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# Blog

2019-09-15
Today, I wrote this LaTeΧ document. It only uses default LaTeΧ functionality, and uses no packages.
If you compile it (twice so that cross references are not ??s), you will get a four page document (pages numbered i to iv), that repeatedly says: "This document ends on page v."
This document ends on page iv and says: "This document ends on page v."
If you then recompile it, you will get a five page document (pages numbered i to v), that repeatedly says: "This document ends on page iv."
This document ends on page v and says: "This document ends on page iv."
Repeatedly compiling it will alternately give these two results, and so the process of compiling this LaTeX document does not converge.
Tags: latex

### Similar posts

 realhats TMiP 2019 treasure punt Big Internet Math-Off stickers 2019 Proving a conjecture

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2019-09-01
This week, I've been in Cambridge for Talking Maths in Public (TMiP). TMiP is a conference for anyone involved in—or interested in getting involved in—any sort of maths outreach, enrichment, or public engagement activity. It was really good, and I highly recommend coming to TMiP 2021.
The Saturday morning at TMiP was filled with a choice of activities, including a treasure punt (a treasure hunt on a punt) written by me. This post contains the puzzle from the treasure punt for anyone who was there and would like to revisit it, or anyone who wasn't there and would like to give it a try. In case you're not current in Cambridge on a punt, the clues that you were meant to spot during the punt are given behing spoiler tags (hover/click to reveal).

### Instructions

Each boat was given a copy of the instructions, and a box that was locked using a combination lock.
Locked boxes.
If you want to make your own treasure punt or similar activity, you can find the LaTeX code used to create the instructions and the Python code I used to check that the puzzle has a unique solution on GitHub. It's licensed with a CC BY 4.0 licence, so you can resuse an edit it in any way you like, as long as you attribute the bits I made that you keep.

### The puzzle

Four mathematicians—Ben, Katie, Kevin, and Sam—each have one of the four clues needed to unlock a great treasure. On a sunny/cloudy/rainy/snowy (delete as appropriate) day, they meet up in Cambridge to go punting, share their clues, work out the code for the lock, and share out the treasure. One or more of the mathematicians, however, has decided to lie about their clue so they can steal all the treasure for themselves. At least one mathematician is telling the truth. (If the mathematicians say multiple sentences about their clue, then they are either all true or all false.)
They meet at Cambridge Chauffeur Punts, and head North under Silver Street Bridge. Ben points out a plaque on the bridge with two years written on it:
"My clue," he says, "tells me that the sum of the digits of the code is equal to the sum of the digits of the earlier year on that plaque (the year is 1702). My clue also tells me that at least one of the digits of the code is 7."
The mathematicians next punt under the Mathematical Bridge, gasping in awe at its tangential trusses, then punt along the river under King's College Bridge and past King's College. Katie points to a sign on the King's College lawn near the river:
"See that sign whose initials are PNM?" says Katie. "My clue states that first digit of the code is equal to the number of vowels on that sign (The sign says "Private: No Mooring"). My clue also tells me that at least one of the digits of the code is 1."
They then reach Clare Bridge. Kevin points out the spheres on Clare Bridge:
"My clue," he says, "states that the total number of spheres on both sides of this bridge is a factor of the code (there are 14 spheres). My clue also tells me that at least one of the digits of the code is 2." (Kevin has not noticed that one of the spheres had a wedge missing, so counts that as a whole sphere.)
They continue past Clare College. Just before they reach Garret Hostel Bridge, Sam points out the Jerwood Library and a sign showing the year it was built (it was built in 1998):
"My clue," she says, "says that the largest prime factor of that year appears in the code (in the same way that you might say the number 18 appears in 1018 or 2189). My clue also says that the smallest prime factor of that year appears in the code. My clue also told me that at least one of the digits of the code is 0."
They then punt under Garret Hostel Bridge, turn around between it and Trinity College Bridge, and head back towards Cambridge Chauffeur Punts. Zut alors, the lies confuse them and they can't unlock the treasure. Can you work out who is lying and claim the treasure for yourself?

### The solution

The solution to the treasure punt is given below. Once you're ready to see it, click "Show solution".

### Similar posts

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2019-07-03
This year's Big Internet Math-Off is now underway with 15 completely new contestants (plus one returning contender). As I'm not the returning contestant, I haven't been spending my time preparing my pitches. Instead, I've spent my time making an unofficial Big Internet Math-Off sticker book.
To complete the sticker book, you will need to collect 162 different stickers. Every day, you will be given a pack of 5 stickers; there are also some bonus packs available if you can find them (Hint: keep reading).

### How many stickers will I need?

Using the same method as I did for last year's World Cup sticker book, you can work out that the expected number of stickers needed to finish the sticker book:
If you have already stuck $$n$$ stickers into your album, then the probability that the next sticker you get is new is
$$\frac{162-n}{162}.$$
The probability that the second sticker you get is the next new sticker is
$$\mathbb{P}(\text{next sticker is not new})\times\mathbb{P}(\text{sticker after next is new})$$ $$=\frac{n}{162}\times\frac{162-n}{162}.$$
Following the same method, we can see that the probability that the $$i$$th sticker you buy is the next new sticker is
$$\left(\frac{n}{162}\right)^{i-1}\times\frac{162-n}{162}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find a new one:
$$\sum_{i=1}^{\infty}i \left(\frac{162-n}{162}\right) \left(\frac{n}{162}\right)^{i-1} = \frac{162}{162-n}$$
Therefore, to get all 162 stickers, you should expect to buy
$$\sum_{n=0}^{161}\frac{162}{162-n} = 918 \text{ stickers}.$$
Using just your daily packs, it will take you until the end of the year to collect this many stickers. Of course, you'll only need to collect this many if you don't swap your duplicate stickers.

### How many stickers will I need if I swap?

To work out the expected number of stickers stickers you'd need if you swap, let's first think about two people who want to complete their stickerbooks together. If there are $$a$$ stickers that both collectors need and $$b$$ stickers that one collector has and the other one needs, then let $$E_{a,b}$$ be the expected number of stickers they need to finish their sticker books. The next sticker they get could be one of three things:
• A sticker they both need (with probability $$\frac{a}{162}$$);
• A sticker one of them needs (with probability $$\frac{b}{162}$$);
• A sticker they both have (with probability $$\frac{162-a-b}{162}$$).
Therefore, the expected number of stickers they need to complete their sticker books is
$$E_{a,b}=1+\frac{a}{162}E_{a-1,b+1}+\frac{b}{162}E_{a,b-1}+\frac{162-a-b}{162}E_{a,b}.$$
This can be rearranged to give
$$E_{a,b}= \frac{162}{a+b}+ \frac{a}{a+b}E_{a-1,b+1} +\frac{b}{a+b}E_{a,b-1}$$
We know that $E_{0,0}=0$ (as if $$a=0$$ and $$b=0$$, both collectors have already finished their sticker books). Using this and the formula above, we can work out that
$$E_{0,1}=162+E_{0,0}=162$$ $$E_{1,0}=162+E_{0,1}=324$$ $$E_{0,2}=\frac{162}2+E_{0,1}=243$$ $$E_{1,1}=\frac{162}2+\frac12E_{0,2}+\frac12E_{1,0}=364.5$$
... and so on until we find that $$E_{162,0}=1269$$, and so our collectors should expect to collect 634 stickers each to complete their sticker books.
For three people, we can work out that if there are $$a$$ stickers that all three need, $$b$$ stickers that two need, and $$c$$ stickers that one needs, then
$$E_{a,b,c} = \frac{162}{a+b+c}+ \frac{a}{a+b+c}E_{a-1,b+1,c} +\frac{b}{a+b+c}E_{a,b-1,c+1} +\frac{c}{a+b+c}E_{a,b,c-1}.$$
In the same way as for two people, we find that $$E_{162,0,0}=1572$$, and so our collectors should expect to collect 524 stickers each to complete their sticker books.
Doing the same thing for four people gives an expected 463 stickers required each.
After four people, however, the Python code I wrote to do these calculations takes too long to run, so instead I approximated the numbers by simulating 500 groups of $$n$$ people collecting stickers, and taking the average number of stickers they needed. The results are shown in the graph below.
The red dots are the expected values we calculated exactly, and the blue crosses are the simulated values. It looks like you'll need to collect at least 250 stickers to finish the album: in order to get this many before the end of the Math-Off, you'll need to find 20 bonus packs...
Of course, these are just the mean values and you could get lucky and need fewer stickers. The next graph shows box plots with the quartiles of the data from the simulations.
So if you're lucky, you could complete the album with fewer stickers or fewer friends.
As a thank you for reading to the end of this blog post, here's a link that will give you two bonus packs and help you on your way to the 250 expected stickers...

### Similar posts

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2019-07-06
@Pat Ashforth: Thanks, fixed

Matthew
2019-07-04
Link to sticker book, in the first paragraph, does not work. It points to mathoffstickbook.com

Pat Ashforth
2019-07-04

Matthew
2019-07-03
minor typo for the 2 collector case

> and so our collectors should expect to collect 364 stickers

should be 634.

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2019-06-19
Last night at MathsJam, Peter Kagey showed me a conjecture about OEIS sequence A308092.
A308092
The sum of the first $$n$$ terms of the sequence is the concatenation of the first $$n$$ bits of the sequence read as binary, with $$a(1) = 1$$.
1, 2, 3, 7, 14, 28, 56, 112, 224, 448, 896, 1791, 3583, 7166, ...
To understand this definition, let's look at the first few terms of this sequence written in binary:
1, 10, 11, 111, 1110, 11100, 111000, 1110000, 11100000, 111000000, ...
By "the concatenation of the first $$n$$ bits of the sequence", it means the first $$n$$ binary digits of the whole sequence written in order: 1, then 11, then 110, then 1101, then 11011, then 110111, and so on. So the definition means:
• The first term is 1, as given in the definition ($$a(1)=1$$).
• The sum of the first 2 terms is the first 2 digits: $$1+10=11$$.
• The sum of the first 3 terms is the first 3 digits: $$1+10+11=110$$.
• The sum of the first 4 terms is the first 4 digits: $$1+10+11+111=1101$$.
• The sum of the first 5 terms is the first 5 digits: $$1+10+11+111+1110=11011$$.
As we know that the sum of the first $$n-1$$ terms is the first $$n-1$$ digits, we can calculate the third term of this sequence onwards using: "$$a(n)$$ is the concatenation of the first $$n$$ bits of the sequence subtract concatenation of the first $$n-1$$ bits of the sequence":
• The third term is $$110 - 11 = 11$$.
• The fourth term is $$1101 - 110 = 111$$.
• The fourth term is $$11011 - 1101 = 1110$$.
• The fifth term is $$110111 - 11011 = 11100$$.

### The conjecture

Peter's conjecture is that the number of 1s in each term is greater than or equal to the number of 1s in the previous term.
I'm going to prove this conjecture. If you'd like to have a try first, stop reading now and come back when you're ready for spoilers. (If you'd like a hint, read the next section then pause again.)

The third term of the sequence onwards can be calculated by subtracting the first $$n-1$$ digits from the first $$n$$ digits. If the first $$n-1$$ digits form a binary number $$x$$, then the first $$n$$ digits will be $$2x+d$$, where $$d$$ is the $$n$$th digit (because moving all the digits to the left one place in binary is multiplying by two).
Therefore the different is $$2x+d-x=x+d$$, and so we can work out the $$n$$th term of the sequence by adding the $$n$$th digit in the sequence to the first $$n-1$$ digits. (Hat tip to Martin Harris, who spotted this first.)

### Carrying

Adding 1 to a binary number the ends in 1 will cause 1 to carry over to the left. This carrying will continue until the 1 is carried into a position containing 0, and after this all the digits to the left of this 0 will remain unchanged.
Therefore adding a digit to the first $$n-1$$ digits can only change the digits from the rightmost 0 onwards.

### Endings

We can therefore disregard all the digits before the rightmost 0, and look at how the $$n$$th term compares to the $$(n-1)$$th term. There are 5 ways in which the first $$n$$ digits could end:
• $$00$$
• $$010$$
• $$01...10$$ (where $$1...1$$ is a string of 2 or more ones)
• $$01$$
• $$01...1$$ (where $$1...1$$ is again a string of 2 or more ones)
We now look at each of these in turn and show that the $$n$$th term will contain at least as many ones at the $$(n-1)$$th term.

#### Case 1: $$00$$

If the first $$n$$ digits of the sequence are $$x00$$ (a binary number $$x$$ followed by two zeros), then the $$(n-1)$$th term of the sequence is $$x+0=x$$, and the $$n$$th term of the sequence is $$x0+0=x0$$. Both $$x$$ and $$x0$$ contain the same number of ones.

#### Case 2: $$010$$

If the first $$n$$ digits of the sequence are $$x010$$, then the $$(n-1)$$th term of the sequence is $$x0+1=x1$$, and the $$n$$th term of the sequence is $$x01+0=x01$$. Both $$x1$$ and $$x01$$ contain the same number of ones.

#### Case 3: $$01...10$$

If the first $$n$$ digits of the sequence are $$x01...10$$, then the $$(n-1)$$th term of the sequence is $$x01...1+1=x10...0$$, and the $$n$$th term of the sequence is $$x01...10+1=x01...1$$. $$x01...1$$ contains more ones than $$x10...0$$.

#### Case 4: $$01$$

If the first $$n$$ digits of the sequence are $$x01$$, then the $$(n-1)$$th term of the sequence is $$x+0=x$$, and the $$n$$th term of the sequence is $$x0+1=x1$$. $$x1$$ contains one more one than $$x$$.

#### Case 5: $$01...1$$

If the first $$n$$ digits of the sequence are $$x01...1$$, then the $$(n-1)$$th term of the sequence is $$x01...1+1=x10...0$$, and the $$n$$th term of the sequence is $$x01...1+1=x10...0$$. Both these contain the same number of ones.

In all five cases, the $$n$$th term contains more ones or an equal number of ones to the $$(n-1)$$th term, and so the conjecture is true.

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2019-04-09
In the latest issue of Chalkdust, I wrote an article with Edmund Harriss about the Harriss spiral that appears on the cover of the magazine. To draw a Harriss spiral, start with a rectangle whose side lengths are in the plastic ratio; that is the ratio $$1:\rho$$ where $$\rho$$ is the real solution of the equation $$x^3=x+1$$, approximately 1.3247179.
A plastic rectangle
This rectangle can be split into a square and two rectangles similar to the original rectangle. These smaller rectangles can then be split up in the same manner.
Splitting a plastic rectangle into a square and two plastic rectangles.
Drawing two curves in each square gives the Harriss spiral.
A Harriss spiral
This spiral was inspired by the golden spiral, which is drawn in a rectangle whose side lengths are in the golden ratio of $$1:\phi$$, where $$\phi$$ is the positive solution of the equation $$x^2=x+1$$ (approximately 1.6180339). This rectangle can be split into a square and one similar rectangle. Drawing one arc in each square gives a golden spiral.
A golden spiral

### Continuing the pattern

The golden and Harriss spirals are both drawn in rectangles that can be split into a square and one or two similar rectangles.
The rectangles in which golden and Harriss spirals can be drawn.
Continuing the pattern of these arrangements suggests the following rectangle, split into a square and three similar rectangles:
Let the side of the square be 1 unit, and let each rectangle have sides in the ratio $$1:x$$. We can then calculate that the lengths of the sides of each rectangle are as shown in the following diagram.
The side lengths of the large rectangle are $$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1$$ and $$\frac1{x^2}+\frac1x+1$$. We want these to also be in the ratio $$1:x$$. Therefore the following equation must hold:
$$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1=x\left(\frac1{x^2}+\frac1x+1\right)$$
Rearranging this gives:
$$x^4-x^2-x-1=0$$ $$(x+1)(x^3-x^2-1)=0$$
This has one positive real solution:
$$x=\frac13\left( 1 +\sqrt[3]{\tfrac12(29-3\sqrt{93})} +\sqrt[3]{\tfrac12(29+3\sqrt{93})} \right).$$
This is equal to 1.4655712... Drawing three arcs in each square allows us to make a spiral from a rectangle with sides in this ratio:
A spiral which may or may not have a name yet.

### Continuing the pattern

The side lengths of the largest rectangle are $$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4}$$ and $$1+\frac2x+\frac1{x^2}+\frac1{x^3}$$. Looking for the largest rectangle to also be in the ratio $$1:x$$ leads to the equation:
$$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4} = x\left(1+\frac2x+\frac1{x^2}+\frac1{x^3}\right)$$ $$x^5+x^4-x^3-2x^2-x-1 = 0$$
This has one real solution, 1.3910491... Although for this rectangle, it's not obvious which arcs to draw to make a spiral (or maybe not possible to do it at all). But at least you get a pretty fractal:

### Continuing the pattern

We could, of course, continue the pattern by repeatedly adding more rectangles. If we do this, we get the following polynomials and solutions:
 Number of rectangles Polynomial Solution 1 $$x^2 - x - 1=0$$ 1.618033988749895 2 $$x^3 - x - 1=0$$ 1.324717957244746 3 $$x^4 - x^2 - x - 1=0$$ 1.465571231876768 4 $$x^5 + x^4 - x^3 - 2x^2 - x - 1=0$$ 1.391049107172349 5 $$x^6 + x^5 - 2x^3 - 3x^2 - x - 1=0$$ 1.426608021669601 6 $$x^7 + 2x^6 - 2x^4 - 3x^3 - 4x^2 - x - 1=0$$ 1.4082770325090774 7 $$x^8 + 2x^7 + 2x^6 - 2x^5 - 5x^4 - 4x^3 - 5x^2 - x - 1=0$$ 1.4172584399350432 8 $$x^9 + 3x^8 + 2x^7 - 5x^5 - 9x^4 - 5x^3 - 6x^2 - x - 1=0$$ 1.412713760332943 9 $$x^{10} + 3x^9 + 5x^8 - 5x^6 - 9x^5 - 14x^4 - 6x^3 - 7x^2 - x - 1=0$$ 1.414969877544769
The numbers in this table appear to be heading towards around 1.414, or $$\sqrt2$$. This shouldn't come as too much of a surprise because $$1:\sqrt2$$ is the ratio of the sides of A$$n$$ paper (for $$n=0,1,2,...$$). A0 paper can be split up like this:
Splitting up a piece of A0 paper
This is a way of splitting up a $$1:\sqrt{2}$$ rectangle into an infinite number of similar rectangles, arranged following the pattern, so it makes sense that the ratios converge to this.

### Other patterns

In this post, we've only looked at splitting up rectangles into squares and similar rectangles following a particular pattern. Thinking about other arrangements leads to the following question:
Given two real numbers $$a$$ and $$b$$, when is it possible to split an $$a:b$$ rectangle into squares and $$a:b$$ rectangles?
If I get anywhere with this question, I'll post it here. Feel free to post your ideas in the comments below.

### Similar posts

 Dragon curves II TMiP 2019 treasure punt Christmas card 2018 World Cup stickers 2018, pt. 3

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2019-05-02
@g0mrb: CORRECTION: There seems to be no way to correct the glaring error in that comment. A senior moment enabled me to reverse the nomenclature for paper sizes. Please read the suffixes as (n+1), (n+2), etc.

(anonymous)
2019-05-02
I shall remain happy in the knowledge that you have shown graphically how an A(n) sheet, which is 2 x A(n-1) rectangles, is also equal to the infinite series : A(n-1) + A(n-2) + A(n-3) + A(n-4) + ... Thank-you, and best wishes for your search for the answer to your question.

g0mrb