Can an arbitrary series of natural numbers always be derived from an algebraic expression? Is there always an algebraic expression from which we can derive any arbitrary and arbitrary large series of natural numbers (greater than zero)? For instance, if we have $1, 2, 4, 8, 16$ a possible expression is $\frac{(2^n)}{2}$ (starting from n=1). Now, let's suppose we have all the natural exponents all the way to $2^{10^6}$ as a series, but the next number is $2^{\frac{(10^6)+1))}{2}}$-3 or that $2^{10^4}$ is missing or that the values are starting to decrease after that. Can we be absolutely certain that there has to be an algebraic expression to derive this series from? Note that I'm not talking about an infinitely large series, since this is not defined. I am talking about an arbitrary large series with any natural numbers. We need an expression to derive an infitely large series from. Thus, it logically follows that if this assumption (I do assume this is the case) is correct there has to be an inifite number of possible algebraic expressions for all such series. Has this been proven true or false? 
A closely related question is whether we know for a fact that there always has to be a polynomial equation for which a given series of natural numbers are the only roots? If this is the case, is this also true for all algebraic numbers? 
 A: There are uncountably many sequences of natural numbers (as many as real numbers), but only countably many formulas that you can write down.
Therefore, there must be uncountably many sequences of natural numbers that have no explicit formula.
A: There is countably many arbitrarily large but finite sequences, and surely there are formulas for each of them. Use the Lagrange interpolation polynomial, for instance.
There are uncountably many infinite sequences, so they cannot be all covered by formulas (written using finitely many different symbols, as we humans do), because there are only countably many such formulas.
A: Concerning your final question (and assuming that by “series” what you mean is “finite set”), the answer is affirmative. If $k_1,\ldots,k_n\in\mathbb N$, then these numbers are the roots of the polynomial$$P(x)=(x-k_1)(x-k_2)\ldots(x-k_n),$$which has no other roots.
And please don't post several questions as a single one.
A: To avoid overlap with other answers, I'm assuming the following:


*

*As an input you're given a finite list of natural numbers- let's say there are $m$ of them in total: $c_1, c_2, \dots, c_m$.

*You want to write down a closed form expression in one variable, $i$, which is a natural number, that takes on each of the values in this finite list as its first $m$ terms.


Indeed this is possible, and furthermore we can always solve it with a polynomial of degree $m-1$. To show this, first write the polynomial:
$$a_{m-1}i^{m-1} + a_{m-2}i^{m-2} + \dots + a_0$$
Now, plug in the values for $i = 1, 2, 3, \dots, m$, and set the RHS equal to $c_i$, and you'll get a set of simultaneous equations i.e. an $m$ by $m$ matrix which you can solve via linear algebra.
Note: I don't think any of your rows will be linearly dependent, but even if some were, that's not a problem! It just means you can do it with an even lower degree polynomial.
