The first animated GIF that I ever made was made with the LaTeX package TikZ and the command line utility ImageMagick. In this post, I’ll give a quick example of how to make a simple GIF that works by layering images with transparent backgrounds on top of each other repeatedly.

TikZ code

In our first step toward making the above GIF, we’ll make a PDF where each frame contains one of the above circles on a transparent background. In this case, each circle is placed uniformly at random with its center in a \(10 \times 10\) box, with a radius in \([0,1]\), and whose color gets progressively more red and less green with a random amount of blue at each step.

\documentclass[tikz]{standalone}
\usepackage{tikz}
\begin{document}
\foreach \red[evaluate=\red as \green using {255 - \red}] in {0,1,...,255} {
\begin{tikzpicture}
\useasboundingbox (0,0) rectangle (10,10);
\pgfmathrandominteger{\blue}{0}{255}
\definecolor{myColor}{RGB}{\red,\green,\blue}
\fill[myColor] ({10*random()},{10*random()}) circle ({1+random()});
\end{tikzpicture}
}
\end{document}

Starting with \documentclass[tikz]{standalone} says to make each tikzpicture its own page in the resulting PDF.

Next we loop through values of \red from 0 to 255, each time setting \green to be equal to 255 - \red so that with each step the amount of red goes up and the amount of green goes down.

The command \useasboundingbox (0,0) rectangle (10,10); gives each frame a \(10 \times 10\) bounding box so that all of the frames are the same size and positioned the same way.

The command \pgfmathrandominteger{\blue}{0}{255} chooses a random blue values between 0 and 255.

The command \fill[myColor] ({10*random()},{10*random()}) circle ({1+random()}); places a circle with its center randomly chosen in the \(10 \times 10\) box and with a radius between \(1\) and \(2\).

ImageMagick

When we compile this code, we get a PDF with one circle on each page. In order do turn this PDF into an animated GIF, we convert the PDF using ImageMagick, a powerful command-line utility for handling images. If we named our PDF bubbles.pdf then running the following code will give us an animated GIF called bubbles.gif.

In this post, I’ll explore the math behind one of my Twitter bots, @xorTriangles. This bot was inspired by the MathOverflow question “Number triangle,” asked by user DSM posted in May 2020.

(I gave an overview of my Twitter bots @oeisTriangles in my post “Parity Bitmaps from the OEIS“. And if you want to build your own bot, I showed the steps for building @BotfonsNeedles in parts I, II, and III of my three-part series “A π-estimating Twitter bot”.)

In May 2020, MathOverflow user DSM posted a question “Number triangle” where they asked about the following construction:

Start with a list of \(n\) bits (\(0\)s and \(1\)s), and to get the next row, combine all adjacent pairs via the XOR operation, \(\oplus\). (A mathematician might call this “addition modulo \(2\).”) That is $$\begin{alignat*}{2} 0 \oplus 0 &= 1 \oplus 1 &&= 0 \quad\text{and} \\ 0 \oplus 1 &= 1 \oplus 0 &&= 1.\end{alignat*}$$

For example, starting with the row \(110101\), the resulting triangle is $$1~~~1~~~0~~~1~~~0~~~1\\ 0~~~1~~~1~~~1~~~1\\ 1~~~0~~~0~~~0\\ 1~~~0~~~0\\ 1~~~0\\ 1$$

We’re especially interested in understanding the triangles with rotationally symmetric boundaries (and consequently insides)! OEIS sequence A334556 enumerates all of the rotationally symmetric triangles with a \(1\) in the upper left corner.

@xorTriangles

My Twitter bot @xorTriangles posts these triangles four times a day, alternating between the triangles described above and a “mod \(3\)” analog. If you want to see how it works, you can check it out on Github.

Here is an illustration by Michael De Vlieger that shows many small XOR triangles and which is available on the OEIS.

Mathematica

One way to find a rotationally symmetric XOR triangle is by setting up a system of equations. Using the convention that the top row of the triangle is labeled \(x_0, x_1, \dots, x_n\), we can use the Pascal’s Triangle-like construction to set up a system of linear equations over \(\mathbb{Z}/2\mathbb{Z}\). The solutions of this system of equations are precisely those that correspond to the XOR triangle we care about! This allows us to write down a matrix whose null space is the set of such triangles:

Because this is the null space of a \(n+1 \times n+1\) matrix over \(\mathbb{Z}/2\mathbb{Z}\), we know that there must be \(2^M\) rotationally symmetric XOR triangles, where \(M\) is an integer—which isn’t obvious from the original construction!

If you’re interested in learning more about this construction, reach out to me or take a look at some of these related OEIS sequences:

A334596: Number rotationally symmetric triangles of size \(n\) with \(1\)s in the corners.

A334591: Largest triangle of \(0\)s in the XOR triangle generated by the binary expansion of \(n\).

A334592: Total number of \(0\)s in the XOR triangle generated by the binary expansion of (n).

A334593: Total number of \(1\)s in the XOR triangle generated by the binary expansion of \(n\).

A334594: Binary interpretation of the \(k\)-th row of the XOR triangle generated by the binary expansion of \(n\).

A334930: Numbers that generate rotationally symmetrical XOR-triangles featuring singleton zero bits in a hexagonal arrangement. (from Michael De Vlieger, see his illustration)

I’ve gotten a lot of mathematical inspiration from Project Euler questions, but perhaps the question that has gotten me thinking the most is Project Euler Problem 208: Robot Walks. In this problem, a robot takes steps either to the right or the left, and at each step, it turns \(\frac 15\) of the way of a circle.

Demo

I started thinking about this problem more seriously after I met Chase Meadors at the 2018 Graduate Student Combinatorics Conference and learned about his Javascript applet which allows a user to control a (virtual) robot 🤖 by using the left and right arrow keys. By repeating the same sequence of moves (e.g. \(4\) steps to the right followed by \(2\) steps to the left) I found that the robot traced out surprising symmetric patterns.

I clonedChase’s Github Repo so that I could customize the robot further. If you go to the URL https://peterkagey.github.io/project-euler-208/?n=8&w=3,2,5,1 you’ll see an example of a robot walk, where n=8 means that each step will be \(\displaystyle \frac 18\) of a circle, and w=3,2,5,1 means that the robot will follow the pattern of \(3\) steps to the right followed by \(2\) steps to the left, followed by \(5\) steps to the right, followed by \(1\) step to the left, and repeating this pattern until it returns to where it began.

Stack Exchange Questions

I’ve asked a number of questions on Math Stack Exchange (MSE) and Code Golf Stack Exchange (CGSE) about these problems.

I’ve also asked about this setup in my Openish Problem Collection in Problem 41 and in Problem 69.

@RobotWalks

Twice each day, my Twitter Bot @RobotWalks tweets a randomly generated Robot Walk cycle. Check out the Github code if you want to see how it works, and read my series onmaking aTwitter Bot if you want to make something like it for yourself.

Jessica’s Robot Walk Prints

I ordered some canvas prints of some numerologically significant walks for my friend Jessica, which she hung behind her TV. There’s no doubt that this caused her to form a deep mental association between me and The Good Place.

What’s next?

Some day, I want to laser cut or 3D print some coasters based on these designs. Reach out to me if that’s something that you’re interested in, and we can do it together!

Sloane [verb not pictured] of my sequence A337655, Peter Kagey, who must have been informed of changes to the encyclopedia, proposed this second way of accounting for addition and multiplication. This has yielded a new entry in the encyclopedia, the following A337946:

Let’s consider the fourth question. The response is negative this time: no, it’s not possible that the tables of addition and multiplication have simultaneously only a few distinct values. This is a result of Paul Erdős and Endre Szemerédi from 1983, which we state precisely here:

“There exist two constants \(C>0\) and \(e > 0\) such that, for the set \(E\) of real numbers of […]

Translation by Alec Jones

A Sum-Product Problem

The article—or at least the part that I could make sense of—is about the opposite of Erdos’s Sum-Product Problem. This article explores questions about the greatest number of different values in the corresponding addition and multiplication tables for different sequences of positive integers.

Jean-Paul’s sequence, A337655, is the lexicographically earliest (greedy) sequence such that for all \(n\) both the addition and multiplication tables of the first \(n\) terms have \(\binom{n+1}{2}\) distinct terms (the greatest number).

The addition and multiplication tables share some values. For example, they both have a \(2\), a \(4\), a \(7\), a \(10\), and so on. My sequence, A337946, is similar to Jean-Paul’s but with the additional restriction that the addition and multiplication tables cannot have any values in common.

Some related sequences

I’ve recently added some new sequences to the OEIS related to these sequences.

One way of thinking of A066720 is that it’s the lexicographically earliest infinite sequence \(S = \(a_i\)_{i=1}^{\infty}\) such that for all \(n\) the set \(S_n = \{a_i\}_{i=1}^{n}\) consisting of the first \(n\) terms of \(S\) has the property that \(S_n \times S_n = \{xy \mid x, y \in S_n\}\) has exactly \(\binom{n+1}{2}\) elements—the most possible.

A347498: minimize the largest term

If instead of minimizing the lexicographic order, we instead minimize the size of the largest element, we get OEIS sequence A347498: for each \(n\), find the least \(m\) such that there exists a subset \(T_n \subseteq \{1, 2, \dots, m\}\) with \(n\) elements such that \(|T_n \times T_n| = \binom{n+1}{2}\). \(A347498(n)\) records the value of the \(m\)s.

For example, \(A347498(8) = 11\) as illustrated by the table below.

A347499: Examples with minimized largest element

We wish to record examples of the sequences described in the above section. There are \(A348481(n)\) subsets of \(\{1,2,…,A347499(n)\}\) with the distinct product property, but we record the lexicographically earliest one in OEIS sequence A347499.

For all of the known values, the lexicographically earliest subset begins with \(1\). Is this true in general?

A347570: Sums with more terms

Perhaps instead of asking when all sums \(x + y\) are distinct for \(x \leq y\), we want to know when all sums \(x_1 + x_2 + \dots + x_n\) are distinct for \(x_1 \leq x_2\leq \dots \leq x_n\). This family of sequences are sometimes called \(B_n\) sequences and have been studied by the likes of Richard Guy.

I’ve recorded the lexicographically earliest \(B_n\) sequences in OEIS sequence A347570.

One of the earliest contributions to the On-Line Encyclopedia of Integer Sequences (OEIS) was a family sequences counting the number of words that begin (or don’t begin) with a palindrome:

Let \(f_k(n)\) be the number of strings of length \(n\) over a \(k\)-letter alphabet that begin with a nontrivial palindrome” for various values of \(k\).

Number of binary strings of length \(n\) that begin with an odd-length palindrome. (A254128)

(If I had known better, I would have published fewer sequences in favor of a table, and I would have requested contiguous blocks of A-numbers.)

I must have written some Python code to compute some small terms of this sequence, and I knew that \(g_k(n) = k^n – f_k(n)\), but I remember being at in my friend Q’s bedroom when the recursion hit me for \(f_k(n)\): $$f_k(n) = kf_k(n-1) + k^{\lceil n/2 \rceil} – f_k\big(\lceil \frac n 2 \rceil \big)$$

“Bifix-free” words

One sequence that I didn’t add to the OEIS was the “Number of binary strings of length n that begin with an even-length palindrome”—that’s because this was already in the Encyclopedia under a different name:

A094536: Number of binary words of length n that are not “bifix-free”.

A “bifix” is a shared prefix and suffix, so a “bifix-free” word is one such that all prefixes are different from all suffixes. More concretely, if the word is \(\alpha_1\alpha_2 \dots \alpha_n\), then \((\alpha_1, \alpha_2, \dots, \alpha_k) \neq (\alpha_{n-k+1},\alpha_{n-k+2},\dots,\alpha_n)\) for all \(k \geq 1\).

The reason why the number of binary words of length \(n\) that begin with an even length palindrome is equal to the number of binary words of length \(n\) that have a bifix is because we have a bijection between the two sets. In particular, find the shortest palindromic prefix, cut it in half, and stick the first half at the end of the word, backward. I’ve asked for a better bijection on Math Stack Exchange, so if you have any ideas, please share them with me!

A Zimin word can be defined recursively, but I think it’s most suggestive to see some examples:

\(Z_1 = A\)

\(Z_2 = ABA\)

\(Z_3 = ABACABA\)

\(Z_4 = ABACABADABACABA\)

\(Z_n = Z_{n-1} X Z_{n-1}\)

All Zimin words \(Z_n\) are examples of “unavoidable patterns”, because every sufficiently long string with letters in any finite alphabet contains a substring that matches the \(Z_n\) pattern.

For example the word \(0100010010111000100111000111001\) contains a substring that matches the Zimin word \(Z_3\). Namely, let \(A = 100\), \(B = 0\), and \(C = 1011\), visualized here with each \(A\) emboldened: \( 0(\mathbf{100}\,0\,\mathbf{100}\,1011\,\mathbf{100}\,0\,\mathbf{100})111000111001\).

I’ve written a Ruby script that generates a random string of length 29 and uses a regular expression to find the first instance of a substring matching the pattern \(Z_3 = ABACABA\). You can run it on TIO, the impressive (and free!) tool from Dennis Mitchell.

# Randomly generates a binary string of length 29.
random_string = 29.times.map { [0,1].sample }.join("")
p random_string
# Finds the first Zimin word ABACABA
p random_string.scan(/(.+)(.+)\1(.+)\1\2\1/)[0]
# Pattern: A B A C A B A

Why 29? Because all binary words of length 29 contain the pattern \(Z_3 = ABACABA\). However, Joshua Cooper and Danny Rorabaugh’s paper provides 48 words of length 28 that avoid that pattern (these and their reversals):

The number of Zimin words of length \(n\) that match the pattern ABA is equal to the number of of words that begin with an odd-length palindrome. Analogously, the number of words with a bifix is equal to the number of words that begin with an even-length palindrome. The number of these agree when \(n\) is odd.

This is the fourth in my installment of My Favorite Sequences. This post discusses sequence A263135 which counts penny-to-penny connections among \(n\) pennies on the vertices of a hexagonal grid. I published this sequence in October 2015 when I was thinking about hexagonal-grid analogs to the “Not Equal” grid. The square-grid analog of this sequence is A123663.

The sequences A047932 and A263135 are about placing pennies on a hexagonal grid in such a way that maximizes the number of penny-to-penny contacts, which occurs when you place the pennies in a spiral. A047932, counts the contacts when the pennies are placed on the faces of the grid; A263135 counts the contacts with the pennies placed on the vertices.

While spiral shapes maximize the number of penny-to-penny contacts, there are sometimes non-spiral shapes that have the same number of contacts. For example, in the case of the square grid, there are \(A100092(n)\) such ways to lay down \(n\) pennies on the square grid with the maximum number of connections. Problem 108 in my Open Problems Collection asks about generalizing this OEIS sequence to other settings such as the hexagonal grid.

Comparing contacts

Notice that the “face” pennies in A047932 can have a maximum of six neighbors, while the “vertex” pennies in A263135 can have a maximum of three. In the limit, most pennies are “interior” pennies with the maximum number of contacts, so \(A047932(n) \sim 3n\) and \(A263135(n) \sim \frac32n\).

Looking at the comparative growth rates, it is natural to ask how the number of connections of \(n\) face pennies compares to the number of connections of \(2n\) vertex pennies. In October 2015 I made a conjecture on the OEIS that this difference grew like sequence A216256.

I believe that the sequence A216256 on the right hand side appears to be the same as the sequence “n appears \(\displaystyle\left\lfloor \frac{2n+1}{3} \right\rfloor\) times,” but I’d have to crack open my Concrete Mathematics book to prove it.

This is Problem 20 in my Open Problem Collection, and I’ve placed a small, $5 bounty on solving this conjecture—so if you have an idea of how to prove this, let me know in exchange for a latte! I’ve asked about this in my Math Stack Exchange question Circle-to-circle contacts on the hexagonal grid—so feel free to answer there or let me know on Twitter, @PeterKagey.

This is the third installment in a recurring series, My Favorite Sequences. This post discusses OEIS sequence A278299, a sequence that took over two years to compute enough terms to add to the OEIS with confidence that it was distinct.

This sequence is discussed in Problem #23 of my Open Problems Collection, which asks for the smallest polyomino (by number of cells) whose cells you can color with \(n\) different colors such that any two different colors are adjacent somewhere in the polyomino. As illustrated below, when there are \(n=5\) colors (say, green, brown, blue, purple, and magenta) there is a \(13\)-cell polyomino which has a green cell adjacent to a blue cell and a purple cell adjacent to a brown cell and so on for every color combination. This is the smallest polyomino with the \(5\)-coloring property.

The Genesis: Unequal Chains

The summer after my third undergraduate year, I decided to switch my major to Math and still try to graduate on time. Due to degree requirements, I had to go back and take some lower-division classes that I was a bit over-prepared for. One of these classes—and surely my favorite—was Bill Bogley‘s linear algebra class, where I half-way paid attention and half-way mused about other things.

Bill wrote something simple on the board that sparked inspiration for me: $$a \neq b \neq c \neq a.$$ He wrote this to indicate that \(a\), \(b\), and \(c\) were all distinct, and this got me thinking: if we have to write a string of four variables in order to say that three variables are distinct, how many would we have to write down to say that four variables were distinct? It turns out that \(8\) will do the trick, with one redundancy: $$a\neq b \neq c \neq d \neq b \color{red}{\neq} c \neq a.$$ Five variables? \(11\): $$a_1 \neq a_2 \neq a_3 \neq a_4 \neq a_5 \neq a_3 \neq a_1 \neq a_4 \neq a_2 \neq a_5.$$ What about \(n\) variables?

My colleague and the then-President of the OSU Math Club, Tommy Pitts, made quick work of this problem. He pointed out that “not equal” is a symmetric, non-transitive, non-reflexive relation. This means that we can model this with a complete graph on \(n\) vertices, where each edge is a relation. Then the number of variables needed in the expression is the number of edges in the complete graph, plus the minimum number of Eulerian paths that we can split the graph into. Searching for this in the OEIS yields sequence A053439. $$A053439^*(n) = \begin{cases} \binom{n}{2} + 1 & n \text{ is odd} \\ \binom{n}{2} + \frac n 2 & n \text{ is even}\end{cases}$$

A Generalization: Unequal Chainmail

This was around 2014, at which time I was writing letters to my friend Alec Jones whenever I—rather frequently!—stumbled upon a new math problem that interested me. In the exchange of letters, he suggested a 2D version of this puzzle. Write the \(n\) variables in the square grid, and say that two variables are unequal if they’re adjacent.

While Tommy solved the 1D version of the problem quickly, the 2D version was much more stubborn! However we were able to make some progress. We found some upper bounds (e.g. the 1D solution) and some lower bounds, and we were able to prove that some small configurations were optimal. Finally, in November 2016, we had ten terms: enough to prove that this sequence was not in the OEIS. We added it as A278299.

\(a(n)\) is the tile count of the smallest polyomino with an \(n\)-coloring such that every color is adjacent to every other distinct color at least once.

(In May 2019, Alec’s student Ryan Lee found the \(11\)th term: \(A278299(11) = 34\). \(A278299(12)\) is still unknown.)

We found these terms by establishing some lower bounds (as explained below) and then implementing a Javascript game (which you can play here) with a Ruby on Rails backend to allow people to submit their hand-crafted attempts. Each solution was constructive proof of an upper bound, so when a user submitted a solution that matched the lower bound, we were able to confirm that term of the sequence.

(One heuristic for making minimal configurations is to start with the construction in OEIS sequence A260643 and add cells as necessary in an ad hoc fashion.)

Lower bounds

There are a few different ways of proving lower bounds.

We know that there needs to be at least \(\binom{n}{2}\) relations, one between each pair of variables. OEIS sequence A123663 gives the “number of shared edges in a spiral of n unit squares,” which can be used to compute a lower bound: $$A039823(n) = \left\lceil \frac{n^2+n+2}{4}\right\rceil$$

Every number needs to be in contact with at least (n-1) other numbers, and each occurrence can be in contact with at most (4) others. So each number needs to occur at least \(\lceil \frac{n-1}{4}\rceil\) times, for a total of \(n\lceil \frac{n-1}{4}\rceil\) occurrences. This bound is usually weaker than the above bound.

For the cases of \(n = 5\) and \(n=9\), the lower bounds were proved using ad hoc methods, by looking at how many cells would need to have a given number of neighbors.

Upper Bounds

Besides the upper bound that comes from the 1-dimensional version of the problem, that only upper bounds that I know of come from hand-crafted submissions on my Javascript game on my website.

Do you have any ideas for an explicit and efficient algorithm for constructing such solutions? If so, let me know on Twitter @PeterKagey.

Asymptotics

The lower and upper bounds show that this is asymptotically bounded between \(n^2/4\) and \(n^2/2\). It’s possible that this doesn’t have a limit at all, but it would be interesting to bound the liminf and limsup further. My intuition is that \(n^2/4\) is the right answer, can you prove or disprove this?

Generalizations

We could play this game on the triangular grid, or in the 3-dimensional cubic grid. Do you have ideas of other graphs that you could do this on?

This game came from Tommy’s analysis of looking at “not equal to” as a symmetric, non-reflexive, non-transitive relation. Can you do a similar analysis on other kinds of relations?

Is there a good way of defining what it means for two solutions to be the same? For a given number of variables, how many essentially different solutions exist? (Related: Open problem #108.)

What if we think of left-right connections as being different from up-down connections, and want both? Or what if we want each variable \(x\) to be neighbors with another \(x\)?

If you have ideas about these questions or any questions of your own, please share them with me by leaving a comment or letting me know on Twitter, @PeterKagey!

In July 2017, I added a mathematically-silly-but-visually-fun sequence, A289523. The sequence works like this: place a circle of area \(\pi\) centered at \((1,1)\), then place a circle of area \(2\pi\) centered at \((2,a(2))\) where \(a(2) = 4\), the least positive integer such that the circle does not overlap with the first circle. Next, place a circle of area \(3\pi\) centered at \((3,a(3))\) where \(a(3) = 7\) is the least positive integer such that the circle does not overlap with the first two circle. Continue this pattern ad infinitum, creating the earliest infinite sequence of positive integers such that no two circles overlap with any others, and a circle centered at \((k, a(k))\) has area \(k\pi\).

I haven’t done much mathematical analysis on this problem, but it would be interesting to see if it’s possible to compute (or put some bounds on) the packing density of the convex hull of the circles. Also, a glance at a plot of the points suggests that the sequence is bounded above by a linear function—is this the case?

The scatter plot of A289523 suggests that the centers of the circles have a linear upper bound. This is to be expected! The areas of the circles increase linearly, and the packing density is (presumably) nonzero.

What is the slope of the upper bound? And what is the packing density of these circles in the limit?

Related Construction

At the end of March, I posted a related puzzle, “Placing Circles Along a Square Spiral”, on Code Golf Stack Exchange. For the post, I made a few animated GIFs that explain the construction and tweeted about them.

Impressively, Code Golf Stack Exchange users tsh, Arnauld, and A username each wrote (deliberately terse) Javascript code that computes the placement of these circles.

In fact, they compute something strictly harder! In the challenge, after laying down all of these circles (in blue), the challenge instructed them to go back to the start and greedily fill the gaps with (red) circles of increasing area. Next, they laid down a third (yellow) generation in the same fashion, and fourth (cyan) generation, and so on.

Related questions

What is the packing density of the first (blue) generation?

What is the packing density of the \(k\)-th generation?

How many “steps” away from the origin is the smallest circle in the \(k\)-th generation?

Do an infinite number of blue circles touch? Do an infinite number of any circles touch? Which ones?

How far can a circle be from its neighbors? Which circles are maximally far from their neighbors?

How does this work if the path the circles follow is not the spiral? Can different paths have significantly different packing densities?

If you have thoughts or ideas about any of this—or if you just want to make animated GIFs together—leave a comment or let me know on Twitter, @PeterKagey!

This is the first installment in a new series, “My Favorite Sequences”. In this series, I will write about sequences from the On-Line Encyclopedia of Integer Sequences that I’ve authored or spent a lot of time thinking about.

I’ve been contributing to the On-Line Encyclopedia of Integer Sequences since I was an undergraduate. In December 2013, I submitted sequence A233421 based on problem A2 from the 2013 Putnam Exam—which is itself based on “Ron Graham’s Sequence” (A006255)—a surprising bijection from the natural numbers to the non-primes. As of today, I’ve authored over 475 sequences based on puzzles that I’ve heard about and problems that I’ve dreamed up.

\(A261865(n)\) is the least integer \(k\) such that some multiple of \(\sqrt k\) falls in the interval \((n, n+1)\).

For example, \(A261865(3) = 3\) because there is no multiple of \(\sqrt 1\) in \((3,4)\) (since \(3 \sqrt{1} \leq 3\) and \(4 \sqrt{1} \geq 4\)); there is no multiple of \(\sqrt{2}\) in \((3,4)\) (since \(2 \sqrt{2} \leq 3\) and \(3 \sqrt 2 \geq 4\)); but there is a multiple of \(\sqrt 3\) in \((3,4)\), namely \(2\sqrt 3\).

As indicated in the picture, the sequence begins $$\color{blue}{ 2,2,3,2,2},\color{red}{3},\color{blue}{2,2,2},\color{red}{3},\color{blue}{2,2},\color{red}{3},\color{blue}{2,2,2},\color{red}{3},\color{blue}{2,2},\color{red}{3},\color{blue}{2,2},\color{magenta}{7},\dots.$$

A conjecture about density

As the example illustrates, \(1\) does not appear in the sequence. And almost by definition, asymptotically \(1/\sqrt 2\) of the values are \(2\)s.

Let’s denote the asymptotic density of terms that are equal to \(n\) by \(d_n\). It’s easy to check that \(d_1 = 0\), (because multiples of \(\sqrt 1\) are never between any integers) and \(d_2 = 1/\sqrt 2\), because multiples of \(\sqrt 2\) are always inserted. I conjecture in Problem 13 of my Open Problem Collection that $$a_n = \begin{cases}\displaystyle\frac{1}{\sqrt n}\left(1 – \sum_{i=1}^{n-1} a_i\right) & n \text{ is squarefree}\\[5mm] 0 & \text{otherwise}\end{cases}$$

If this conjecture is true, then the following table gives approximate densities.

I’m interested in values of \(n\) such that \(A261865(n)\) is large, and I reckon that there are clever ways to construct these, perhaps by looking at some Diophantine approximations of \(\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \dots\). In February, I posted a challenge on Code Golf Stack Exchange to have folks compete in writing programs that can quickly find large values of \(A261865(n)\).

Impressively, Noodle9’s C++ program won the challenge. In under a minute, this program found that the input \(n=1001313673399\) makes \(A261865\) particularly large: \(A261865(1001313673399) = 399\). Within the time limit, no other programs could find a value of \(n\) that makes \(A261865(n)\) larger.

\(n\)

Order of magnitude

\(A261865(n)\)

Time

1

\(1 \times 10^{0}\)

2

(0s)

3

\(3 \times 10^{0}\)

3

(0s)

23

\(2.3 \times 10^{1}\)

7

(0s)

30

\(3.0 \times 10^{1}\)

15

(0s)

184

\(1.84 \times 10^{2}\)

38

(0s)

8091

\(8.091 \times 10^{3}\)

43

(0s)

16060

\(1.606 \times 10^{4}\)

46

(0s)

16907

\(1.691 \times 10^{4}\)

58

(0s)

20993

\(2.099 \times 10^{4}\)

61

(0s)

26286

\(2.629 \times 10^{4}\)

97

(0s)

130375

\(1.304 \times 10^{5}\)

118

(0s)

169819

\(1.698 \times 10^{5}\)

127

(0s)

2135662

\(2.136 \times 10^{6}\)

130

(0s)

2345213

\(2.345 \times 10^{6}\)

187

(0s)

46272966

\(4.627 \times 10^{7}\)

193

(1s)

222125822

\(2.221 \times 10^{8}\)

210

(5.2s)

237941698

\(2.379 \times 10^{8}\)

217

(5.7s)

257240414

\(2.572 \times 10^{8}\)

227

(6.2s)

1205703469

\(1.206 \times 10^{9}\)

267

(31s)

1558293414

\(1.558 \times 10^{9}\)

299

(41.8s)

4641799364

\(4.642 \times 10^{9}\)

303

(2.1m)

6600656102

\(6.601 \times 10^{9}\)

323

(3m)

11145613453

\(1.115 \times 10^{10}\)

335

(5.2m)

20641456345

\(2.064 \times 10^{10}\)

354

(9.8m)

47964301877

\(4.796 \times 10^{10}\)

358

(22.9m)

105991039757

\(1.06 \times 10^{11}\)

385

(52m)

119034690206

\(1.19 \times 10^{11}\)

397

(59.1m)

734197670865

\(7.342 \times 10^{11}\)

455

(6.4h)

931392113477

\(9.314 \times 10^{11}\)

501

(8.4h)

1560674332481

\(1.561 \times 10^{12}\)

505

(14.2h)

A table of record values as computed by Code Golf Stack Exchange user Neil. The first 16 values agree with Jon E. Schoenfield’s computations that were added to the OEIS in September 2015

Related Ideas

Sequence \(A327953(n)\) counts the number of positive integers \(k\) such that there is some integer \(\alpha^{(n)}_k > 2\) where \(\alpha^{(n)}_k\sqrt{k} \in (n, n+1)\). It appears to grow roughly linearly like \(A327953(n) \sim 1.3n\), but I don’t know how to prove this.

Take any function \(f\colon\mathbb N \rightarrow \mathbb R\) that is positive, has positive first derivative, and has negative second derivative. Then, what is the least \(k\) such that some multiple of \(f(k)\) is in \((n,n+1)\)?

For example, what is the least integer \(k \geq 3\) such that there is a multiple of \(\ln(k)\) in \((n, n+1)\)?

What is the least \(k \in \mathbb N\) such that there exists \(m \in \mathbb N\) with \(k2^{1/m} \in (n,n+1)\)?

What is the least \(m \in \mathbb N\) such that there exists \(k \in \mathbb N\) with \(k2^{1/m} \in (n,n+1)\)?

A343205 is the auxiliary sequence that gives the value \(m\) such that \(m\sqrt{A261865(n)} \in (n, n+1)\). Does this sequence have an infinite limit inferior?

If you can answer any of these questions, or if you spend time thinking about this, please let me know on Twitter, @PeterKagey!

Neil Sloane is the founder of the On-Line Encyclopedia of Integer Sequences (OEIS). Every year or so, he gives a talk at Rutgers in which he discusses some of his favorite recent sequences. In 2017, he spent some time talking about a 1971 letter that he got from Richard Guy, and some questions that went along with it. In response to the talk, I investigated the letter and was able to sort out Richard’s 45-year-old idea, and correct and compute some more terms of his sequence.

Richard Guy and his sequences

Richard Guy was a remarkable mathematician who lived to the remarkable age of 103 years, 5 months, and 9 days! His life was filled with friendships and collaborations with many of the giants of recreational math: folks like John Conway, Paul Erdős, Martin Gardner, Donald Knuth, and Neil Sloane. But what I love most about Richard is how much joy and wonder he found in math. (Well, that and his life-long infatuation with his wife Louise.)

[I’m] an amateur [mathematician], I mean I’m not a professional mathematician. I’m an amateur in the more genuine sense of the word in that I love mathematics and I would like everybody in the world to like mathematics.

In January 2017, Neil Sloane gave a talk at Doron Zeilberger’s Experimental Mathematics Seminar, and about six minutes in, Neil discusses a letter that Richard sent to him at Cornell—which was the forwarded to Bell Labs—in June 1971.

When I was working on the book, the 1973 Handbook of Integer Sequences, I would get letters from Richard Guy from all over the world. As he traveled around, he would collect sequences and send them to me.

At 11:30, Neil discusses “sequence I” from Richard’s letter, which he added to the OEIS as sequence A279197:

Number of self-conjugate inseparable solutions of \(X + Y = 2Z\) (integer, disjoint triples from \(\{1,2,3,\dots,3n\}\)).

Neil mentioned in the seminar that he didn’t really know exactly what the definition meant. With some sleuthing and programming, I was able to make sense of the definition, write a Haskell program, correct the 7th term, and extend the sequence by a bit. The solutions for \(A279197(1)\) through \(A279197(10)\) are listed in a file I uploaded to the OEIS, and Fausto A. C. Cariboni was able to extend the sequence even further, submitting terms \(A279197(11)\)–\(A279197(17)\).

How the sequence works.

The idea here is to partition \(\{1,2,3,\dots,3n\}\) into length-3 arithmetic progressions, \(\bigl\{\{X_i,Z_i,Y_i\}\bigr\}_{i=1}^{n}\). And in particular, we want them to be inseparable and self-conjugate.

An inseparable partition is one whose “smallest” subsets are not a solution for a smaller case. For example, if \(n=3\), then the partition \[\bigl\{ \{1,3,5\}, \{2,4,6\}, \{7,8,9\} \bigr\}\] is separable, because if the subset \(\bigl\{ \{1,3,5\}, \{2,4,6\} \bigr\}\) is a solution to the \(n=2\) case.

A self-conjugate partition is one in which swapping each \(i\) with each \(3n+1-i\) gets back to what we started with. For example, \(\bigl\{\{1,3,5\}, \{2,4,6\}\bigr\}\) is self-congugate, because if we replace the \(1\) with a \(6\) and the \(2\) with a \(5\), and the \(i\) with a \(7-i\), then we get the same set: \(\bigl\{\{6,4,2\}, \{5,3,1\} \bigr\}\)

Generalizing Richard Guy’s idea

Of course, it’s natural to wonder about the separable solutions, or what happens if the self-conjugate restriction is dropped. In exploring these cases, I found four cases already in the OEIS, and I computed five more: A282615–A282619.

Table of sequences counting the ways of partitioning a set into length-3 arithmetic progressions, subject to various restrictions.

Generalizing further

There are lots of other generalizations that might be interesting to explore. Here’s a quick list:

Look at partitions of \(\{1,2,\dots,kn\}\) into \(n\) parts, all of which are an arithmetic sequence of length \(k\).

Count partitions of \(\{1,2,\dots,n\}\) into any number of parts of (un)equal size in a way that is (non-)self-conjugate and/or (in)separable.

Consider at partitions of \(\{1,2,\dots,3n\}\) into \(n\) parts, all of which are an arithmetic sequence of length \(3\), and whose diagram is “non-crossing”, that is, none of the line segments overlap anywhere. (See the 6th and 11th cases in the example for \(A279197(6) = 11\).)

If explore any generalizations of this problem on your own, if you’d like to explore together, or if you have an anecdotes about Richard Guy that you’d like to share, let me know on Twitter!