# 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}]]

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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

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

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That is not a necessary condition for the rows to converge to a symmetrical distribution. The rows may remain asymmetric, but the relative asymmetry may get smaller and smaller. –  Rahul Jan 27 '12 at 7:52
I did work where detector counts were used. The counts follow the Poisson distribution. As the number of counts gets higher the normal distribution provides a "good enough" approximation so that the normal distribution can be used instead of the Poisson because of experimental error. But there is no error in the expression of the distribution function itself. If you tried to fit the normal to the Mahonian numbers the mean and standard deviation would get converge, but the third and fourth moments would never converge. –  MaxW Jan 27 '12 at 18:00
In order to get convergence you'd have to chop a symmetric part of the Mahonian numbers out of a whole row. So the Mahonian numbers would better approximate a skewed distribution like the Poisson distribution. –  MaxW Jan 27 '12 at 18:07
Just noticed from comments and looking up sequence that the function proposed is not actually the Mahonian numbers which are symmetric. –  MaxW Jan 27 '12 at 18:22
The function provided is sequence A179748. See: oeis.org/A179748 –  MaxW Jan 27 '12 at 18:25
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