I know that the math folks here will know this, but I found it interesting and thought others might like to learn it. Besides, I find when I post a rudimentary question about math on this board, I learn at least three knew things I hadn’t thought to ask about. So here goes.
So, I watched this video from the Numberphile, and he shows that any quasi-Fibonacci series will predict the golden ratio. That is, randomly choose two starting numbers, make the third their sum, and then continue along. You get a good approximation of the golden ratio, and in fact, it will regress to it faster than Fibonacci will.
MPSIMS (mathematical, pointless stuff I must share)
Take any two numbers, say x and y and start a Fibonacci series starting with x and y and use the rule that each term is the sum of the two previous ones. It goes
x, y, x+y, x+2y, 2x+3y, 3x+5y, 5x+8y,…, f_{n-1}x+f_{n}y,… where the f_n are the standard Fibonacci numbers. That the ratio approaches the same golden ratio is now easy to prove.
It’s true unless you start with two numbers whose ratio is exactly the reciprocal of the golden ratio. In that case all subsequent numbers will be in that same ratio. Even the slightest amount of round off error will destroy this and the golden ratio will rapidly prevail.
It is easy to prove that, with any initial numbers, the series is the sum of an exponentially growing series and an exponentially decaying series, a*(phi)^n and b*(phi)^(-n), where phi is the golden ratio and n is the index of this term in the series. The coefficients a and b are determined by the starting numbers. Since phi is bigger than one, (phi)^n gets bigger exponentially, while (phi)^(-n) decays to zero exponentially. Thus, the a*(phi)^n term always wins, with only one exception. If a is exactly zero, there is only the decaying term and the ratio of consecutive terms is always the 1/phi and not phi.
