# Does this table fit the normal distribution?

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]


• It seems that there is a symmetrical triangle of Mahonian numbers, as given in OEIS A008302 (scroll down to "EXAMPLE" to see the triangle). The rows of your triangle correspond to a >-shaped traversal of this one. Maybe that helps? – Rahul Jan 27 '12 at 8:09
• I intended to write that I had not mentioned the Mahonian numbers in the question, but I now noticed that I had done so. – Mats Granvik Jan 27 '12 at 10:23