While the explicit formula for the n-th Fibbonacci number is indeed F_n = 1/sqrt5 * [ ((1+sqrt5)/2)^n - ((1-sqrt5)/2)^n ] or F_n = (1.6180^n + 0.6180^n)/2.2361 you can notice that the -((1-sqrt5)/2)^n = 0.6180^n term tends to 0, and is in fact smaller than 0.5 even for n=2, which gives us an even simpler formula for F_n = ceil(1.6180^n/2.2361).
This would also take care of the limited precission of the float point numbers.
n = [1, 3, 5, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17]
Although, wouldn't variable overflow be the case for any algorithm, since already the 100th or so Fibonacci number takes more space than 2^64
And if you were to use the full float precision of sqr5 (and not just 4 digits like I did), you would run into 64bit float overflow way before that algorithm stops being accurate
If you were dealing with fixed-length integers, just use a lookup table.
Also, you're wrong regarding floating-point. On my machine, fib(101) is off from your approximation by ~2 million (!). Well before hitting the limits of a 64-bit float. It starts erroring at n=71, and only gets worse from there. Worse, the relative delta grows as n increases.
Thanks for actually testing it