Definition: The falling factorial $(x)_n$ is defined as $(x)_n=x(x-1)\cdots(x-(n-1))$, or recursively as $(x)_0=1$ and $(x)_{n+1}= (x)_n(x-n)$.
Theorem: $(x)_{n+m}=(x)_n(x-n)_m$
Proof: Obvious.
Theorem: $(x+y)_n = \sum_{k=0}^{n}{{n \choose k} (x)_k(y)_{n-k}}$
Proof: You could prove this algebraically, but this is really a combinatorial theorem. $(x)_n$ is equal to the number of ways of selecting an ordered sequence of $n$ distinct elements out of a set of $x$ elements. Using this reasoning, a simple counting argument can prove this theorem, with the terms in the right sum representing the number of ways to choose a sequence of $k$ distinct elements from a set of $x$ elements, a sequence of $n-k$ distinct elements from a set of $y$ elements, and $n \choose k$ different ways to interleave the two sequences.Theorem:
$(x)_n(x)_m = \sum_{k=0}^{m}{{m \choose k} (n)_{m-k}(x)_{n+k}}$
Proof:
We'll use the previous theorems: $$ (x)_m=((x-n)+n)_m = \sum_{k=0}^{m}{{m \choose k} (x-n)_k(n)_{m-k}}$$ So: \[\begin{aligned} (x)_n(x)_m &= (x)_n((x-n)+n)_m \\ &= \sum {{m \choose k} (x)_n(x-n)_k(n)_{m-k}} \\ &= \sum {{m \choose k}(x)_{n+k}(n)_{m-k}} \end{aligned}\]
Theorem: Let $\operatorname{lcm}[m]=\operatorname{lcm}(1,2,3,4,...,m)$ be the least common multiple of the numbers from 1 to m.
(a) $(x)_m$ is divisible by $m!$ for all $m\geq 0$ and integer $x$.
(b) $(x+D)_m-(x)_m$ is divisible by $\frac{m!D}{\operatorname{lcm}[m]}$
(c)$\frac{(x)_m\operatorname{lcm}[m]}{m!}$ is a congruent function.
Proof:
(a)This follows directly from: $\frac{(x)_m}{m!} ={x \choose m}$
(b)
\[\begin{aligned}
(x+D)_m-(x)_m & = \sum_{k>0}{m\choose k}(D)_k(x)_{m-k} \\
& = \sum_{k>0}{m \choose k}D(D-1)_{k-1}(x)_{m-k} \\
& = \sum_{k>0}{ \frac{m!D}{k} \frac{(D-1)_{k-1}}{(k-1)!}
\frac{(x)_{m-k}}{(m-k)!} }
\end{aligned}\]
Part (a) shows that the second and third terms in the product of the summand are integers. But
$\frac{Dm!}{k}$ is divisible by
$\frac{Dm!}{\operatorname{lcm}[m]}$
for $k=1...m$, so we have shown that all terms are divisible by this term.
(c) This follows easily from (b).
Any element of $\operatorname{Cong}(\mathbb{N})$ can be written uniquely as: $$f(n)=\sum_{k=0}^{\infty} a_k \frac{\operatorname{lcm}[k]}{k!} (n)_k$$
This has interesting implications, since the product of two congruent functions is congruent, it means that the product of $f(x)=\frac{\operatorname{lcm}[n]}{n!}(x)_n$ and $g(x)=\frac{\operatorname{lcm}[m]}{m!}(x)_m$ can be written in this form.
Using our third theorem above, we know we can expand $f(x)g(x)$ as: $$f(x)g(x) = \frac{\operatorname{lcm}[m]\operatorname{lcm}[n]}{m!n!} \sum_{k=0}^m {m \choose k}(n)_{m-k} (x)_{n+k}$$ $$= \sum_{j=n}^{m+n} c_{m,n,j} \frac{\operatorname{lcm}[j]}{j!} (x)_j$$ Where $$c_{m,n,j} = {\frac{\operatorname{lcm}[m]\operatorname{lcm}[n]j!}{m!n!lcm[j]}} {m \choose {j-n}} (n)_{m+n-j}$$ $$= {\frac{\operatorname{lcm}[m]\operatorname{lcm}[n]}{lcm[j]}}{j \choose {j-n,j-m,n+m-j}}$$
That these terms are integers is not hard to prove directly.
Note that the $c_{m,n,j}$ are symmetric in $n$ and $m$, as they should be. Also, this formulta for $c_{m,n,j}$ is zero if $j<\max(m,n)$ or $j>m+n$, so we can define $c_{m,n,j}$ for all triples $m,n,j$ using this formula.
Now, if $f$ is represented by the sequence $\{a_k\}$ and $g$ is represented by the sequence $\{b_k\}$, then their product can be represented by $\{p_k\}$ where: $$p_k = \sum_{m,n} a_m b_n c_{m,n,k}$$ This is a finite sum, since $c_{m,n,k}$ is zero unless $k$ is less than $m+n$ and greater than both $m$ and $n$.
This means we can represent the ring of congruent functions by representing then as a sequence of integers, and then defining the product using this formula.