Could there be a sum of polynomials for the nth prime?

Question

All I am wondering is if there could be a formula of the following form for the nth prime number:

$$\sum_{i=0}^n f(i,n)$$

where f(i,n) is a series of terms only featuring a constant and integer powers of n and i.

Answer 1

By grouping terms with equal degrees in $i$, we can write $$f(i,n)=f_d(n)i^d+f_{d-1}(n)i^{d-1}+\dots+f_1(n)i+f_0(n)$$ for $f_k(n)$ some polynomials in $n$. Summing this over $i$ gives $$\sum_{i=0}^n f(i,n)=\sum_{k=0}^d f_k(n)\sum_{i=0}^n i^k$$. The sum of $k$-th powers of numbers from $0$ to $n$ can be written as a polynomial $P_k(n)$, so the expression in your question is a polynomial $\sum_{k=0}^d f_k(n)P_k(n)$. As I mention in a comment, this polynomial grows either linearly or at least quadratically, which means that this cannot be an expression for the $n$-th prime, since that grows line $n\log n$.