The one-in-million Crystal

Counting a Pyramid, One Layer at a Time

One of the pleasures of recreational mathematics is discovering that numbers often have ambitions beyond mere counting. Some want to arrange themselves into shapes. Others insist on occupying space. And a few—rather impertinently—do both.

Let us begin, as so many good mathematical stories do, with a handful of dots.

Triangular Numbers: Arithmetic with Good Posture

Place a single dot on the page. Beneath it, place two dots. Beneath those, three. Beneath those, four, and so forth. Continue this process (i.e., placing a consecutive number of dots below the previous ones) for as long as your patience or your paper allows. You will soon find that your dots have conspired to form an equilateral triangle.

Now count them.

The totals you obtain—after one row, after two rows, after three, after four—are called triangular numbers. They have been known since antiquity and were studied by the Pythagoreans, who delighted in the idea that numbers could look like some shape (they called them "figurative numbers"). These numbers are the arithmetic counterparts of bowling-pin arrangements and billiard-ball racks.

There is a pleasing inevitability to how triangular numbers grow. Each new triangle is formed by adding a row that contains exactly one more dot than the previous row. Can you find a formula for the $n^{th}$ triangular number based on the geometry of these numbers?

From Flatland to the Third Dimension

Now, suppose, emboldened by success, you decide to stack your triangles.

Start with a single dot. On top of it, place nothing—it is already complete. Beneath it, place a triangle of three dots. Beneath that, a larger triangle of six dots. Continue, always making sure that each layer is itself perfectly triangular.

What you have constructed is a small pyramid (a tetrahedron, to be precise, whose faces are all triangles). If you count the total number of dots in the entire stack, you obtain a new and more mysterious sequence of numbers known as tetrahedral numbers.

Where triangular numbers are flat and courteous, tetrahedral numbers are spatial and slightly arrogant. They insist on occupying volume. You cannot fully appreciate them without either a sketch, a pile of marbles, or a tolerant desk.

A Pattern Built from Patterns

There is a delightful recursive quality to tetrahedral numbers. Each one is formed by taking everything that came before and adding a new triangular layer on the bottom. In other words, tetrahedral numbers are built out of triangular numbers.

This nesting of ideas, numbers made of shapes made of numbers, is exactly the sort of thing that caused Lewis Carroll to write letters to mathematicians and led Charles Dodgson to invent logic puzzles that still torment undergraduates.

If you list the tetrahedral numbers in order, you may notice that the differences between them are not random. They whisper the sequence of triangular numbers. The mathematics is doing what it often does best: repeating itself in a slightly higher dimension.

Cannonballs and Other Serious Matters

The classic illustration of tetrahedral numbers involves cannonballs stacked in a pyramid—a favorite problem in old mathematical texts and a surprisingly common sight in Renaissance paintings. (One suspects artists enjoyed the excuse to paint spheres.)

How many cannonballs are required to build such a stack with a given number of layers? The answer, of course, is a tetrahedral number. The same question arises when stacking oranges, marbles, or when one is feeling especially abstract points in space.

Nature, too, seems fond of this arrangement. Certain crystal lattices and molecular structures adopt tetrahedral patterns, as if matter itself has an appreciation for tidy arithmetic.

A Puzzle, Not a Prescription

At this point, some readers will begin asking for a formula. Resist the urge.

Tetrahedral numbers are best met gradually, like a stranger at a dinner party who turns out to be far more interesting than expected. You can understand them perfectly well by thinking in layers, by counting carefully, and by observing how each new number grows from the last.

As a challenge, try this:

A crystal consists of $100,000,000$ layers of atoms, such that there is $1$ atom in the first layer, $3$ in the second, $6$ in the third, $10$ in the fourth, $15$ in the fifth, and so forth, as illustrated below. Exactly how many atoms are there in the entire crystal?$^{1}$

Solution

Our objective is to determine the total number of atoms in the entire crystal, rather than just the number of atoms in a specific layer. When the first layer is placed on top of the second layer, the second on the third, the third on the fourth, and so on, the resulting geometric structure is a , which is equivalent to a (See Fig.2). A number is classified as tetrahedral if it represents the total number of spheres arranged to form a tetrahedron. It's important to note that the number $1$ is also considered tetrahedral. Moreover, each successive tetrahedral number or stack of atoms is obtained simply by adding a greater triangular base to the sum. This is a crucial observation, since it entails that any given tetrahedral number is nothing but the sum of , for which young famously provided a well-known formula (which can be proved by ).

Fig.2: The left side of this diagram shows the formation of the tetrahedral stack on the right side.

We can examine this concept more closely by looking at the concept of triangular numbers. A triangular number counts objects (in this case, our spherical atoms) arranged in an equilateral triangle. The first five consecutive triangular numbers are $1$, $3$, $6$, $10$, and $15$ (See Fig. 1). Following the analogy given in the problem, note that the $n^{th}$ triangular number is the number of atoms in the triangular arrangement with $n$ atoms on each side, and is equal to the sum of the $n$ integers from $1$ to $n$ (Start by adding the atoms from the top vertex down to the base of the triangle until all the atoms are summed together). Numerically, this pattern is made evident by decomposing the first five tetrahedral numbers in sums of triangular numbers, and these in sums of consecutive numbers, as follows:

$$1=1$$

$$4=1+3=1+(1+2)$$
$$10=1+3+6=1+(1+2)+(1+2+3)$$
$$20=1+3+6+10=1+(1+2)+(1+2+3)+(1+2+3+4)$$
$$35=1+3+6+10+15=1+(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)$$

Since the formula for the nth triangular number is $\frac{n(n+1)}{2}$, we can express the cumulative sum up to a crystal layer $a_{n}$ as

$$a_{n}=\sum_{k=1}^{n}\frac{k(k+1)}{2}=\frac{1(1+1)}{2}+\frac{2(2+1)}{2}+...+\frac{n(n+1)}{2}$$

$$=\frac{1}{2}\left( 1\cdot2+2\cdot3+...+n(n+1) \right)=\frac{1}{2}\left( \frac{n(n+1)(n+2)}{3} \right)\quad^{1}$$

$$= \frac{n(n+1)(n+2)}{6} $$

1. The formula for the sum $1\cdot2+2\cdot3+...+n(n+1)$ can be proved by mathematical induction. Strictly speaking, this development is not necessary, but it is how I happened to arrive at the correct expression.

Now, we may wish to verify the correctness of the explicit formula we have just found. To do this, we need to make sure that it satisfies the sequence of tetrahedral numbers (the number of atoms $n^{th}$) layer for any arbitrary term, always remembering that this sequence is just the sum of the first $k$ consecutive triangular numbers.

Proof of Correctness

Let $a_{1}, a_{2}, a_{3}, ...$ be the sequence defined by specifying that $a_{1}=1$ and $a_{k}=a_{k-1}+\frac{k(k+1)}{2}$ for all integers $k\ge2$, and let $P(n)$ be the equation

$$a_{n}=\frac{n\left( n+1 \right)\left( n+2 \right)}{6}$$

We will use mathematical induction to prove that for all integers $n\ge1$, $P(n)$ is true.

Basis step; $P(1)$ is true:

To establish $P(1)$, we must show that

$$a_{1}=\frac{1\cdot \left( 1+1 \right)\left( 1+2 \right)}{6}$$

But the left-hand side of $P(1)$ is

$$a_{1}=1$$

and the right-hand side of $P(1)$ is

$$\frac{1\cdot \left( 1+1 \right)\left( 1+2 \right)}{6}=\frac{\left( 2 \right)\left( 3 \right)}{6}=\frac{6}{6}=1$$

Thus, the two sides of $P(1)$ are equal to the same quantity, and hence $P(1)$ is true.

Inductive step; for all integers $k\ge1$, if $P(k)$ is true, then $P(k+1)$ is also true:

Suppose that $k$ is any integer with $k\ge1$ such that

$$a_{k}=\frac{k\left(k+1 \right)\left(k+2 \right)}{6} \:\:\gets P(k)\:\:\text{inductive hypothesis}$$

We must show that

$$a_{k+1}=\frac{(k+1)\left(k+2 \right)\left(k+3 \right)}{6}\:\:\gets P(k+1)$$

But the left-hand side of $P(k+1)$ is

$$a_{k+1}=a_{(k+1)-1}+\frac{(k+1)(k+2)}{2} \:\:\:\:\:\: \text{by definition of} \:\:a_{1}, a_{2}, a_{3}...$$

$$=a_{k}+\frac{(k+1)(k+2)}{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$

$$=\frac{k\left(k+1 \right)\left(k+2 \right)}{6}+\frac{(k+1)(k+2)}{2} \:\:\:\:\:\: \text{by substitution from the inductive hypothesis} $$

$$=\frac{k\left(k+1 \right)\left(k+2 \right)}{6}+\frac{(k+1)(k+2)}{2}\:\:\:\:\:\: $$

$$=(k+1)(k+2)\left( \frac{k}{6}+ \frac{1}{2}\right)$$

$$=(k+1)(k+2)\left( \frac{k}{6}+ \frac{3}{6}\right)$$

$$=\frac{(k+1)\left(k+2 \right)\left(k+3 \right)}{6}$$

which equals the right-hand side of $P(k+1)$. As was to be shown, which leads us to our final calculation. The total number of atoms in the entire crystal $A$ is, therefore

$$A=\frac{100,000,000(100,000,001)(100,000,002)}{6}=166,666,671,666,666,700,000,000$$

The result of this calculation seems quite impressively large, but it's only approximately $27.68\% $ of . For quantities of substances, compounds, crystals, and the atomic world, this is a hyperbole of understatement!

1. From Hoeflin, R. K. (1985, April). Problem 39 in The Mega Test. , 7(4), p. 132.

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