Have I discovered a new fractal / what is this fractal called?

Question

So I was experimenting with fractals and created this. The equations are

$$ A \leftarrow A^2 + x - B \\ B \leftarrow B^2 - y + A $$ and the set is defined as the points (x,y) on a graph which when the equations are iterated, $A$ AND $B$ remain bounded.

(The colours in the graph are from the number of iterations required to surpass the threshold of $-2$ for either $A$ or $B$.)

Initial values for $A$ and $B$ are 0

if you are wandering about computation, it is iterated 216 times and the threshold for being considered not bound is $A$ or $B$ being over 10,000,000,000.

Answer 1

Not an answer but too long for a comment.

I reproduced the image with this code for shader editor app from fdroid for Android:

#ifdef GL_FRAGMENT_PRECISION_HIGH
precision highp float;
#else
precision mediump float;
#endif

uniform vec2 resolution;

void main(void) {
  float R = 8.0;
  vec2 uv = R * (gl_FragCoord.xy / resolution.xy - vec2(0.5)) * resolution.xy/resolution.y;
  vec4 x = vec4(uv.x) + vec4(0,0,1,1) / resolution.y * R;
  vec4 y = vec4(uv.y) + vec4(0,1,0,1) / resolution.y * R;
  vec4 A = vec4(0.0);
  vec4 B = vec4(0.0);
  vec4 m = vec4(0.0);
  bool escaped = false;
  for (int i = 0; i < 256; ++i)
  {
    vec4 A1 = A*A + x - B;
    vec4 B1 = B*B + y + A;
    A = A1;
    B = B1;
    if ((A*A+B*B).x > 1.0e6)
    {
      m = vec4(i) - log2(log2(A*A+B*B));
      escaped = true;
      break;
    }
  }
  m = floor(m);
  bool axis = abs(x.x) < 0.1 || abs(y.x) < 0.1;
  x = floor(x);
  y = floor(y);
  bool e = m.x == m.y && m.y == m.z && m.z == m.w;
  bool grid = x.x == x.w && y.x == y.w;
  gl_FragColor = vec4(
    (grid ? vec3(1) : vec3(axis ? 0.0 : 0.8)) *
    (escaped ? vec3(e) : vec3(0.8))
  , 1.0);
}

The grid spacing is 1, centered on the origin. Note the signs in the formula: $$ A_{n+1} = A_n^2 + x - B_n \\ B_{n+1} = B_n^2 + y + A_n $$