The Arithmetic Triangle is Analytical too

Pascal’s triangle is one of the most famous of all mathematical diagrams. It is simple to construct and rich in mathematical patterns. There is always a chance of finding something never seen before, and the discovery of new patterns is very satisfying.

Not too long ago, Harlan Brothers found Euler’s number ${e}$ in the triangle (Brothers, 2012(a),(b)). This is indeed surprising. The number ${e}$ is ubiquitous in analysis but it is far from obvious why it should turn up in the arithmetic triangle.

The Arithmetic Triangle

We are all familiar with Pascal’s Triangle, also known as the Arithmetic Triangle (AT). It is named after French mathematician Blaise Pascal. Pascal’s Traité du triangle arithmétique (Treatise on Arithmetical Triangle) was published in 1665. Of course, the triangle was known many centuries before him.

Many well-known sequences can be found in the AT. For example, the diagonals, or lines parallel to an edge, produce the constant sequence (all 1’s), natural numbers (1, 2, 3, 4, …), triangular numbers (1, 3, 6, 10, … ), tetrahedral numbers (1, 4, 10, 20, … ) and so on. Sums of the rows yield the powers of 2 (1, 2, 4, 8, … ) while slanting sums produce the Fibonacci numbers (1, 1, 2, 3, 5, … ). Many more familiar sequences can be found.

The Entries in the AT are Binomial Coefficients

Each entry in the arithmetic triangle is the sum of the two entries closest to it in the row above. The ${k}$ -th entry in row ${n}$ is the binomial coefficient ${\binom{n}{k}}$ (read ${n}$ -choose- ${k}$ ), the number of ways of selecting ${k}$ elements from a set of ${n}$ distinct elements. These binomial coefficients are defined as follows:

$\displaystyle { n \choose k } = \displaystyle{ \frac{n!}{k!(n-k)!}}$

This symbol represents the number of combinations of ${k}$ objects selected from a set of ${n}$ . The entry in the ${n}$ -th row and ${k}$ -th column of Pascal’s triangle is ${{\tbinom {n}{k}}}$ . The entry in the topmost row is ${{\tbinom {0}{0} = 1}}$ . The entire triangle can be constructed using the relationship known as Pascal’s Identity:

$\displaystyle {\displaystyle {n \choose k}={n-1 \choose k-1}+{n-1 \choose k}}$

The triangle may then be represented as in the Figure above.

Row Products Produce a Surprise

Harlan Brothers considered the products of numbers in each row:

$\displaystyle p_n = \binom{n}{0} \binom{n}{1} \binom{n}{2} \cdots \binom{n}{n} = \prod_{k=0}^n \binom{n}{k} \,.$

Using the expression of binomial coefficients in terms of factorials, he easily deduced an explicit expression for the product:

$\displaystyle p_n = (n!)^{n+1}\prod_{k=0}^n (k!)^{-2}$

This gives the rapidly-growing sequence ${1, 1, 2, 9, 96, 2500, \dots}$ . He then considered the ratios of successive terms, ${r_n = p_n/p_{n-1}}$ , which also grow rapidly. However, when he considered the second order ratios ${r_n/r_{n-1}}$ , he found that they converged towards a finite limit. By direct computation, he discovered that ${r_{n} = n^n/n!}$ and so

$\displaystyle \frac{r_{n+1}}{r_{n}} = \frac{ {n^{n+1}}/{(n+1)!} }{ n^{n}/n!}$

which leads to

$\displaystyle \frac{ p_{n+1}/p_{n} }{ p_{n}/p_{n-1} } = \left( 1 + \frac{1}{n} \right)^{n}$

Of course, the limit of this expression is well known:

$\displaystyle \lim_{n\rightarrow\infty} \left[\frac{ p_{n+1}/p_{n} }{ p_{n}/p_{n-1} } \right] = \lim_{n\rightarrow\infty} \left( 1 + \frac{1}{n} \right)^{n} = e \,.$

This is the remarkable result discovered by Harlan Brothers. His investigations along these lines are continuing, and we look forward to further results in due course.

Sources

${\bullet}$ Brothers, Harlan J., 2012(a): Math Bite: Finding ${e}$ in Pascal’s Triangle. Math. Mag. 85, pg 51. \url{doi:10.4169/math.mag.85.1.51}

${\bullet}$ Brothers, Harlan J., 2012(b): Pascal’s triangle: the hidden stor-e. Math. Gazette, March 2012, 145–148.

${\bullet}$ Brothers, Harlan J., Publications.

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