Here’s a surprise: A pendulum can be made stable in its inverted position if its support is oscillated rapidly up and down.

Even more improbably, in 1993 David Acheson and Tom Mullin showed that a double and even a triple pendulum can be made to stand up vertically like this if the pivot vibrates at the right frequency.

“The ‘trick’ really did work, and it worked, in fact, far better than we could ever have imagined,” Acheson wrote in 1089 and All That. “We were quite taken aback by just how stable the inverted state could be, and provided the pendulums were kept roughly aligned with one another, we could push them over by as much as 40 degrees or so and they would still gradually wobble back to the upward vertical.”

They published their results in Nature and later appeared on the BBC program Tomorrow’s World, but their demonstration didn’t impress everyone — afterward, an outraged caller upbraided the producers for “lowering their usually high standards” and “falling prey to two tricksters from Oxford.”

Here’s the double pendulum (thanks, Eccles):

In 1951, Arthur B. Brown of Queens College noted that the number 3 can be expressed as the sum of one or more positive integers in four ways (taking the order of terms into account):

3
1 + 2
2 + 1
1 + 1 + 1

As it turns out, any positive integer n can be so expressed in 2^{n – 1} ways. Brown asked, how can this be proved?

William Moser of the University of Toronto offered this insightful solution:

Imagine the digit 1 written n times in a row. For example, if n = 4:

1 1 1 1

This is a picket fence, with n pickets and n – 1 spaces between them. At each space we can choose either to insert a plus sign or leave it blank. So that gives us n – 1 tasks to perform (i.e., making this choice for each space) and two options for each choice. Thus the total number of expressions for n as a sum is 2^{n – 1}, or, in the case of n = 4, eight:

1 1 1 1 = 4
1 + 1 1 1 = 1 + 3
1 1 + 1 1 = 2 + 2
1 1 1 + 1 = 3 + 1
1 + 1 + 1 1 = 1 + 1 + 2
1 + 1 1 + 1 = 1 + 2 + 1
1 1 + 1 + 1 = 2 + 1 + 1
1 + 1 + 1 + 1 = 1 + 1 + 1 + 1

(Pi Mu Epsilon Journal 1:5 [November 1951], 186.)

The fifth power of any one-digit number ends with that number:

0⁵ = 0
1⁵ = 1
2⁵ = 32
3⁵ = 243
4⁵ = 1024
5⁵ = 3125
6⁵ = 7776
7⁵ = 16807
8⁵ = 32768
9⁵ = 59049

11/26/2016: UPDATE, after hearing from some readers who are thinking more deeply than I am:

First, this immediately implies that any integer raised to the fifth power ends with the same digit as the original number.

Second, the same effect occurs regularly at higher powers, specifically 9, 13, 17, and x = 1 + 4n where n = {0, 1, 2, 3, …}.

Does anyone know what this rule is called? I found it in Reuben Hersh and Vera John-Steiner’s 2011 book Loving + Hating Mathematics — Eugene Wigner writes of falling in love with numbers at his school in Budapest: “After a few years in the gymnasium I noticed what mathematicians call the Rule of Fifth Powers: That the fifth power of any one-digit number ends with that same number. Thus, 2 to the fifth power is 32, 3 to the fifth power is 243, and so on. At first I had no idea that this phenomenon was called the Rule of Fifth Powers; nor could I see why it should be true. But I saw that it was true, and I was enchanted.”

I actually can’t find a rule by that name. Perhaps it goes by a different name in English-speaking countries?

12/08/2016 UPDATE: It’s a consequence of Fermat’s little theorem, as explained in this extraordinarily helpful PDF by reader Stijn van Dongen.

(Thanks to Evan, Dave, Sid, and Stijn.)

Lee Sallows has been working on a new experiment in self-reference that he calls self-descriptive squares, arrays of numbers that inventory their own contents. Here’s an example of a 4×4 square:

The sums of the rows and columns are listed to the right and below the square. These sums also tally the number of times that each row’s rightmost entry, or each column’s lowermost entry, appears in the square. So, for example, the sum of the top row is 3, and that row’s rightmost entry is 1; correspondingly, the number 1 appears three times in the square. Likewise, the sum of the rightmost column is 2, and the lowermost entry in that column, 4, appears twice in the square.

In this example this property extends to the diagonals — and, pleasingly, each sum applies to both ends of its diagonal. The northwest-southeast diagonal totals 2, and both -2 and 4 appear twice in the square. And the southwest-northeast diagonal totals 3, and both 1 and 0 appear three times.

“Easy to understand, but not so easy to produce!” he writes. “I’m still in the throes of figuring out the surprisingly complicated theory of such squares. It turns out there are just two basic squares of 3×3. One of them can be found at the centre of this 5×5 example, which is therefore a concentric self-descriptive square:”

(Thanks, Lee.)

“Eight complete perfect dovetail shuffles, breaking pack exactly in center; that is, cutting off just 26 cards each time and dropping cards from each half alternately, brings the pack to its original order.”

— T. Nelson Downs, in a letter to fellow magician Edward G. “Tex” McGuire, 1923

11/14/2016 UPDATE: Sid Hollander and Harold VanAken sent this demonstration:

Here’s what it looks like in the hands of a skilled shuffler (thanks to reader Sascha Müller):