Number Sequence Announcer I would like to ask if anyone could please suggest me accurate ( ! ) sequence number announcer tool. It can be web based application, working within website, or desktop software. It can be free of charge (preferable) or paid. Would be willing to pay for it but it has to announce the number correctly. There are no math formulas behind the sequence. Numbers are ordered from left to right - the first number is the oldest one and the last number is the most recent one. I would like to use it mainly for binary digits (0 or 1 only) but sometimes also for ''normal'' numbers. So far the only website I know is:
http://oeis.org
But everything is announced wrong. I am testing like 50+ situations and being aware of the most recent number (the one listed on the most right side so the last one). Obviously I am testing the said website without typing in the most recent number, even if I know it. But it is all the time giving me wrong announcement.
I am Not asking for assistance regarding accurately announcing first next number in the sequence but, to clarify again and to avoid misunderstanding, I am asking for suggestion about which other website/tool/software could I use for this. So it will tell me which number will occur next.
Your reply would be highly appreciated, thank you in advance.
 A: Claim. If you give me an arbitrary sequence, I can produce a random number and prove that it is, in fact, the next term of the sequence. Indeed, suppose your terms are $a_1, a_2, a_3, ..., a_n$. Then I choose a random number $X$. We can construct a polynomial of degree $n$ that satisfies the sequence $a_1, a_2, ..., a_n, X$; that is, a polynomial $p$ such that $p(k) = a_k$.
Proof sketch. First, we fit the points $(1, a_1)$ and $(2, a_2)$ with a line:
$$f_1(x) = (x-1)(a_2-a_1) + a_2$$
Now consider the quadratic function
$$f_2(x) = Q_1 \cdot (x-1)(x-2) + f_1(x)$$
Observe that regardless of the value of $Q_1$, $f(1) = a_1$ and $f(2) = a_2$, as needed. Consider $$f_2(3) = 2Q_1 + f_1(3)$$ which we need to equal $a_3$. Then let  $$Q_1 = \frac{a_3 - f_1(3)}2$$ and so $f_2(x)$ satisfies the sequence for $a_1, a_2, a_3$.
Now consider the cubic function
$$f_3(x) = Q_2 \cdot (x-1)(x-2)(x-3) + f_2(x)$$
We can solve for $Q_2$ such that $f_3(4) = a_4$:
$$Q_2 = \frac{a_4 - f_2(4)}6$$
Notice a pattern? We can do this indefinitely for any sequence, which means regardless of the $X$ chosen, we can create a function that gives the next term in the sequence $a_1, a_2, a_3, ..., a_n, X$.
More succinctly: Suppose your terms are $a_1, a_2, a_3, ..., a_n$. Then I choose a random number $X$. Then the polynomial that satisfies the sequence $a_1, a_2, ..., a_n, X$ is $f_n(x)$, where $f$ satisfies the recurrence relation
$$f_n(x) = \frac{a_{n+1} - f_{n-1}(n+1)}{n!} \prod_{k=1}^n (x-k) + f_{n-1}(x)$$ with the basis $$f_1(x) = (x-1)(a_2-a_1) + a_2$$

Because the next term in an arbitrary sequence can be proven to be arbitrary, a more meaningful request might be to find the next entry in a well-known sequence of numbers (e.g. 1, 1, 2, 3, 5, 8, X), and OEIS is almost certainly the best tool for this.
