So the equation comes out to f(x) = sin^2(A * B^x) for particular choices of A and B. But I don't think you actually need the squaring - you can do the same

So the equation comes out to f(x) = sin^2(A * B^x) for particular choices of A and B. But I don't think you actually need the squaring - you can do the same thing with just f(x) = sin(A * B^x), and there's no particular need to talk about iterated logistic maps. For example, let B = 256, and let A = 2π * A', where A' is a number from 0 to 1. Then for some positive integer n, - f(n) ignores the first 8n binary digits of A', because sin(2π * q) ignores the integer part of q, and f(n) = sin(2π * A' * 2^(8n)). - But f(n) can be determined to a high degree of precision from the following 8 binary digits. This is just because sin is continuous and doesn't have a high slope anywhere, so if you split [0, 2π] into 256 intervals, knowing which interval q is in gives you a suitably small range for sin(q). Therefore, you can put whatever you want in the digits after those 8. - Thus, the procedure for encoding a series of y values into A' is just to set each 8 binary digits in A' to an integer from 0 to 255 proportional to arcsin(y_value). Of course, more than 8 bits can be used to achieve higher precision. Indeed, you can replace sin with any continuous periodic function, though the required precision depends on how steep the function gets. Here's a simple demo: https://drive.google.com/file/d/1SPdsHCZjH9wY0xjUvrX9ga1TTU2Zpz7P/view?usp=sharing https://drive.google.com/file/d/1SPdsHCZjH9wY0xjUvrX9ga1TTU2...