Given non-consecutive points in a sequence find relationship Given the following values $236, 290, 1420$ and their positions $12, 13, 20$ in an unknown sequence. Is it possible to find the starting value and (non-linear) interval?
Edit:
If I also at that the $0$th value is $0$ and the interval always increases, does that narrow it?
Edit 2:
To explain I am trying to reverse-engineer something. So far I've found the following $0, 20, 25, 236, 290, 370, 1420$ for positions $0, 1, 2, 12, 13, 14, 20$. The $0$th value may not fit the sequence however.
 A: If you know that the interval always increases, and always increases by the same amount, then you can find it. For then you can say that the $i$th interval, $g(i) = a + bi$ for some unknown $ a$ and $b$, or better still for this problem, write
$$
g(i) = a + b(i-13)
$$
Then you know that 
$$
g(13) = a = 290 - 236 = 54
$$
and that 
$$
g(13) + g(14) + \ldots + g(20) = 1420 - 236 = 1184. 
$$
On the other hand, you also have 
$$
g(13) + g(14) + \ldots + g(20) = a + (a+b) + \ldots + (a+7b) = 8a + 28b
$$
So 
$$
8a + 28b = 1184 \\
a = 54\\
432 + 28b = 1184 \\
28b = 752\\ 
b \approx 26.86
$$
Not a very satisfying result. :(
It's probably also not consistent with the claim that the 0th value is 0....I didn't check that. 
A: If you don't need to have an undrelying algebraic structure then I think you can set the problem up as a Linear Programming problem.
Say the values are $x_0, x_1, ...$
Then your constraints are that $x_{i+1}>x_i$, that $x_{12}=236, x_{13}=290, x_{20}=1420$.
You can solve to maximise $x_0$ and to minimise $x_0$ to give you a range of values for $x_0$
