First Contact

https://commons.wikimedia.org/wiki/File:1961_CPA_2562.jpg

On April 12, 1961, witnesses saw a spaceship enter Earth’s atmosphere and descend to the ground in a ploughed field in the Leninsky Put Collective Farm near the Soviet village of Smelovka. At a height of 7 kilometers, a spaceman left the ship and drifted to earth on a parachute. The spaceman later reported:

As I stepped on the firm soil, I saw a woman and a girl. They were standing beside a spotted calf and gazing at me with curiosity. I started walking towards them and they began walking towards me. But the nearer they got to me the slower their steps became. I was still wearing my flaming orange spacesuit and they were probably frightened by it. They had never seen anything like it before.

‘I’m a Russian, comrades. I’m a Russian,’ I shouted, taking off my helmet.

The woman was Anna Takhtarova, wife of the local forester, and the girl, Rita, was her granddaughter.

‘Have you really come from outer space?’ she asked a little uncertainly.

‘Just imagine, I certainly have,’ I replied.

He was Yuri Gagarin, and the site would soon receive a permanent monument marking the landing place of Vostok-1.

See To Whom It May Concern.

Aronson’s Sequence

In 1982, J.K. Aronson of Oxford, England, sent this mysterious fragment to Douglas Hofstadter:

‘T’ is the first, fourth, eleventh, sixteenth, twenty-fourth, twenty-ninth, thirty-third …

The context of their discussion was self-reference, so presumably the intended conclusion of Aronson’s sentence was … letter in this sentence. If one ignores spaces and punctuation, then T does indeed occupy those positions in Aronson’s fragment; the next few terms would be 35, 39, 45, 47, 51, 56, 58, 62, and 64. The Online Encyclopedia of Integer Sequences gives a picture:

1234567890 1234567890 1234567890 1234567890 1234567890
Tisthefirs tfourthele venthsixte enthtwenty fourthtwen
tyninththi rtythirdth irtyfiftht hirtyninth fortyfifth
fortyseven thfiftyfir stfiftysix thfiftyeig hthsixtyse
condsixtyf ourthsixty ninthseven tythirdsev entyeighth
eightiethe ightyfourt heightynin thninetyfo urthninety
ninthonehu ndredfourt honehundre deleventho nehundreds
ixteenthon ehundredtw entysecond onehundred twentysixt
honehundre dthirtyfir stonehundr edthirtysi xthonehund
redfortyse cond...

But there’s a catch: In English, most ordinal adjectives (FIRST, FOURTH, etc.) themselves contain at least one T, so the sentence continually creates more work for itself even as it lists the locations of its Ts. There are a few T-less ordinals (NINE BILLION ONE MILLION SECOND), but these don’t arrange themselves to mop up all the incoming Ts. This means that the sentence must be infinitely long.

And, strangely, that throws our initial presumption into confusion. We had supposed that the sentence would end with … letter in this sentence. But an infinite sentence has no end — so it’s not clear whether we ought to be counting Ts at all!

The Lonely Runner

https://commons.wikimedia.org/wiki/File:Lonely_runner.gif
Image: Wikimedia Commons

Suppose k runners are running around a circular track that’s 1 unit long. All the runners start at the same point, but they run at different speeds. A runner is said to be “lonely” if he’s at least distance 1/k along the track from every other runner. The lonely runner conjecture states that each of the runners will be lonely at some point.

This is obviously true for low values of k. If there’s a single runner, then he’s lonely before he even leaves the starting line. And if there are two runners, at some point they’ll occupy diametrically opposite points on the track, at which point both will be lonely.

But whether it’s always true, for any number of runners, remains an unsolved problem in mathematics.

Conway’s RATS Sequence

Write down an integer. Remove any zeros and sort the digits in increasing order. Now add this number to its reversal to produce a new number, and perform the same operations on that:

conway's rats sequence

In the example above, we’ve arrived at a pattern, alternating between 12333334444 and 5566667777 but adding a 3 and a 6 (respectively) with each iteration. (So the next two sums, after their digits are sorted, will be 123333334444 and 55666667777, and so on.)

Princeton mathematician John Horton Conway calls this the RATS sequence (for “reverse, add, then sort”) and in 1989 conjectured that no matter what number you start with (in base 10), you’ll either enter the divergent pattern above or find yourself in some cycle. For example, starting with 3 gives:

3, 6, 12, 33, 66, 123, 444, 888, 1677, 3489, 12333, 44556, 111, 222, 444, …

… and now we’re in a loop — the last eight terms will just repeat forever.

Conway’s colleague at Princeton, Curt McMullen, showed that the conjecture is true for all numbers less than a hundred million, and himself conjectured that every RATS sequence in bases smaller than 10 is eventually periodic. Are they right? So far neither conjecture has been disproved.

(Richard K. Guy, “Conway’s RATS and Other Reversals,” American Mathematical Monthly 96:5 [May 1989], 425-428.)

On Foot

On May 20, 2005, to convey its size, Italian artist Gianni Motti walked the length of the nascent Large Hadron Collider, followed by a cameraman.

At an average speed of 5 kph, it took him 5 hours 50 minutes to walk all 27 kilometers of the underground ring. Today a proton covers the same distance 11,000 times in 1 second.

Motti dubbed his effort “Higgs: In Search of Anti-Motti.” I don’t think he found it.

Pandigital Pi

In the July/August 2006 issue of MIT Technology Review, Richard Hess noted that this expression:

 3 + \frac{16 - 8^{-5}}{97 + 2^{4}} \approx \pi - 3.3 \times 10^{-9}

provides a good approximation to π using each of the digits 1-9 once. He challenged readers to do better, limiting themselves to the operators +, -, ×, ÷, exponents, decimal points, and parentheses.

The best solution received was from Joel Karnofsky:

 3.14 + \left ( 7^{-.9^{-6}} + 2/8 \right )^{5} \approx \pi - 9.3 \times 10^{-11}

But Karnofsky noted that this is probably not the best possible. “Unfortunately, my estimate is that there are on the order of 1016 unique values that can be generated under the given conditions and I cannot see how to avoid checking essentially all of them to fnd a guaranteed best. With maybe a thousand computers I think this could be done in my lifetime.”

Indeed, eight months later Sergey Ioffe sent this solution, which he found “using a genetic-like algorithm applying mutations to a population of parse trees and keeping some number of best ones”:

 3 + 5^{-\left ( 7^{.1} \right )} + \frac{.49^{8}}{2^{6}}  \approx \pi - 3.8 \times 10^{-13}

Even that has now been surpassed — on the Contest Center’s ongoing pi approximation page, Oleg Vlasii offers this expression:

 \left ( \frac{2}{.98} - .3 \right ) \times \left (.4 + 5^{(7^{-.6}-.1)}  \right )  \approx \pi - 4.1 \times 10^{-14}

And it’s possible to do even better than this if zero is added as a tenth digit.

03/25/2017 UPDATE: Reader Danesh Forouhari wondered whether there’s a “unidigital” formula for pi. There is — Viète’s formula:

\displaystyle \frac{2}{\pi } = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{{2 + \sqrt{2}}}}{2} \cdot \frac{\sqrt{2 + {\sqrt{{2 + \sqrt{2}}}}}}{2} \cdots

Matched Set

It’s well known that the sum of the cubes of the first n integers equals the square of their sum:

13 + 23 + 33 + 43 + 53 = (1 + 2 + 3 + 4 + 5)2

California State University mathematician David Pagni found another case in which the sum of cubes equals the square of a sum. Take any whole number:

28

List all its divisors:

1, 2, 4, 7, 14, 28

Count the number of divisors of each of these:

1 has 1 divisor
2 has 2 divisors
4 has 3 divisors
7 has 2 divisors
14 has 4 divisors
28 has 6 divisors

Now cube these numbers and sum the cubes:

13 + 23 + 33 + 23 + 43 + 63 = 324

And sum the same set of numbers and square the sum:

(1 + 2 + 3 + 2 + 4 + 6)2 = 324

The two results are the same: The sum of the cubes of these numbers will always equal the square of their sum.

(David Pagni, “An Interesting Number Fact,” Mathematical Gazette 82:494 [July 1998], 271-273.)

03/10/2017 UPDATE: Reader Kurt Bachtold points out that this was originally discovered by Joseph Liouville, a fact that I should have recalled, as I’d written about it in 2011. (Thanks, Kurt.)

The Music Animation Machine

Berkeley software engineer Stephen Malinowski creates animated graphical scores of musical works.

“The vertical positions of the bars on the screen represent the relative pitches, while the color can represent instruments or voices, thematic material or tonality,” explains Crétien van Campen in The Hidden Sense. “When they are synchronized, the sound and image are easily linked in our perception. Musical structures like Bach’s canons or his many-voiced compositions thus become understood and accessible by means of a visual aid.”

There’s much more on Malinowski’s YouTube channel; here are some of his favorites.

For the Record

At one point in Samuel Beckett’s 1951 novel Molloy, the title character finds himself at the seaside and “lays in a store of sucking stones”:

They were pebbles but I call them stones. Yes, on this occasion I laid in a considerable store. I distributed them equally between my four pockets, and sucked them turn and turn about. This raised a problem which I first solved in the following way. I had say sixteen stones, four in each of my four pockets these being the two pockets of my trousers and the two pockets of my greatcoat. Taking a stone from the right pocket of my greatcoat, and putting it in my mouth, I replaced it in the right pocket of my greatcoat by a stone from the right pocket of my trousers, which I replaced by a stone from the left pocket of my trousers, which I replaced by a stone from the left pocket of my greatcoat, which I replaced by the stone which was in my mouth, as soon as I had finished sucking it. Thus there were still four stones in each of my four pockets, but not quite the same stones.

It occurs to him that this method won’t ensure that every stone is eventually sucked, and he works out a plan that will achieve this. This takes eight pages, “one of the longest and most detailed accounts of someone working at a mathematical problem in a work of fiction,” according to Richard Phillips in Numbers: Facts, Figures and Fiction.

Maddeningly, in the end Molloy throws away all the stones but one, “for they all tasted exactly the same.”