Is there a name for this kind of "Pascal's Triangle"?

I've been working on a problem and came across an interesting triangle that functions almost exactly like a Pascal Triangle. I'd like to see if I am able to find more about the properties of these triangles. I've found information on Pascal's simplexes, but that is not what this is.

Pascal's Triangle can be constructed by the fact that each cell is the sum of the $2$ above it. But I'm interested in the resulting triangles when you sum the above $3$ cells, $4$ cells, or more.

$n=3$:

$$ 1\\ 1\ \ 1\ \ 1\\ 1\ \ 2\ \ 3\ \ 2\ \ 1\\ 1\ \ 3\ \ 6\ \ 7\ \ 6\ \ 3\ \ 1\\ $$

$n=5$:

$$ 1\\ 1\ \ \ 1\ \ \ 1\ \ \ 1\ \ \ 1\\ 1\ \ \ 2\ \ \ 3\ \ \ 4\ \ \ 5\ \ \ 4\ \ \ 3\ \ \ 2\ \ \ 1\\ 1\ \ \ 3\ \ \ 6\ \ 10\ 15\ 18\ 19\ 18\ 15\ 10\ \ 6\ \ \ 3\ \ \ 1 $$

In particular what I'd like to know is how to generate an arbitrary row, like the $20$th row when $n=11$.

Solution 1:

They don't have names to my knowledge, but they're easy to calculate. For example, you can find the third row of the order-3 triangle by expanding:

$$(1 + x + x^2)^3 = 1 + 3x + 6x^2 + 7x^3 + 6x^4 + 3x^5 + x^6$$

In general, the $n$th row of the order-$k$ triangle is given by expanding

$$\left( \sum_{i=0}^{k-1} x^i \right)^n$$

and reading off the coefficients. The proof (it's not hard) is left as an exercise.

Solution 2:

The version with $n=3$ is called the "trinomial triangle"; here is its Wikipedia page. Here is a previous math.SE question about it.

In general, these triangles are calculable via multinomial coefficients. The expression $$\binom{n}{k_1,\;k_2,\;\ldots,\;k_m}=\frac{n!}{k_1!k_2!\cdots k_m!}$$ is the coefficient of $a_1^{k_1}a_2^{k_2}\cdots a_m^{k_m}$ in the expansion of $$(a_1+a_2+\cdots+a_m)^n$$ where $k_1+k_2+\cdots+k_m=n$. Then substitute $a_i=x^{i-1}$, and collect the appropriate terms.

Solution 3:

Though I rather like the previous two answers, this configuration can also be thought of as a cellular automaton. Thus, your $n=3$ version can be computed in Mathematica like so:

radius = 1;
duration = 9;
ca = CellularAutomaton[{Total[##] &, {}, radius}, {{1}, 0}, duration];
Grid[ca /. 0 -> ""]

Simply increase radius and/or duration as desired. We can generate your 20th row for $n=11$ and check it against the first, accepted answer as follows:

ca = First[CellularAutomaton[{Total[##] &, {}, 5}, {{1}, 0}, {{20}}]];
nom = CoefficientList[Expand[Sum[x^i, {i, 0, 10}]^20], x];
ca == nom
(* Out: True *)