The Pascal triangle can be described by the recurrence:
$P(n,1)=1, k>1: P(n,k) = P(n-i,k-1) + P(n-i,k)$
This well known triangle has the basic properties that the ratios of consecutive anti-diagonal sums (Fibonacci numbers) tend to the golden ratio, and that the ratios of consecutive row sums certainly tend to $2$.
By the central limit theorem, wikipedia quote: "When divided by $2^n$, the nth row of Pascal's triangle becomes the binomial distribution in the symmetric case where $p = 1/2$. By the central limit theorem, this distribution approaches the normal distribution as n increases."
Another triangle with the same golden ratio - and consecutive rows sums ratio tending to $2$ properties, albeit with somewhat slower convergence, is the cumulative column sums of the Mahonian numbers with the recurrence:
$T(n,1)=1, k>1: T(n,k) = \sum\limits_{i=1}^{k-1} T(n-i,k-1)$
starting:
$\begin{bmatrix} 1&0&0&0&0&0&0 \\ 1&1&0&0&0&0&0 \\ 1&1&1&0&0&0&0 \\ 1&1&2&1&0&0&0 \\ 1&1&2&3&1&0&0 \\ 1&1&2&5&4&1&0 \\ 1&1&2&6&9&5&1 \end{bmatrix}$
Does this later table also by some argument fit the normal distribution as n gets large?
Arguments against this seems to be that the right-hand half of the table gets bigger and bigger while the left-hand side remains constant tending to the factorial numbers.
On the other hand by plotting the values of the first few rows it looks like a normal distribution.
Edit 4.2.2012:
The Mathematica 8 program for the table is:
Clear[T]; nn = 15; T[n_, 1] = 1; T[n_, k_] := T[n, k] = If[n >= k, Sum[T[n - i, k - 1], {i, 1, k - 1}], 0] MatrixForm[Table[T[n, k], {n, nn}, {k, nn}]]
Edit 20.10.2014:
Clear[T]; width = 20 height = 35000; T[n_, 1] = 1; T[n_, k_] := T[n, k] = If[n >= k, Sum[T[n - i, k - 1], {i, 1, k - 1}], 0] Table[ListLinePlot[Flatten[Table[T[n, k], {n, nn, nn}, {k, width}]], DataRange -> {0, width}, PlotRange -> {0, height}, InterpolationOrder -> If[nn - 1 >= 11, 11, nn - 1]], {nn, 1, width}]; Show[%, ImageSize -> Large]