While I fully agree with the premise that any sequence of numbers can have an essentially infinite choice of 'next number', depending on how you define your seq

While I fully agree with the premise that any sequence of numbers can have an essentially infinite choice of 'next number', depending on how you define your sequence, and the simplest way to prove that is simply to given a sequence of _n_ numbers solve the _n+1_ polynomial equation, given that it is incompletely specified then you have an infinite number of solutions. Or just drop in a heavy side step function (https://en.wikipedia.org/wiki/Heaviside_step_function https://en.wikipedia.org/wiki/Heaviside_step_function). However, i'm not sure that I understand your second example. If N=2^n (assuming 2 to the power of n, not 2 xor n) then one would expect the sequence to be 1, 2, 4, 8, 16, etc. If N=2^n using the C xor notation, then we would expect a sequence 3, 0, 1, 6, 7, 4, etc Your third example is pure evil, and I'm glad that I never had you as a maths teacher ;-)