This is a guest post by David Benjamin, who’s previously written several other guest posts on various topics.
It’s unavoidable that part of doing mathematics will always involve arithmetic: the simple calculations, additions and multiplications that so much else is built on. But the beauty of mathematics is that even these basic operations can be performed in a multitude of interesting ways.
This is the fourth in a series of guest posts by David Benjamin, exploring the secrets of Pascal’s Triangle.
If we highlight the multiples of any of the Natural numbers $\geq 2$ in Pascal’s triangle then they create a pattern of inverted triangles.
The images above are evocative of the Sierpinski sieve (also known as the Sierpinski gasket or Sierpinski’s triangle), a fractal described in 1915 by the Polish mathematician Waclaw Sierpiński (1882-1969).
Fractals are beautiful geometric shapes. Small, even down to (theoretically) infinitesimal areas of a fractal are identical to the entire shape. The Koch snowflake, generated geometrically by successive iterations on an equilateral triangle, is an example of a fractal. Julia sets and Mandelbrot sets are examples of fractals generated using recursion on complex functions. Many examples of fractals appear in nature, and the Polish-born French-American polymath Benoit Mandelbrot (1924-2010) suggested that fully developed turbulent flows are fractals.
It is a lovely surprise to discover that a simple fractal can be found inside Pascal’s triangle. It is achieved by considering all the numbers in the triangle modulo 2 – equivalent to colouring in only the multiples of 2, as in the first diagram at the top of the post. In this version, every odd number becomes $1$ and every even number becomes $0$, and by considering sufficiently many lines of the triangle, the Sierpinski pattern emerges.
If we consider the first 32 rows of the mod$(2)$ version of the triangle as binary numbers: $1, 11, 101, 1111, 10001,…$ and convert them into decimal numbers, we obtain the sequence:
\[1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, \]
\[196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, \]
\[50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295\]
Interestingly, all members of this sequence are factors of the final term, $4294967295 = 2^{32} – 1$. Since this is one less than a power of two, it’s a Mersenne number. Why the first $31$ terms are all factors of the 32nd term is difficult to summarise here but there is a thread on StackExchange discussing what happens to the pattern after the $32nd$ term.
$4294967295$ has prime factorisation $3 \times 5 \times 17 \times 257 \times 65537$. These five prime factors are Fermat numbers – numbers of the form $2^{2^{n}}+1$ – in this case with $n = 0, 1, 2, 3$ and $4$. As of the time of writing these are the only known Fermat numbers which are also prime.
These patterns in the rows of the triangle are intriguing, and my own efforts to understand them have uncovered a few other interesting discoveries – notably, that while the 32nd term is not divisible by the 33rd, the 34th term is exactly 3 times the 33rd. The pairs of terms after that seem to alternate, as they do from the start of the sequence, between a non-integer ratio and a ratio of exactly 3, which I conjecture is a pattern that will continue.
$e$ and $\pi$ are two of the most used transcendental numbers. The Swiss mathematician Leonhard Euler (1707-1783) connected them with the most beautiful equation, called Euler’s identity:
\[e^{i\pi}+1=0\]
There are many approximations connecting $e$, $\pi$ and other irrational numbers to be found here.
In 2012 Harlan J. Brothers proved that
\[\displaystyle\lim_{n\to \infty} \frac{\ \displaystyle\frac{s_{n+1}}{s_n}\ }{\displaystyle\frac{s_n}{s_{n-1}}}=e\]
where $s_n$ is the product of the numbers on row $n$ of Pascal’s triangle. The proof can be found on Cut the Knot, part of the wonderful website of Dr Ron Knott.
In 2007 Jonas Castillo Toloza discovered a connection between $\pi$ and the reciprocals of the triangular numbers (which can be found on one of the diagonals of Pascal’s triangle) by proving
\[\pi= 2 + \frac{1}{1} + \frac{1}{3} – \frac{1}{6} – \frac{1}{10} + \frac{1}{15} + \frac{1}{21} – \frac{1}{28} – \frac{1}{36} + \frac{1}{45} + \frac{1}{55} – \ldots\]
Three proofs are given on Cut the Knot.
The infinite sum of the reciprocals of the Natural numbers is called the harmonic series, $H_n$, where
$H_n = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \ldots$
The series is divergent, but it crawls its way towards infinity, and takes $15092688622113788323693563264538101449859497$ terms just to pass a total of $100$.
The harmonic series can be used to create a version of Pascal’s triangle – the series itself is placed along the two leading diagonals, and the entries are then related by each being the difference of the fraction to its left, and the one diagonally above it and to its left. For example, $\frac{1}{30} = \frac{1}{5}-\frac{1}{6}$.
Dividing the first term in the $n^{th}$ row by every other term in that row creates the $n^{th}$ row of Pascal’s triangle. The table below shows the calculations for the $5^{th}$ row:
| $\frac{1}{5}$ | $\frac{1}{20}$ | $\frac{1}{30}$ | $\frac{1}{20}$ | $\frac{1}{5}$ |
| $\frac{1}{5}\div \frac{1}{5} =1$ | $\frac{1}{5}\div \frac{1}{20} =4$ | $\frac{1}{5}\div \frac{1}{30} =6$ | $\frac{1}{5}\div \frac{1}{20} =4$ | $\frac{1}{5}\div \frac{1}{5} =1$ |
In our next post, we’ll talk about probability and statistics in Pascal’s triangle, and consider some of Pascal’s other contributions.
Watch mathematician and entrepreneur Anne-Marie Imafidon MBE explain binary numbers. Anne-Marie studied for an MSc in mathematics at Oxford University, and founded the social enterprise Stemettes to encourage more women and girls into STEM careers.
This week, Katie and Paul are blogging from the Heidelberg Laureate Forum – a week-long maths conference where current young researchers in maths and computer science can meet and hear talks by top-level prize-winning researchers. For more information about the HLF, visit the Heidelberg Laureate Forum website.
To make all the HLF attendees’ lives easier this week, the organisers have put together a marvelous app which includes the programme for the event, information about the conference, a full delegate list with a built-in messaging system, and a few other bells and whistles. There’s also an intriguing mystery, which has been eating away at me since I installed the app at the start of the conference.
Previously, we posted about Katie’s Binary Nail varnish tutorial video, and how you can use glitter (1) and no glitter (0) to encode binary messages in your nail varnish. We also posted an accompanying puzzle, stated as:
Suppose I want to paint my nails on one hand differently every day for a month – so I need to use all 31 combinations involving glitter. Assuming that a nail painted with plain varnish can have glitter added, but a nail with glitter needs to be nail-varnish-removed before it can become a plain nail again, what order do I apply the different combinations so that you minimise the amount of nail varnish remover I’ll need to use?
Here’s our take on the solution.
I’ve just posted my latest YouTube video, in which I explain how to use binary numbers to jazz up your nail varnish:
Alongside this video, I also have an associated puzzle for you to think about.
One of the many jobs we’re gradually getting round to in our new flat is that of tiling a small section of the kitchen surface, which for some reason was left blank by the original builders and all intervening owners. And what better thing to tile it with than binary numbers?