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Any number that you choose can be the nth number for any given n. It is easy to find a rule based on a polynomial of order 4 such that the first four numbers are as listed in the question and the nth is the chosen next number. There are also non-polynomial solutions. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.

The simplest soution, based on a polynomial of order 3, is

t(n) = (n^2 - 3*n + 2)/2 or (n - 1)*(n - 2)/2 for n = 1, 2, 3, ...

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Q: What is the nth term for 0 0 1 3?
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