<s>[INST] Product rule for simplex numbers The $n$th triangular number is defined as $T_2(n) = n(n+1)/2$, and there is an interesting product rule for triangular numbers: 
$$T_2(mn) = T_2(m)\,T_2(n) + T_2(m-1)\,T_2(n-1).$$
The tetrahedral numbers $T_3(n) = n(n+1)(n+2)/6$ satisfy a similar product rule:
$$T_3(mn) = T_3(m)\,T_3(n) + 4 T_3(m-1)\,T_3(n-1) + T_3(m-2)\,T_3(n-2).$$
Can these product rules be generalized to higher dimensions? Are there analogous formulas for $T_k(mn)$, where $$T_k(n) = \frac1{k!} n(n+1)(n+2)\cdots(n+k-1)\ ?$$
 [/INST] By denoting with $(n)_k$ the falling Pochhammer symbol
$$(n)_k = n\cdot(n-1)\cdot\ldots\cdot(n-k+1)$$
we have:
$$ (mn+1)_2 = \frac{1}{2}\left((m+1)_2(n+1)_2+(m)_2(n)_2\right)\tag{1}$$
that leads to the first identity. Since:
$$ (mn+2) = \alpha(n+2)(m+2)+\beta(n-1)(m-1) $$
is solved by $\alpha=\frac{1}{3}$ and $\beta=\frac{2}{3}$ and
$$ (mn+2) = \gamma(n+1)(m+1)+\delta(n-2)(m-2)$$
is solved by $\gamma=\frac{2}{3}$ and $\delta=\frac{1}{3}$, multiplying both sides of $(1)$ by $(mn+2)$ we get:
$$ (mn+2)_3 = \frac{1}{6}\left((m+2)_3(n+2)_3+4(m+1)_3(n+1)_3+(m)_3(n)_3\right)\tag{2}$$
that leads to the second identity. 
Trying to reproduce the same argument, we may look for $\alpha$ and $\beta$ such that:
$$(mn+3)=\alpha(m+3)(n+3)+\beta(m-1)(n-1),$$
so equating the coefficients of $mn$ we get $\alpha+\beta=1$ and equating the coefficients of $(m+n)$ we get $3\alpha-\beta=0$, from which $\alpha=\frac{1}{4},\beta=\frac{3}{4}$. The constant terms match since $\frac{9}{4}+\frac{3}{4}=3$. So we go on and try to find $\gamma,\delta$ such that:
$$(mn+3)=\gamma(m+2)(n+2)+\delta(m-2)(n-2).$$
However, by choosing $\gamma=\delta=\frac{1}{2}$ the constant term does not match. Ouch. This gives that the formula for pentatope numbers depends on tetrahedral numbers, too. For istance, by multiplying both sides of $(2)$ by $(mn+3)$ we get:
$$\begin{array}{rcl}(mn+3)_4 &=& \frac{1}{24}\left((m+3)_4(n+3)_4+11(m+2)_4(n+2)_4\right.
\\&+&\left.11(m+1)_4(n+1)_4+(m)_4(n)_4\color{red}{-16(m+1)_3(n+1)_3}\right).\end{array}\tag{3}$$
If we don't care introducing further terms, we can simply go on as we did for producing $(3)$ from $(2)$.
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