I don't know why I took an interest in this. But I did. I'm not sharing this post because it's revolutionary. I'm sharing it because I went through the exercise of counting the distinct knots myself, rigorously(but not quite formally), and maybe someone else will find my process interesting.

I started by NOT defining the standing ends. I left all ends equal at first to define "base knots". Why? Well it helps realize later which knots are actually the same knot form BUT just with different ends tucked, and makes it easy to associate an ashley-like knot with its Hunter-like sister. The first part of that turns out to be mostly irrelevant, which maybe I could have figured out at first, but the irrelevance is very interesting too.

Anyway, in this way I can describe any of these base layouts with three specifiers each being something like over or under, u or o, etc.

I'll use -1 for under and 1 for over so I can indicate reversals/negations easily.

If we label the top left end, and the top middle left loop-segment, and the top right end, then for instance A1b is 1,-1,1 and A2b is -1, -1, -1 etc. Check visually here and you'll get my scheme:

http://igkt.net/sm/index.php?topic=2826.msg19395#msg19395The lower three crossings are fully determined by the upper three so I don't need to consider them.

I will assume that any knot that I can see in a mirror is identical for practical purposes of form and function to the knot that I'm holding in front of the mirror, so I'll count those as the same knot. By this symmetry, any knot that can be written A,B,C can be written instead as its mirror C,-B,A. The middle flips because the middle indicator is defined as the over/underness of the LEFT top middle loop segment which becomes the right top middle loop segment in the mirror. So if the left bit was on top, in the mirror the right bit is.

This means that I can ignore all A,1,C combinations and just use combinations with -1 in the middle (just as has been done in this thread which is why I use -1 and not 1)

Let's look more at what happens with rotations and mirror reflections. All of these assume that all ends are equal (no working ends chosen yet).

Rx: We can see that a top to bottom rotation over the x axis never changes a layout. : ABC --> ABC

Rz: We can see that a rotation of 180 degrees in plane (around z axis) creates: -C,B,-A.

1,1,1 --> -1,1,-1 but 1,1,-1 remains unchanged.

Ry: A left to right flip over the vertical axis does the same thing as the 180 rotation.. because it's just like a rotation with a top to bottom flip.

-C, B, -A

Mx: A left to right mirror reflection (x axis inversion) -----> C,-B,A (already explained)

My: A top to bottom mirror reflection (y axis inversion) -----> -A,-B,-C

Mz: A Z-axis inversion (knot viewed on end with a mirror) ---> -A,-B,-C (same as My... for now)

So let's start with all the A -1 C combos and go from there.

1) 1, -1, 1 A1b-like

2) 1, -1, -1 B1a-like

3) -1, -1, 1 B2a-like

4) -1, -1,- 1 A2b-like

Number 4 is just number 1 with Rz applied, same base knot.

These( 1 and 4) are A1b and A2b, the two Alpine butterfly forms. More later.

..and so we just have

1) 1, -1, 1 A1b-like

2) 1, -1, -1 B1a-like

3) -1, -1, 1 B2a-like

I can't find a way to reduce the base set further. Knots 2 and three textually look l/r mirror symmetric but they aren't because of the asymmetric nature of the middle index as explained before. Rotation symmetry won't change them and a combination of Mx and My produces no change.

Now, if we want ashley-like (opposing ends) knots, then once we select the left standing end there is only one ashley-like choice for the right. So we have two opposing choices of standing ends for each knot, which I will label as P=1 for left end up and P= -1 for left end down.

I can now rewrite all the knots with a 4th descriptor A, B, C : P My designations are not exactly like those used by xarax.

1a) 1, -1, 1 : -1 A1b (not just A1a-like, this now IS the A1b knot because the working ends are now selected)

1b) 1, -1, 1 : 1

(I'll call P=-1 the a versions and P=1 the b version)

etc.

So I get 6 ashley-like knots at this stage (this will get fixed)

Now the odd thing is that xarax listed my 1b in his list of four, but not my 2b and 3b. If you rotate 1b by 180 degrees you see(again) it's the same as 4a, which he did list. So why list 1b at all and not 2b and 3b? He did then point out that really it's the same though, so here I'll prove it (I didn't YET prove that 1a and 1b are equal only that 1b and 4a are).

2b and 3b are also identical to their partners by symmetry, we'll see, but so is 1b (which we also still have to prove) so I guess it was just odd to me to see 4 in the "base set"

Remember how that vertical flip (rotation) over the x axis did nothing to the knot before? Well, now it still does nothing ... EXCEPT flip the working ends!

Now:

Rx: A,B,C: P ---> A,B,C : -P

So 1a=1b and 2a=2b and 3a=3b. And indeed, there are 3 distinct ashley like knots, and there should be 3 distinct hunter type knots.

So here they are again with their names:

1) 1, -1, 1 A1b-like : Alpine Butterfly / ?? (no clear Hunter's-like name)

2) 1, -1, -1 B1a-like : Ashley "evil imposter" / ?? (no clear Hunter's-like name

3) -1, -1, 1 B2a-like : Ashley / Hunter's

I'm not saying anything really new, and I'm sure knot theory papers have this all in it too with S01 symmetry and the like. Like I said, just went through the process so thought I'd share my experience and notes.

Just for fun here are the rest of the transformations.

Rx: A,B,C: P ---> A,B,C : -P

Rz: -C,B,-A : -P*A*C (Diagonal styles don't change P under this rotation. A*C is 1 for both ropes up or down, -1 for diagonal)

Ry: -C,B,-A: P*A*C (so Rz is different from Ry now, diagonal style knots DO flip P under this rotation.)

Mx: C,-B,A : P (already explained)

My: -A,-B,-C : -P

Mz: -A,-B,-C: P (no longer same as My)

So now all 6 transformations are different now.

None of these will ever change knot 1 into knot 2 or 3. Though. If they were different before selecting ends, they are after too.