Nov 8, 2025
Descendelbrot
z_(n+1) = (z_n)^(2 - sin(arg(c)/2 - t)^20) + c
In most of the image, the power is (very close to) 2, but in a narrow, rotating pie-slice it decreases smoothly to 1.
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Descendelbrot
z_(n+1) = (z_n)^(2 - sin(arg(c)/2 - t)^20) + c
In most of the image, the power is (very close to) 2, but in a narrow, rotating pie-slice it decreases smoothly to 1.
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Mandelbrot set with escape time coloring, but the red, green, and blue channels use different iteration maximums, ranging between 64 and 6400.
Smoothly transitioning from the Mandelbrot fractal to the Burning Ship fractal by adjusting just how absolute those values are.
Maybe you've noticed that the graph of y = e^-x + e^x looks rather similar to a parabola. Does that mean that using that formula as the basis of an escape-time algorithm will result in a fractal that looks similar to the Mandelbrot set?
Yes. Yes it does.
z_(n+1) = z_n^(6 - t) * e^(z_n^t + z_n^-t) + (-1)^n * c - 1
t grows from 1 to 2. Rotated 90° clockwise.
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z_(n+1) = e^(z_n) + e^(z_n^(i^(4t))) - 2 + c for t from 0 to 1
z_(n+1) = i^(4t) * sin(z_n) + c if n is even
z_(n+1) = i^(-4t) * sinh(z_n) + c if n is odd
t grows from 0 to 1.
z_(n+1) = e^(1 + a₁(z_n) + a₂(z_n)² + a₃(z_n)³ + a₄(z_n)⁴) + c
All the as start at 0, then a₁ begins to grow. When it reaches 1, it stops, and a₂ begins to grow, and on up the chain until all four as are 1.