My Favorite Sequences Archives

My Favorite Sequences: A263135

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.

A263135: Placing Pennies

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.

Conjecture: For $n > 0$, \[A263135(2n) – A047932(n) = \lceil\sqrt{3n – 3/4} – 1/2\rceil = A216256(n).\]

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.

My Favorite Sequences: “Not Equal” Grid

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.

Five colors of blocks, where any two different colors of blocks are adjacent somewhere in the polyomino.

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.
OEIS sequence A278299.

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

A screenshot from my game illustrating the largest known term: $A278299(14) = 56$. Every number is connected to every other number. The red edges refer to redundant connections.

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!

My Favorite Sequences: A289523

This the second post in a recurring series, My Favorite Sequences. If you like this sort of thing, check out the Integer Sequence Review from The Aperiodical!

A289523: Packing Circles of Increasing Area

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 sequence begins 1, 4, 7, 1, 11, 16, 5, 21, 27, 34, 10, 1, 41, 17, 49, 25, 57, 6, 33, 66, 43, 14, ....

Finding an upper bound

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.

https://twitter.com/PeterKagey/status/1375960530863628290

I made some animated GIFs and posted about this on Code Golf Stack Exchange. https://t.co/VksT5lCe5K pic.twitter.com/mDzkwlxKbF
— Peter Kagey (@PeterKagey) March 28, 2021

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.

An animation illustrating successive generations of circles on a square spiral.

My Favorite Sequences: A261865

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: Multiples of square roots

(This problem is closely related to Problem 13 in my Open Problems Collection.)

In September 2015, I submitted sequence A261865:

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

An illustration of the first dozen terms of A261865

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 scatterplot of $A261865(n)$. Notice the records at $A261865(184)=38$ and $A261865(8091)=43$.

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.

$i$	$d_i$
$1$	$d_1 = 0\%$
$2$	$d_2 = 70.7\%$
$3$	$d_3 = 16.9\%$
$4$	$d_4 = 0\%$
$5$	$d_5 = 5.54\%$
$6$	$d_6 = 2.79\%$
$7$	$d_7 = 1.53\%$
$10$	$d_{10} = 0.797\%$
$11$	$d_{11} = 0.519\% $
$399$	$d_{399} = 3.53 \times 10^{-11} \%$

This was computed with the Mathematica code:

d[i_] := (d[i] = If[
  SquareFreeQ[i], 
  N[(1 - Sum[d[j], {j, 2, i - 1}])/Sqrt[i], 50], 
  0
])

Finding Large Values

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?

Scatterplot of A343205, generated in Mathematica. If the main conjecture is true, then this is not bounded below by $\alpha n$ for any positive value of $\alpha$.

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

\(i\)	\(d_i\)
\(1\)	\(d_1 = 0\%\)
\(2\)	\(d_2 = 70.7\%\)
\(3\)	\(d_3 = 16.9\%\)
\(4\)	\(d_4 = 0\%\)
\(5\)	\(d_5 = 5.54\%\)
\(6\)	\(d_6 = 2.79\%\)
\(7\)	\(d_7 = 1.53\%\)
\(10\)	\(d_{10} = 0.797\%\)
\(11\)	\(d_{11} = 0.519\% \)
\(399\)	\(d_{399} = 3.53 \times 10^{-11} \%\)

Category Archives: My Favorite Sequences