TITLE: Is there a name for this kind of "Pascal's Triangle"?
QUESTION [14 upvotes]: 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 $n=11$.

REPLY [6 votes]: 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 *)