If 0 1 1 1 2 2 3 6 then 5?

Answer 1

The question is ambiguous since it is not clear what the question means.

If there is a mapping such that

t(0) = 1

t(1) = 1

t(2) = 2

t(3) = 6

and you require t(5), then the answer is anynumber. It is easy to find a rule based on a polynomial of order 4 such that the first four numbers are 1, 1, 2 and 6 followed by 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.

Here, fitting the 3rd degree polynomial,

t(n) = (2n^3 - 3n^2 + n + 6)/6 for n = 0, 1, 2, ...

gives t(5) = 31.