They discuss in the article that the key thing is for the next Fibonacci number to not overflow, so you need f(n+1) < b hence b > φ^{n+1}/√5. So to calculate t

They discuss in the article that the key thing is for the next Fibonacci number to not overflow, so you need f(n+1) < b hence b > φ^{n+1}/√5. So to calculate the best base for the n'th Fibonacci number, compute the (n+1)th Fibonacci number and then add one to that, should do the trick. (Whoops, stack is overflowing, whyyyyyy) I'm assuming that if 2^n doesn't work it's only because you get a wrong result for n=0? Then 1+2^n should beat both of these. And the issue there seems to be unrelated to precision, it is just that the denominator 1 – b^-1 – b^-2 < 0, which requires b > phi at all times, but in that case I am surprised that 3^n works at n=0. But, if you want to stay above phi, maybe b := 1 + 13*n // 8*n wastes a bit less memory.