Proving an integer sequence is defined by a polynomial

Question

Let's say you have some integer sequence, $a_1, a_2, . . ., a_n$, and you can prove that there exists some $r$ such that the $r$th difference of the sequence is a sequence of just constants. How could we prove that $a_n$ can be defined by some polynomial in $n$?

By $r$th difference I mean you keep on subtracting consecutive terms from each other to form a new sequence. So, the first difference of the sequence $a_n = n^2$ (which is $1$, $4$, $9,$ $16,$ $25,$ $36,...$) would be $3$, $5$, $7,$ $9,$ $11,...$ The second difference of $a_n$ would be $2,$ $2,$ $2,$ $2,...$

The Brilliant article proves that "If a polynomial $f(x)$ has degree $k$, then the $k$th difference is constant," but it doesn't prove the converse. What's the theorem that shows the converse of the statement is true? — Eli Yablon, May 18 '23 at 00:12
Did you see how to "reconstruct the polynomial" using a row of the difference table, alongside the proof after it? In particular, $ \begin{array} {r c l l l } f(n) = & & \frac{D_{d}(1)}{d!} & \times & (x-1)(x-2)\ldots (x-d+1) (x-d) \ & + & \frac{D_{d-1}(1)}{(d-1)!} & \times & (x-1)(x-2)\ldots (x-d+1 ) \ & + & \ldots \ & + & \frac{D_1(1)}{1!} & \times & (x-1) \ & + & \frac{D_0 (1)}{0!} & \times & 1 \end{array} $ — Calvin Lin, May 18 '23 at 00:18
@eliyablon FYI, using a site search, there's Proof that common differences imply a polynomial-generated sequence, which I believe deals with your question. There's also other related questions, e.g., Use the method of finite differences to determine a formula, Help with a different approach to extracting a polynomial equation from differences, etc. — John Omielan, May 18 '23 at 00:46

Proving an integer sequence is defined by a polynomial

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