Does this table fit the normal distribution? The Pascal triangle can be described by the recurrence:

$P(n,1)=1, k&gt;1: P(n,k) = P(n-i,k-1) + P(n-i,k)$

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:

$T(n,1)=1, k&gt;1: T(n,k) = \sum\limits_{i=1}^{k-1} T(n-i,k-1)$

starting:

$\begin{bmatrix} 1&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0 \\ 1&amp;1&amp;0&amp;0&amp;0&amp;0&amp;0 \\ 1&amp;1&amp;1&amp;0&amp;0&amp;0&amp;0 \\ 1&amp;1&amp;2&amp;1&amp;0&amp;0&amp;0 \\ 1&amp;1&amp;2&amp;3&amp;1&amp;0&amp;0 \\ 1&amp;1&amp;2&amp;5&amp;4&amp;1&amp;0 \\ 1&amp;1&amp;2&amp;6&amp;9&amp;5&amp;1 \end{bmatrix}$

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 &gt;= 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 &gt;= 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 -&gt; {0, width}, PlotRange -&gt; {0, height}, 
   InterpolationOrder -&gt; If[nn - 1 &gt;= 11, 11, nn - 1]], {nn, 1, 
   width}];
Show[%, ImageSize -&gt; Large]
No. The second table isn't symmetrical about the center value in the row. The Pascal triangle has this property.

The function provided is sequence A179748. See: http://oeis.org/A179748

I'd also point out again that this is not the Mahonian numbers as has been pointed out in the comments. The Mahonian numbers are symmetric. The Mahonian numbers are sequence A008302. See:
http://oeis.org/A008302
No. The second table isn't symmetrical about the center value in the row. The Pascal triangle has this property.

The function provided is sequence A179748. See: http://oeis.org/A179748

I'd also point out again that this is not the Mahonian numbers as has been pointed out in the comments. The Mahonian numbers are symmetric. The Mahonian numbers are sequence A008302. See:
http://oeis.org/A008302