an infinite set of possible {e,n,d,x,a,b}s. In by far the majority of these cases (i.e. there are a finite number that don’t follow this and an infinite number that do), n = 1. I found this through outputting {e,n,d,x,a,b} when x+n = several different incremental odd numbers. I found something pretty crazy so I hope someone reads through this.

There are some obvious patterns that I noticed and I don’t think I saw anyone else notice, so I’ll go over them, but first, here are the outputs for the first 5 odd numbers 1, 3, 5, 7 and 9. As you can probably see, for each odd (x+n), you can map out every possible value just by incrementing a and b. So although it’s a confusing infinite set, all you need to do is print out the stuff that goes in the grid to see it. Every incrementing a value has infinite odd (x+n) values based on incrementing b by 2 (i.e. a=1 has infinite odd (x+n)s, as do a=2, a=3 etc, and for a=1, one of those odd (x+n)s comes from b=3, another comes from b=5, b=7 etc).

>>5581

So firstly we need to think about what x and n could possibly equal for (x+n) to be odd. n will only ever be 0 if c is a square (i.e. n = ((a+b)/2)-d = ((4+4)/2)-4 = 0). If that’s the case, x = 0 too (i.e. x = d-a = 4-4 = 0). So if (x+n) is odd, the lowest possible n value will be 1. In the case of (x+n) = 1 (the first picture), that means the only possible way to make (x+n) is with n=1 and x=0. For every other (x+n) value, e.g. (x+n)=5, there are more possibilities. (x+n)=5 when x=0 and n=5, x=1 and n=4, x=2 and n=3, x=3 and n=2, and x=4 and n=1. So if we were to try to find mathematical patterns, we would have to do so for every different possible combination of x and n values, right?

Well, as you’ll see in the pictures for every (x+n) above 1, there are a finite number of cells for which n=/=1, followed by an infinite set of cells for which n=1. I’m going to focus on the infinite sets for now in my explanation.

The main points of comparison are f, d and (n-1), as VQC said. Within the infinite sets where n=1, f will always equal (x+n)^2 and (n-1) will equal 0. This will be true for all odd values of (x+n) where n=1 (which is an infinite set). In the case of nn+2d(n-1)+f-1, the value of d is irrelevant, because 2d(1-1) is 0. Also, nn-1 will be 0. That means, for the infinite set for which (x+n) equals an odd number and n=1 (which will begin after a certain increment of a and b), (x+n)(x+n) = f. That means we can very easily calculate x and n, since, if f=(x+n)(x+n) and n=1, x will be sqrt(f)-1. This is true for an infinite set of numbers. Either I’m somehow wrong (I don’t see how I could be) or this has been staring us in the face the whole time.

Now, the obvious flaw in this finding is that it doesn’t represent the values of (x+n) where n=/=1. For each odd (x+n), this is a finite set, so we should be able to play around with them and find a pattern. It does look very noisy though (if you look at f in the first several outputs for 3, 5, 7 and 9, you'll see it doesn't seem very obvious how to straight calculate it). I only just put this all together, so I haven't messed with it myself. Could anyone double check my working? Because it seems too easy for (x+n)(x+n) to equal f for the vast majority of cases for it to be entirely true or useful.

>>5583

I don't know about a maximum possible (n-1) value but I have been working on figuring out where the finite set where n=/=1 ends and the infinite set where n=1 begins. Each universal set begins with the finite set, which begins with a=1, and a just increments by 1 the whole time. That means the length of the finite set is also the highest a value for which n=/=1. The finite sets end like this:

>(x+n)=3, finite set length is 1

>(x+n)=5, finite set length is 7

>(x+n)=7, finite set length is 17

>(x+n)=9, finite set length is 31

>(x+n)=11, finite set length is 49

This is going to look a bit convoluted, and I'll try to figure out how to explain myself, but you can calculate the highest a for n=/=1 / the length of the finite set from (x+n) as follows:

((((x+n)-1)/2)-1)*2 + 4(the ((((x+n)-1)/2)-1)th triangle number) + 1

Here are some examples so it maybe doesn't look so complex.

Say (x+n) = 5

(((5-1)/2)-1)*2 + 4(the (((5-1)/2)-1)th triangle number) + 1 = 2 + 4 + 1 = 7

Say (x+n) = 7

(((7-1)/2)-1)*2 + 4(the (((7-1)/2)-1)th triangle number) + 1 = 4 + 12 + 1 = 17

Say (x+n) = 9

(((9-1)/2)-1)*2 + 4(the (((9-1)/2)-1)th triangle number) + 1 = 6 + 24 + 1 = 31

So if you have a particular odd (x+n) value, if you plug it into that formula, all of the infinite a values above that point that creates your given odd (x+n) value will have n=1 and f=(x+n)(x+n).

>>5585

I did try to see if there was an obvious way to calculate the maximum n value for a given (x+n) based on each of these universal sets beginning with a=1. b in every case for a given odd (x+n) is b=a+2(x+n). n is on the right end of this spreadsheet. It's obviously incrementing, but it isn't linear. Any ideas?

>>5588

>>5587

Here it is updated with the maximum n value's corresponding x value (which is obviously going to be the minimum x since it has to add with the maximum n to make (x+n)). It's incrementing by 1 in groups of two. Each group is one number long (i.e. 4, 4, 4, 5, 5, 5), and the next is one longer than that (i.e. 6, 6, 6, 6, 7, 7, 7, 7). That's the only obvious pattern I can see.

>>5589

I just noticed d also has that pattern but it's 1 more than x since a is always 1 when n is max and x is min. Not that it changes anything.