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Chip-Firing on Infinite kk-ary Trees

Dillan Agrawal PRIMES STEP Selena Ge PRIMES STEP Jate Greene PRIMES STEP Dohun Kim PRIMES STEP Rajarshi Mandal PRIMES STEP Tanish Parida PRIMES STEP Anirudh Pulugurtha PRIMES STEP Gordon Redwine PRIMES STEP Soham Samanta PRIMES STEP Albert Xu PRIMES STEP Tanya Khovanova MIT
Abstract

We use an infinite kk-ary tree with a self-loop at the root as our underlying graph. We consider a chip-firing process starting with NN chips at the root. We describe the stable configurations. We calculate the number of fires for each vertex and the total number of fires. We study a sequence of the number of root fires for a given kk as a function of NN and study its properties. We do the same for the total number of fires.

1 Introduction

1.1 History and Background

In 1975, Arthur Engel [3] introduced the probabilistic abacus, an early version of chip-firing. In 1986, Joel Spencer [9] independently discovered one-dimensional chip-firing. In 1991, Björner, Lovász, and Shor introduced a chip-firing game on graphs.

The chip-firing game is played on a graph and is an important part of structural combinatorics. In this game, each vertex has a certain number of chips and can fire if the number of chips is greater than or equal to its degree. When fired, a vertex sends a single chip to each neighbor. The game ends when we can no longer fire any vertices. The resulting configuration is called a stable configuration.

Chip-firing is similar to Abelian sandpiles, which were introduced in 1987 by Bak, Tang, and Wiesenfeld [1]. In Abelian sandpiles, each place on a finite grid has a slope. Each slope eventually increases as grains of sand are randomly placed on the grid. When the slope reaches a certain height, they collapse and split the sand grains evenly among their neighbors on the grid. Abelian sandpiles were the first dynamical system discovered to have self-organized criticality.

Chip-firing has deep connections to the Tutte polynomial and group theory. More details can be found in Merino’s survey from 2005 [6].

More recently, Musiker and Nguyen [7] have discovered many of the properties of chip-firing on infinite binary trees when starting with 2n12^{n}-1 chips at the root in both labeled and unlabeled chip-firing. Inagaki, Khovanova, and Luo [4] have also examined infinite binary trees. They found formulae for the total number of times each vertex fires, the total number of times the root fires, and the total number of fires.

Continuing the work of Inagaki, Khovanova, and Luo, we generalize some of the formulae they found to infinite kk-ary trees. We calculate the number of times each vertex fires, the number of times the tree’s root fires, and the total number of fires in a kk-ary tree. We also calculate and study many old and new integer sequences.

In Section 2, we provide preliminary definitions and results that are important for the rest of the paper. The graph we use is a rooted, undirected infinite kk-ary tree with a self-loop at the root, and our chip-firing game starts with NN chips at the root.

In Section 3, we show that if we start with NN chips at the root of our tree, the stable configuration will form a finite tree of height nn, where nn is the integer such that kn1k1Nkn+1kk1\frac{k^{n}-1}{k-1}\leq N\leq\frac{k^{n+1}-k}{k-1}. The number of chips on a vertex in the stable configuration is dependent only on its layer and is a function of the kk-ary expansion of Nkn1k1N-\frac{k^{n}-1}{k-1}.

In Section 4, we calculate the number of times each vertex fires, given the total number of chips (NN). In Section 4.2, we use the formula from the previous section to explore the number of root fires in more detail. Given kk, the number of root fires as a function of NN—the total number of chips—has an interesting structure. Its first and second difference sequences exhibit fractal properties, which we study in more detail.

In Section 5, we calculate the total number of fires and the corresponding difference sequences. We consider a case for N=kn1k1N=\frac{k^{n}-1}{k-1}, which plays a special role in our paper. We explain the connection between our sequences and schizophrenic numbers, which are numbers combining properties of rational and irrational numbers.

2 Preliminaries

This paper studies a chip-firing game on a specific graph: an infinite undirected rooted kk-ary tree with a self-loop at the root.

2.1 The underlying graph

In a rooted tree, we distinguish one vertex as the root. Every other vertex vv has exactly one parent vertex vpv_{p}, defined as the first vertex after vv in the shortest path from vv to the root. The vertex vv is said to be a child of the vertex vpv_{p}. Any vertex other than vertex vv on the shortest path from vv to root is an ancestor of vv, while vv is a descendant of each of its ancestors.

Refer to caption
Figure 1: An infinite undirected rooted 5-ary tree with a self-loop at the root

We define a vertex vv as being on layer i+1i+1 if the shortest path from the root to vv traverses ii edges. By this definition, the root rr is on layer 1.

Sometimes, we need to refer to a subgraph of our underlying graph. For this, we need some more definitions.

The height of a tree is the length of the longest path from the root to a leaf. In a perfect kk-ary tree, every non-leaf vertex has kk children, and all leaves are the same distance from the root. In other words, all leaves are located on the same layer h+1h+1, where hh is the height of such a tree.

2.2 Chip-firing process

We consider a standard unlabeled chip-firing game [5], where some number of indistinguishable chips are placed at each vertex. A vertex can fire if it has at least the same number of chips as its degree. In our case, a vertex can fire when it has at least k+1k+1 chips. When a vertex fires, it sends one chip along each edge. We consider the chip-firing game where the initial state has NN chips at the root and no chips anywhere else. The vertices fire until no firing is possible.

A stable configuration is the distribution of chips on the graph where no vertex can fire; in other words, in a stable configuration, every vertex has fewer chips than its degree. In our case, we reach a stable configuration when every vertex has at most kk chips.

The following theorem, proved in [2], describes when the chip-firing game may stabilize. It applies to any connected undirected graph G=(V,E)G=(V,E), where VV is the set of vertices and EE is the set of edges. We denote the number of vertices and edges by |V||V| and |E||E|, respectively.

Theorem 1 (Theorem 2.3 in [2]).

Let NN be the total number of chips. We have:

  • If N>2|E||V|N>2|E|-|V|, then the game is infinite and never reaches a stable configuration.

  • If N<|E|N<|E|, then the game is finite, reaching a stable configuration.

  • If |E|N2|E||V||E|\leq N\leq 2|E|-|V|, then there exists an initial chip configuration that leads to an infinite game and a configuration that leads to a finite game.

In our case, we have an infinite kk-ary tree with a finite number of chips; therefore, we have N<|E|N<|E|, implying that the configuration leads to a finite game.

Klivans [5] studied many properties of unlabeled chip-firing. In particular, we use the following property from this work called the Global Confluence.

Theorem 2 (Theorem 2.2.2 in [5]).

If we can reach a stable configuration with a finite number of fires, then the stable configuration is unique.

This theorem implies that the order of fires does not matter. In our case, the game is finite, implying that we always reach a unique stable configuration.

2.3 Base kk

Our formulae related to kk-ary trees can be better expressed in base kk. Hence, we use the notation (m)k(m)_{k} or mkm_{k} to indicate that a string of digits mm is written in base kk. For example, 1112=(111)2=7111_{2}=(111)_{2}=7. Correspondingly, we will write [n]k[n]_{k} to denote the string of digits representing a decimal integer nn in base kk. Thus, [n]k=m[n]_{k}=m is equivalent to n=(m)kn=(m)_{k}.

We denote by νk(n)\nu_{k}(n) the largest power of kk that divides nn. In other words, νk(n)\nu_{k}(n) is the number of trailing zeros in [n]k[n]_{k}. When kk is prime, the value νk(n)\nu_{k}(n) is called the kk-adic valuation of nn. We will extend the terminology to composite numbers as well. In other words,

kνk(n)|n and kνk(n)+1n.k^{\nu_{k}(n)}|n\quad\textrm{ and }\quad k^{\nu_{k}(n)+1}\nmid n.

3 Stable configurations

In Proposition 3.3 from [7], the stable configurations are described for the case k=2k=2. We describe the stable configurations for any k2k\geq 2.

Proposition 3.

If we start with NN chips at the root, where kn1k1Nkn+1kk1\frac{k^{n}-1}{k-1}\leq N\leq\frac{k^{n+1}-k}{k-1}, then the vertices containing chips in the stable configuration form a perfect kk-ary tree with height n1n-1. Furthermore, every vertex on the same layer has the same number of chips.

Proof.

We prove this proposition by induction on nn. When n=1n=1, the number of chips at the root ranges from 11 to kk, which is already a stable configuration, so this is a perfect kk-ary tree with a height of 0. As only the root has chips in a stable configuration, every vertex on the same layer has the same number of chips.

Suppose the proposition holds for n=in=i; we will prove that it holds for n=i+1n=i+1. Suppose we start with NN chips at the root, where ki+11k1Nki+2kk1\frac{k^{i+1}-1}{k-1}\leq N\leq\frac{k^{i+2}-k}{k-1}. We deploy the following firing strategy.

  1. 1.

    Fire the root repeatedly until it cannot fire anymore.

  2. 2.

    Fire the root’s children and their subtrees in parallel.

  3. 3.

    Whenever the kk children of the root fire, fire the root.

  4. 4.

    Repeat steps 2 and 3 until the stable configuration is reached.

As this process is parallel, the root’s children contain the same number of chips at any step. Every time the root fires, it sends 1 chip to itself. Therefore, the root always possesses at least 1 chip. After step 1, each child of the root contains between ki1k1\frac{k^{i}-1}{k-1} and ki+1kk1\frac{k^{i+1}-k}{k-1} chips. In step 3, when the kk children of the root fire, they each send 1 chip to the root. After that, the root gets kk chips, allowing it to fire, causing the root to return 1 chip to each child immediately. After step 1, we can interpret each subtree rooted at layer 2 as an independent tree with a self-loop at its root.

Therefore, the firing process of each subtree rooted at layer 2 is isomorphic to chip-firing on a rooted tree where the number of chips initially placed at the root is between ki1k1\frac{k^{i}-1}{k-1} and ki+1kk1\frac{k^{i+1}-k}{k-1}, inclusive. By the induction hypothesis, once in the stable configuration, each subtree has the same number of chips on every vertex of the same layer. Moreover, vertices with chips form a perfect kk-ary tree with height n=i1n=i-1. As the firing processes in all kk root’s subtrees are the same, the stable configurations of the subtrees are the same. Thus, after the process stabilizes, the vertices containing chips in the whole tree form a perfect kk-ary tree with height ii, and every vertex on the same layer has the same number of chips. Since we are considering unlabeled chip-firing, the global confluence property applies, so the stable configuration is always as described above, finishing the proof. ∎

Proposition 4.

If we start with NN chips at the root, where kn1k1Nkn+1kk1\frac{k^{n}-1}{k-1}\leq N\leq\frac{k^{n+1}-k}{k-1}, then for 0in10\leq i\leq n-1, the resulting stable configuration has ai+1a_{i}+1 chips on each vertex on layer i+1i+1, where an1a2a1a0a_{n-1}\dots a_{2}a_{1}a_{0} is the base kk expansion of Nkn1k1N-\frac{k^{n}-1}{k-1} with possible leading zeros.

Proof.

We know that all vertices with chips form a perfect kk-ary tree, where each vertex with chips has from 1 to kk chips, inclusive, and each vertex in the same layer has the same number of chips. Suppose each vertex on layer i+1i+1 has jij_{i} chips. As there are kik^{i} vertices on layer i+1i+1, the total number of chips is i=0n1jiki\sum_{i=0}^{n-1}j_{i}k^{i}, which equals NN. Thus, the distribution of chips into layers in the final configuration implies representing NN in base kk with digits between 11 and kk. Since this representation is unique, to prove that the above formula describes the correct distribution of chips, we need to demonstrate that the total number of chips equals NN. When we place ai+1a_{i}+1 chips on each vertex, we are placing a total of

i=0n1ki(ai+1)\displaystyle\sum_{i=0}^{n-1}k^{i}(a_{i}+1) =i=0n1kiai+i=0n1ki\displaystyle=\sum_{i=0}^{n-1}k^{i}a_{i}+\sum_{i=0}^{n-1}k^{i}
=Nkn1k1+kn1k1\displaystyle=N-\frac{k^{n}-1}{k-1}+\frac{k^{n}-1}{k-1}
=N\displaystyle=N

chips, as desired. ∎

We see that numbers of the form N=kn1k1N=\frac{k^{n}-1}{k-1} play a special role, as the stable configuration reaches layer nn for the first time when we start with the number of chips in this form placed at the root.

From now on, let an1a2a1a0a_{n-1}\ldots a_{2}a_{1}a_{0} be the base kk expansion of Nkn1k1N-\frac{k^{n}-1}{k-1} with possible leading zeros. When we need to emphasize the dependency of these digits aia_{i} on the base KK and the number NN, we will use the notation ai(N,k)a_{i}(N,k) instead of aia_{i}. Similar to [7, 4], for the stable configuration starting with NN chips at the root in a kk-ary tree, let us denote the number of chips at each vertex on layer i+1i+1 as ci(N,k)c_{i}(N,k) for 0in10\leq i\leq n-1. Proposition 4 states that ci(N,k)=ai(N,k)+1c_{i}(N,k)=a_{i}(N,k)+1. When the values of NN and kk are fixed, we will use the notations aia_{i} and cic_{i} to simplify the calculations.

The number nn also plays a special role in our paper. It was defined through inequalities

kn1k1Nkn+1kk1.\frac{k^{n}-1}{k-1}\leq N\leq\frac{k^{n+1}-k}{k-1}.

After multiplying every term by k1k-1 and adding 1, we get

knN(k1)+1kn+1k+1.k^{n}\leq N(k-1)+1\leq k^{n+1}-k+1.

Taking the log base kk of both sides gives us

nlogk(N(k1)+1)logk(kn+1k+1)<n+1.n\leq\log_{k}(N(k-1)+1)\leq\log_{k}(k^{n+1}-k+1)<n+1.

Thus, we can define nn as

n=logk(N(k1)+1).n=\lfloor\log_{k}(N(k-1)+1)\rfloor.

4 The number of times each vertex fires

In [7], Musiker and Nguyen counted the number of times each vertex fires for a binary tree when starting with N=21N=2^{\ell}-1 unlabeled chips at the root. In [4], this was extended to any number of chips at the root; however, it was limited to binary trees. Here, we generalize their results for a kk-ary tree.

4.1 The number of fires for a vertex on layer i+1i+1

Let fi(N,k)f_{i}(N,k) denote the total number of fires of each vertex on layer i+1i+1. When NN and kk are fixed, we will sometimes use the notation fif_{i} to simplify the calculations. We first look at the difference in the number of fires between layers i1i-1 and ii.

Lemma 5.

The difference in the number of fires between vertices on layers i+1i+1 and i+2i+2 equals the number of chips in the stable configuration belonging to the descendants of one particular vertex on layer i+1i+1 divided by kk. In other words,

fi(N,k)fi+1(N,k)=j=i+1n1kji1cj(N,k),f_{i}(N,k)-f_{i+1}(N,k)=\sum\limits_{j=i+1}^{n-1}k^{j-i-1}c_{j}(N,k),

where n=logk(N(k1)+1)n=\lfloor\log_{k}(N(k-1)+1)\rfloor.

Proof.

The number of times a vertex fires equals the total number of chips sent to all children of the vertex divided by kk. For a given vertex, the number of chips sent to its children is equal to the number of chips placed at the vertex’s descendants plus the number of chips sent back from its children. The number of chips sent back to the vertex from its children is kk times the number of fires of each child. Thus, the difference between the number of fires of a vertex vv and its child is the number of chips belonging to the descendants of vv in the stable configuration divided by kk.

Let vertex vv be on layer i+1i+1, and let j>ij>i. The number of vertices on layer j+1j+1 that are the descendants of vv is kjik^{j-i}. Each descendant contains cj(N,k)c_{j}(N,k) chips by Proposition 4. Adding them up across all of the vertices, the number of chips placed at the descendants of vv is j=i+1n1kjicj(N,k)\sum\limits_{j=i+1}^{n-1}k^{j-i}c_{j}(N,k), and the result follows after dividing this by kk. ∎

Example 1.

As the chips on the last layer, nn, never fire, we have fn1=0f_{n-1}=0. From Lemma 5, we get fn2=cn1f_{n-2}=c_{n-1}.

Now, we can calculate the number of fires for each vertex on layer i+1i+1.

Theorem 6.

Given the total number of chips NN and the index n=logk(N(k1)+1)n=\lfloor\log_{k}(N(k-1)+1)\rfloor, the number of fires for each vertex on layer i+1i+1 is:

fi(N,k)=j=1ni1(kj1k1)ci+j(N,k).f_{i}(N,k)=\sum_{j=1}^{n-i-1}\left(\frac{k^{j}-1}{k-1}\right)c_{i+j}(N,k).
Proof.

We prove this formula using backward induction, starting with i=n1i=n-1. As the last layer never fires, we have fn1=0f_{n-1}=0. Our formula gives us the same: fn1=j=10(kj1k1)ci+j=0f_{n-1}=\sum_{j=1}^{0}(\frac{k^{j}-1}{k-1})c_{i+j}=0. Thus, the base of the induction is established.

Suppose the formula is true for a given i+1i+1. We want to calculate the number of fires for ii. In other words, we want to find fif_{i} assuming that fi+1=j=1ni2(kj1k1)ci+j+1f_{i+1}=\sum_{j=1}^{n-i-2}(\frac{k^{j}-1}{k-1})c_{i+j+1}.

From Lemma 5, we have that

fifi+1=j=i+1n1kji1cj.f_{i}-f_{i+1}=\sum_{j=i+1}^{n-1}k^{j-i-1}c_{j}.

Thus, after reordering terms and keeping track of indices, we get

fi\displaystyle f_{i} =fi+1+j=i+1n1kji1cj=j=1ni2(kj1k1)ci+j+1+j=i+1n1kji1cj\displaystyle=f_{i+1}+\sum_{j=i+1}^{n-1}k^{j-i-1}c_{j}=\sum_{j=1}^{n-i-2}\left(\frac{k^{j}-1}{k-1}\right)c_{i+j+1}+\sum_{j=i+1}^{n-1}k^{j-i-1}c_{j}
=j=1ni2(kj1k1)ci+j+1+j=0ni2kjci+j+1=ci+1+j=1ni2(kj+11k1)ci+j+1\displaystyle=\sum_{j=1}^{n-i-2}\left(\frac{k^{j}-1}{k-1}\right)c_{i+j+1}+\sum_{j=0}^{n-i-2}k^{j}c_{i+j+1}=c_{i+1}+\sum_{j=1}^{n-i-2}\left(\frac{k^{j+1}-1}{k-1}\right)c_{i+j+1}
=j=1ni1(kj1k1)ci+j(N,k),\displaystyle=\sum_{j=1}^{n-i-1}\left(\frac{k^{j}-1}{k-1}\right)c_{i+j}(N,k),

concluding the proof. ∎

The following corollary allows us to express the number of fires of each vertex on layer i+1i+1 through the number of fires of the root.

Corollary 7.

We have

fi(N,k)=f0(Nkn1k1ki+kni1k1,k),f_{i}(N,k)=f_{0}\left(\left\lfloor\frac{N-\frac{k^{n}-1}{k-1}}{k^{i}}\right\rfloor+\frac{k^{n-i}-1}{k-1},k\right),

where n=logk(N(k1)+1)n=\lfloor\log_{k}(N(k-1)+1)\rfloor.

Proof.

The formula for the number of fires of each vertex on layer i+1i+1 is

fi(N,k)=j=1ni1(kj1k1)ci+j(N,k).f_{i}(N,k)=\sum_{j=1}^{n-i-1}\left(\frac{k^{j}-1}{k-1}\right)c_{i+j}(N,k).

This matches up with the total number of fires for the root of a tree if we start with NN^{\prime} chips, where [N]k[N^{\prime}]_{k} is the first nin-i digits of [N]k[N]_{k}. In other words,

cj(N,k)=ci+j(N,k).c_{j}(N^{\prime},k)=c_{i+j}(N,k).

In this case,

f0(N,k)=j=1ni1(kj1k1)cj(N,k)=fi(N,k).f_{0}(N^{\prime},k)=\sum_{j=1}^{n-i-1}\left(\frac{k^{j}-1}{k-1}\right)c_{j}(N^{\prime},k)=f_{i}(N,k).

From Proposition 4, we know that aj(N,k)=ai+j(N,k)a_{j}(N^{\prime},k)=a_{i+j}(N,k). This implies that

Nkni1k1=Nkn1k1ki,N^{\prime}-\frac{k^{n-i}-1}{k-1}=\left\lfloor\frac{N-\frac{k^{n}-1}{k-1}}{k^{i}}\right\rfloor,

and the corollary follows. ∎

The corollary above implies that we can focus only on the number of root fires to understand the properties of the number of vertex fires on other layers.

4.2 The number of root fires

As a particular case of Theorem 6, we get the number of root fires.

Corollary 8.

Given the total number of chips NN and the index n=logk(N(k1)+1)n=\lfloor\log_{k}(N(k-1)+1)\rfloor, the number of times the root fires is

f0(N,k)=1k1j=1n1(kj1)cj(N,k).f_{0}(N,k)=\frac{1}{k-1}\sum_{j=1}^{n-1}(k^{j}-1)c_{j}(N,k).
Example 2.

Suppose we have a ternary tree (k=3)(k=3), and the total number of chips is 9 (N=9)(N=9). This means that we have index n=log3(92+1)=2n=\lfloor\log_{3}(9\cdot 2+1)\rfloor=2. The ternary representation of Nk21k1=94=5N-\frac{k^{2}-1}{k-1}=9-4=5 is 12312_{3}, implying that a1=1a_{1}=1 and c1=2c_{1}=2. This means that the total number of times the root fires is j=11(3j122)=2\sum_{j=1}^{1}\left(\frac{3^{j}-1}{2}\cdot 2\right)=2 times.

As in [4], we are looking for a recursive formula.

Corollary 9.

We have

f0(N)=Nk1+f0(Nk1).f_{0}(N)=\left\lceil\frac{N}{k}\right\rceil-1+f_{0}\left(\left\lceil\frac{N}{k}\right\rceil-1\right).
Proof.

The number of times the root fires is equal to the number of times it fires initially plus the number of times its children fire. ∎

4.3 Special case of N=kn1k1N=\frac{k^{n}-1}{k-1}

The case of N=kn1k1N=\frac{k^{n}-1}{k-1} is important throughout our paper, as the stable configuration reaches layer nn for the first time when we start with kn1k1\frac{k^{n}-1}{k-1} chips in this form placed at the root. Moreover, if we start with kn1k1\frac{k^{n}-1}{k-1} chips at the root, then ci=1c_{i}=1 for 0i<n0\leq i<n. Thus, the number of fires for a vertex on level i+1i+1 is

fi(N,k)=j=1ni1(kj1k1)=knik(ni)+(ni)1(k1)2.f_{i}(N,k)=\sum_{j=1}^{n-i-1}\left(\frac{k^{j}-1}{k-1}\right)=\frac{k^{n-i}-k(n-i)+(n-i)-1}{(k-1)^{2}}.

Plugging in k=2k=2, we get

fi(N,2)=2ni2(ni)+(ni)1=2nin+i1,f_{i}(N,2)=2^{n-i}-2(n-i)+(n-i)-1=2^{n-i}-n+i-1,

matching the answer from [7].

The total number of root fires when N=kn1k1N=\frac{k^{n}-1}{k-1} is

f0(N,k)=1k1j=1n1(kj1)=kn1k1nk1=knnk+(n1)(k1)2.f_{0}(N,k)=\frac{1}{k-1}\sum_{j=1}^{n-1}(k^{j}-1)=\frac{\frac{k^{n}-1}{k-1}-n}{k-1}=\frac{k^{n}-nk+(n-1)}{(k-1)^{2}}.

Plugging in k=2k=2, we see that for 2n12^{n}-1 chips on a binary tree, the root fires 2nn12^{n}-n-1 times, which is the same result as in [7]. These numbers form sequence A000295 in the OEIS [8]. Starting from index 1, the first few terms are

0, 1, 4, 11, 26, 57, 120, 247, 502, 1013,.0,\ 1,\ 4,\ 11,\ 26,\ 57,\ 120,\ 247,\ 502,\ 1013,\ \ldots.

When we plug in k=3k=3, we get that there are 3n2n14\frac{3^{n}-2n-1}{4} root fires, which corresponds to sequence A000340 in the OEIS with a shifted index. Similarly, when we plug in k=4k=4 and k=5k=5, we get 4n3n19\frac{4^{n}-3n-1}{9} and 5n4n116\frac{5^{n}-4n-1}{16} root fires, which correspond to sequences A014825 and A014827 in the OEIS [8]. These sequences appear later in our paper in Table 4.

4.4 Differences

Notice that for all N{ak+1,ak+2,,(a+1)k}N\in\{ak+1,ak+2,\ldots,(a+1)k\}, the same vertex of the tree fires the same number of times. Thus, it makes sense to consider the number of fires as a function of Nk\left\lceil\frac{N}{k}\right\rceil. Therefore, we introduce a new set of functions:

gi(m,k)=fi(mk,k).g_{i}(m,k)=f_{i}(mk,k).

Table 1 shows values of g0(m,k)g_{0}(m,k) for 1m101\leq m\leq 10 and 2k62\leq k\leq 6.

kk\mm 1 2 3 4 5 6 7 8 9 10
2 0 1 2 4 5 7 8 11 12 14
3 0 1 2 3 5 6 7 9 10 11
4 0 1 2 3 4 6 7 8 9 11
5 0 1 2 3 4 5 7 8 9 10
6 0 1 2 3 4 5 6 8 9 10
Table 1: Values of g0(m,k)g_{0}(m,k) for 1m101\leq m\leq 10 and 2k62\leq k\leq 6.

The values for g0(m,2)g_{0}(m,2) were calculated in [4], and they form sequence A376116 in the OEIS [8].

The sequence g0(m,3)g_{0}(m,3) has been submitted to the OEIS as sequence A378724 and starts as

0, 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 18.0,\ 1,\ 2,\ 3,\ 5,\ 6,\ 7,\ 9,\ 10,\ 11,\ 13,\ 14,\ 15,\ 18.

Consider the difference sequence di(m,k)=gi(m+1,k)gi(m,k)d_{i}(m,k)=g_{i}(m+1,k)-g_{i}(m,k). In this section, we are interested in the case i=0i=0. The following proposition describes the recursion for d0(m,k)d_{0}(m,k).

Proposition 10.

The difference sequence d0(m,k)=g0(m+1,k)g0(m,k)d_{0}(m,k)=g_{0}(m+1,k)-g_{0}(m,k) satisfies the recurrence

d0(m,k)={d0(m1k,k)+1, if m1 is a multiple of k,1, otherwise.d_{0}(m,k)=\begin{cases}d_{0}(\frac{m-1}{k},k)+1,&\text{ if $m-1$ is a multiple of $k$},\\ 1,&\text{ otherwise}.\end{cases}
Proof.

We know that

d0(m,k)\displaystyle d_{0}(m,k) =g0(m+1,k)g0(m,k)=f0(km+k,k)f0(km,k)\displaystyle=g_{0}(m+1,k)-g_{0}(m,k)=f_{0}(km+k,k)-f_{0}(km,k)
=m+f0(m)(m1)f0(m1)\displaystyle=m+f_{0}(m)-(m-1)-f_{0}(m-1)
=1+f0(m)f0(m1)\displaystyle=1+f_{0}(m)-f_{0}(m-1)

by Corollary 9. Now, since f0(x)f_{0}(x) depends only on xk\left\lceil\frac{x}{k}\right\rceil, if we have mk=m1k\left\lceil\frac{m}{k}\right\rceil=\left\lceil\frac{m-1}{k}\right\rceil, it means that m1m-1 is not a multiple of kk. This implies that d0(m,k)=1d_{0}(m,k)=1. If m1m-1 is a multiple of kk, then we need to calculate f0(m)f0(m1)f_{0}(m)-f_{0}(m-1). We have f0(m)=g0(m1k+1)f_{0}(m)=g_{0}(\frac{m-1}{k}+1) and f0(m1)=g0(m1k)f_{0}(m-1)=g_{0}(\frac{m-1}{k}). Therefore, f0(m)f0(m1)=g0(m1k+1)g0(m1k)=d0(m1k,k)f_{0}(m)-f_{0}(m-1)=g_{0}(\frac{m-1}{k}+1)-g_{0}(\frac{m-1}{k})=d_{0}(\frac{m-1}{k},k), and d0(m,k)=d0(m1k,k)+1d_{0}(m,k)=d_{0}(\frac{m-1}{k},k)+1, as claimed. ∎

Now, we can state the formula.

Theorem 11.

The difference sequence d0(m,k)=g0(m+1,k)g0(m,k)d_{0}(m,k)=g_{0}(m+1,k)-g_{0}(m,k) satisfies the formula

d0(m,k)={n, if m=kn1k1 for some integer n,νk((k1)m+1)+1, otherwise.d_{0}(m,k)=\begin{cases}n,&\text{ if $m=\frac{k^{n}-1}{k-1}$ for some integer $n$},\\ \nu_{k}((k-1)m+1)+1,&\text{ otherwise}.\end{cases}
Proof.

If m=kn1k1m=\frac{k^{n}-1}{k-1}, then we can prove the formula based on the recurrence and using induction on nn. When n=1n=1, we have m=1m=1 and d0(m,k)=1d_{0}(m,k)=1,. Thus, the base of the induction is established. Assume that d0(kn1k1,k)=nd_{0}(\frac{k^{n}-1}{k-1},k)=n. By the recurrence from Proposition 10, we have

d0(kn+11k1,k)=d0(kn1k1,k)+1=n+1.\displaystyle d_{0}\left(\frac{k^{n+1}-1}{k-1},k\right)=d_{0}\left(\frac{k^{n}-1}{k-1},k\right)+1=n+1.

Now, if mm cannot be represented as kn1k1\frac{k^{n}-1}{k-1} for any integer nn, then we can prove the formula based on the recurrence and using induction. Our base case is when νk(m(k1)+1)=0\nu_{k}(m(k-1)+1)=0. Therefore, we see that m1m-1 is not a multiple of kk. The recurrence gives us d0(m,k)=1=νk(m(k1)+1)+1d_{0}(m,k)=1=\nu_{k}(m(k-1)+1)+1. Thus, the base of induction is established.

Assume for all mkn11k1m\neq\frac{k^{n-1}-1}{k-1}, such that νk(m(k1)+1)=n1\nu_{k}\left(m(k-1)+1\right)=n-1 is true, then d0(m,k)=nd_{0}(m,k)=n satisfies the recurrence. Consider some mkn1k1m\neq\frac{k^{n}-1}{k-1} such that νk(m(k1)+1)=n\nu_{k}\left(m(k-1)+1\right)=n. Since n>0n>0, we know that m1m-1 must be a multiple of kk. By the recurrence we have d0(m,k)=d0(m1k)+1d_{0}(m,k)=d_{0}(\frac{m-1}{k})+1. Since mkn1k1m\neq\frac{k^{n}-1}{k-1}, we have m1kkn1k1\frac{m-1}{k}\neq\frac{k^{n-1}}{k-1}. Notice that νk((m1)(k1)k+1)=νk(m(k1)+1k)=n1\nu_{k}(\frac{(m-1)(k-1)}{k}+1)=\nu_{k}(\frac{m(k-1)+1}{k})=n-1. Thus, we have that d0(m,k)=d0(m1k)+1=n+1d_{0}(m,k)=d_{0}(\frac{m-1}{k})+1=n+1. ∎

In other words, excluding the first digit, if [m]k[m]_{k} ends with ii ones, we have d0(m,k)=g0(m+1,k)g0(m,k)=i+1d_{0}(m,k)=g_{0}(m+1,k)-g_{0}(m,k)=i+1.

Remark.

If Nkn1k1N\neq\frac{k^{n}-1}{k-1}, then the number of trailing ones in the base kk representation of NN, without the first digit, is equal to the number of trailing zeros of Nkn1k1N-\frac{k^{n}-1}{k-1} in base kk or νk(Nkn1k1)\nu_{k}(N-\frac{k^{n}-1}{k-1}). This is the same as νk(N(k1)+1)\nu_{k}(N(k-1)+1).

Example 3.

The difference sequence d0(m,2)d_{0}(m,2) is sequence A091090 in the OEIS [8], as shown in [4]. The latter sequence is defined in the OEIS as the number of editing steps (delete, insert, or substitute) to transform the binary representation of nn into the binary representation of n+1n+1. Sequence A091090 starts from index 1 as

1, 1, 2, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1,.1,\ 1,\ 2,\ 1,\ 2,\ 1,\ 3,\ 1,\ 2,\ 1,\ 3,\ 1,\ 2,\ 1,\ 4,\ 1,\ 2,\ 1,\ \ldots.

The difference sequence d0(m,3)d_{0}(m,3) has been submitted to the OEIS as sequence A378725 and starts as

1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1,.1,\ 1,\ 1,\ 2,\ 1,\ 1,\ 2,\ 1,\ 1,\ 2,\ 1,\ 1,\ 3,\ 1,\ 1,\ 2,\ 1,\ 1,\ 2,\ 1,\ 1,\ 3,\ 1,\ \ldots.

One can notice that these sequences all have a similar structure due to a similar repetition pattern. Each sequence begins with kk ones. Afterward, there are k1k-1 ones between every two numbers that are not equal to 1. Furthermore, for any m>1m>1, if we remove all numbers less than mm, the resulting sequence starts with kk occurrences of mm, followed by k1k-1 occurrences of mm between every two numbers that are not equal to mm.

In other words, we can construct the difference sequence d0(m,k)d_{0}(m,k) in the following way. Start by creating a sequence of all 11’s. Then, replace every kkth 11 with a 22 beginning with the (k+1)(k+1)th 11. Next, replace every kkth 22 with a 33, starting with the (k+1)(k+1)th 22. Continue this process indefinitely, replacing every kkth occurrence of x1x-1 with xx, beginning with the (k+1)(k+1)th x1x-1. This fractal structure of sequences d0d_{0} follows from Proposition 10.

5 The total number of fires

5.1 The total number of fires

We can also calculate the total number of fires on an infinite undirected kk-ary tree with a self-loop at the root. Let F(N,k)F(N,k) denote the total number of fires in a kk-ary tree when we start with NN chips placed at the root. As we have done previously, we sometimes remove the parameters in the function FF when NN and kk are fixed.

Theorem 12.

The total number of fires is

F(N,k)=1(k1)2m=1n1(mkm+1(m+1)km+1)cm(N,K),F(N,k)=\frac{1}{(k-1)^{2}}\sum_{m=1}^{n-1}(mk^{m+1}-(m+1)k^{m}+1)c_{m}(N,K),

where n=logk(N(k1)+1)n=\lfloor\log_{k}(N(k-1)+1)\rfloor.

Proof.

Totaling the number of fires for all the vertices, we have

F(N,k)=i=0n2kifi.F(N,k)=\sum_{i=0}^{n-2}k^{i}f_{i}.

Substituting fif_{i} from Theorem 6, we get

F(N,k)=i=0n2kij=1ni1kj1k1ci+j.F(N,k)=\sum_{i=0}^{n-2}k^{i}\sum_{j=1}^{n-i-1}\frac{k^{j}-1}{k-1}c_{i+j}.

Rearranging and replacing i+ji+j with mm, we get

F(N,k)\displaystyle F(N,k) =i=0n2j=1ni1ki+jkik1ci+j\displaystyle=\sum_{i=0}^{n-2}\sum_{j=1}^{n-i-1}\frac{k^{i+j}-k^{i}}{k-1}c_{i+j}
=1k1(i=0n2j=1ni1ki+jci+ji=0n2j=1ni1kici+j)\displaystyle=\frac{1}{k-1}\left(\sum_{i=0}^{n-2}\sum_{j=1}^{n-i-1}k^{i+j}c_{i+j}-\sum_{i=0}^{n-2}\sum_{j=1}^{n-i-1}k^{i}c_{i+j}\right)
=1k1(m=1n1mkmcmm=1n1(1+k+k2++km1)cm)\displaystyle=\frac{1}{k-1}\left(\sum_{m=1}^{n-1}mk^{m}c_{m}-\sum_{m=1}^{n-1}(1+k+k^{2}+\cdots+k^{m-1})c_{m}\right)
=1k1(m=1n1mkmcmm=1n1km1k1cm)\displaystyle=\frac{1}{k-1}\left(\sum_{m=1}^{n-1}mk^{m}c_{m}-\sum_{m=1}^{n-1}\frac{k^{m}-1}{k-1}c_{m}\right)
=1(k1)2m=1n1(mkm+1(m+1)km+1)cm.\displaystyle=\frac{1}{(k-1)^{2}}\sum_{m=1}^{n-1}(mk^{m+1}-(m+1)k^{m}+1)c_{m}.

This concludes the proof. ∎

For k=2k=2, the sequence of the total number of fires was calculated in [4]. It is sequence A376131 in the OEIS. The sequence starts from index 1 as

0, 1, 2, 6, 7, 11, 12, 23, 24, 28, 29, 40, 41, 45,.0,\ 1,\ 2,\ 6,\ 7,\ 11,\ 12,\ 23,\ 24,\ 28,\ 29,\ 40,\ 41,\ 45,\ \ldots.

For k=3k=3, the sequence of the total number of fires starts from index 1 as

0, 1, 2, 3, 8, 9, 10, 15, 16, 17, 22, 23, 24, 42, 43, 44, 49, 50, 51,.0,\ 1,\ 2,\ 3,\ 8,\ 9,\ 10,\ 15,\ 16,\ 17,\ 22,\ 23,\ 24,\ 42,\ 43,\ 44,\ 49,\ 50,\ 51,\ \ldots.

This sequence has been submitted to the OEIS as sequence A378726. We have a recursive formula for function FF, similar to the formula for f0f_{0} found in Corollary 9.

Proposition 13.

We have

F(N)=f0(N)+kF(Nk1).F(N)=f_{0}(N)+kF\bigg{(}\bigg{\lceil}\frac{N}{k}\bigg{\rceil}-1\bigg{)}.
Proof.

The total number of fires is equal to the number of root fires plus the number of fires in all of the kk subtrees with roots on level 2. Each subtree receives Nk1\left\lceil\frac{N}{k}\right\rceil-1 chips from the root, so each would fire F(Nk1)F\left(\left\lceil\frac{N}{k}\right\rceil-1\right) times. ∎

5.2 Special case of N=kn1k1N=\frac{k^{n}-1}{k-1}

If we start with N=kn1k1N=\frac{k^{n}-1}{k-1} chips at the root, we have ci=1c_{i}=1 for 0i<n0\leq i<n. This case is special, so we want to look at the total number of fires for such NN separately.

Proposition 14.

We have

F(kj+11k1)=(k(n1)n1)kn+k(n+1)n+1(k1)3.F\left(\frac{k^{j+1}-1}{k-1}\right)=\dfrac{(k(n-1)-n-1)k^{n}+k(n+1)-n+1}{(k-1)^{3}}.
Proof.

Using the formula for the total number of fires, we get that there are a total of

m=1n1mkm+1mkmkm+1(k1)2\displaystyle\sum^{n-1}_{m=1}\dfrac{mk^{m+1}-mk^{m}-k^{m}+1}{(k-1)^{2}}
=1(k1)2m=1n1(mkm+1mkmkm+1)\displaystyle=\dfrac{1}{(k-1)^{2}}\sum^{n-1}_{m=1}(mk^{m+1}-mk^{m}-k^{m}+1)
=1(k1)2(m=1n1mkm+1m=1n1mkmm=1n1km+m=1n11)\displaystyle=\dfrac{1}{(k-1)^{2}}\left(\sum^{n-1}_{m=1}mk^{m+1}-\sum^{n-1}_{m=1}mk^{m}-\sum^{n-1}_{m=1}k^{m}+\sum^{n-1}_{m=1}1\right)
=1(k1)2((n1)kn2knkk1+n1)\displaystyle=\dfrac{1}{(k-1)^{2}}\left((n-1)k^{n}-2\dfrac{k^{n}-k}{k-1}+n-1\right)
=nkn+1nknkn+1kn+k+nkn+1(k1)3\displaystyle=\dfrac{nk^{n+1}-nk^{n}-k^{n+1}-k^{n}+k+nk-n+1}{(k-1)^{3}}
=(knnk1)kn+nk+kn+1(k1)3\displaystyle=\dfrac{(kn-n-k-1)k^{n}+nk+k-n+1}{(k-1)^{3}}
=(k(n1)n1)kn+k(n+1)n+1(k1)3\displaystyle=\dfrac{(k(n-1)-n-1)k^{n}+k(n+1)-n+1}{(k-1)^{3}}

fires. ∎

For k=2k=2, we get

F(2n1)=(2n2n1)2n+2n+2n+1=(n3)2n+n+3.F(2^{n}-1)=(2n-2-n-1)2^{n}+2n+2-n+1=(n-3)2^{n}+n+3.

This matches Corollary 3.7 in [7], and it is sequence A045618 in the OEIS, with an index shift.

For k=3k=3, we get

F(3n12)=18((3n3n1)3n+3n+3n+1)=18((2n4)3n+2n+4)=14((n2)3n+n+2),F\left(\frac{3^{n}-1}{2}\right)=\frac{1}{8}((3n-3-n-1)3^{n}+3n+3-n+1)=\frac{1}{8}((2n-4)3^{n}+2n+4)=\frac{1}{4}((n-2)3^{n}+n+2),

which is sequence A212337 in the OEIS, with an index shift.

As is often the case in this paper, the difference sequences of our sequences are easier to analyze. For example, for k=2k=2, our sequence A045618(n+1)(n+1) is the sum of the first nn terms of sequence A000337, which is a much older sequence that is simpler. Similarly, for k=3k=3, the sequence A212337(n+1)(n+1) is the sum of the first nn terms of A014915.

Here, we make a general statement.

Proposition 15.

The term F(kj+11k1)F\left(\frac{k^{j+1}-1}{k-1}\right) is the sum of the first jj terms of the sequence b(n,k)b(n,k), which is defined as b(1,k)=1b(1,k)=1 and b(n,k)=nkn1+b(n1,k)b(n,k)=nk^{n-1}+b(n-1,k).

Proof.

First, we show that the closed form of b(n,k)b(n,k) is

b(n,k)=kn((k1)n1)+1(k1)2.b(n,k)=\frac{k^{n}((k-1)n-1)+1}{(k-1)^{2}}.

The closed form satisfies b(1,k)=k((k1)1)+1(k1)2=1b(1,k)=\frac{k((k-1)-1)+1}{(k-1)^{2}}=1, so it is left to prove that it satisfies the recurrence. Indeed, the expression nkn1+b(n1,k)nk^{n-1}+b(n-1,k) is equal to

nkn1+\displaystyle nk^{n-1}+ kn1((k1)(n1)1)+1(k1)2=kn1((k1)2n+(k1)(n1)1)+1(k1)2\displaystyle\frac{k^{n-1}((k-1)(n-1)-1)+1}{(k-1)^{2}}=\frac{k^{n-1}((k-1)^{2}n+(k-1)(n-1)-1)+1}{(k-1)^{2}}
=kn1((k2k)nk)+1(k1)2=kn((k1)n1)+1(k1)2.\displaystyle=\frac{k^{n-1}((k^{2}-k)n-k)+1}{(k-1)^{2}}=\frac{k^{n}((k-1)n-1)+1}{(k-1)^{2}}.

The partial sums of sequence b(n,k)b(n,k) are given by

j=1nkj((k1)j1)+1(k1)2=(k1)j=1njkjj=1nkj+n(k1)2.\sum_{j=1}^{n}\frac{k^{j}((k-1)j-1)+1}{(k-1)^{2}}=\frac{(k-1)\sum_{j=1}^{n}jk^{j}-\sum_{j=1}^{n}k^{j}+n}{(k-1)^{2}}.

Now, we have

j=1njkj=j=1nkn+1kjk1=kn+1((k1)n1)+k(k1)2 and j=1nkj=kn+1kk1.\sum_{j=1}^{n}jk^{j}=\sum_{j=1}^{n}\frac{k^{n+1}-k^{j}}{k-1}=\frac{k^{n+1}((k-1)n-1)+k}{(k-1)^{2}}\quad\text{ and }\quad\sum_{j=1}^{n}k^{j}=\frac{k^{n+1}-k}{k-1}.

Substituting this gives

(k1)kn+1((k1)n1)+k(k1)2kn+1kk1+n(k1)2=kn+1((k1)n1)+kkn+1+k+n(k1)(k1)3.\frac{(k-1)\cdot\frac{k^{n+1}((k-1)n-1)+k}{(k-1)^{2}}-\frac{k^{n+1}-k}{k-1}+n}{(k-1)^{2}}=\frac{k^{n+1}((k-1)n-1)+k-k^{n+1}+k+n(k-1)}{(k-1)^{3}}.

This simplifies to

kn+1(knn2)+k(n+2)n(k1)3,\frac{k^{n+1}(kn-n-2)+k(n+2)-n}{(k-1)^{3}},

which is F(N,k)F(N,k) for N=kn+11k1N=\frac{k^{n+1}-1}{k-1}. This is the same as the sequences we study for index n+1n+1. ∎

For k=4k=4, we get

F(4n13)=127((3n5)4n+3n+5).F\left(\frac{4^{n}-1}{3}\right)=\frac{1}{27}((3n-5)4^{n}+3n+5).

This sequence has been submitted to the OEIS as sequence A378727. Starting from index 1, the first few terms are

0, 1, 10, 67, 380, 1973, 9710, 46119, 213600, 970905, 4349650, 19262731,.0,\ 1,\ 10,\ 67,\ 380,\ 1973,\ 9710,\ 46119,\ 213600,\ 970905,\ 4349650,\ 19262731,\ \ldots.

These terms are the partial sums of sequence A014916.

One can notice that every odd-indexed term is divisible by 10. Here, we make a general statement.

Proposition 16.

We have that F(k2j+11k1)F\left(\frac{k^{2j+1}-1}{k-1}\right) is divisible by 2(k+1)2(k+1) for any non-negative jj.

Proof.

By Proposition 14,

F(k2j+11k1)=(k(2n2)2n)k2n1+k(2n)2n+2(k1)3.F\left(\frac{k^{2j+1}-1}{k-1}\right)=\frac{(k(2n-2)-2n)k^{2n-1}+k(2n)-2n+2}{(k-1)^{3}}.

We want to prove that the above is divisible by 2(k+1)2(k+1). We have

12(k+1)F(k2j+11k1)\displaystyle\frac{1}{2(k+1)}F\left(\frac{k^{2j+1}-1}{k-1}\right) =(k(2n2)2n)k2n1+2nk2n+22(k1)3(k+1)\displaystyle=\frac{(k(2n-2)-2n)k^{2n-1}+2nk-2n+2}{2(k-1)^{3}(k+1)}
=(n1)(k2n1)nk(k2n21)(k1)3(k+1).\displaystyle=\frac{(n-1)(k^{2n}-1)-nk(k^{2n-2}-1)}{(k-1)^{3}(k+1)}.

Dividing the numerator by k21=(k1)(k+1)k^{2}-1=(k-1)(k+1) gives

(n1)(k2n2+k2n4++1)nk(k2n4+k2n6++1).(n-1)(k^{2n-2}+k^{2n-4}+\cdots+1)-nk(k^{2n-4}+k^{2n-6}+\cdots+1).

We want to prove that this expression is divisible by (k1)2(k-1)^{2}. It would be sufficient to consider this expression as a function of kk and to prove that this expression and its derivative equals 0 at k=1k=1. When k=1k=1, this expression is equal to (n1)nn(n1)=0(n-1)n-n(n-1)=0. The derivative of this expression is

(n1)i=0n12ik2i1ni=0n2(2i+1)k2i.(n-1)\sum_{i=0}^{n-1}2ik^{2i-1}-n\sum_{i=0}^{n-2}(2i+1)k^{2i}.

When k=1k=1, the derivative is

2(n1)i=0n1ini=0n2(2i+1)=(n1)(n1)nn(n1)2=0,2(n-1)\sum_{i=0}^{n-1}i-n\sum_{i=0}^{n-2}(2i+1)=(n-1)(n-1)n-n(n-1)^{2}=0,

finishing the proof. ∎

For our next example, we consider k=5k=5.

Example 4.

For k=5k=5, we get

F(5n14)=164((4n6)5n+4n+6)=132((2n3)5n+2n+3).F\left(\frac{5^{n}-1}{4}\right)=\frac{1}{64}((4n-6)5^{n}+4n+6)=\frac{1}{32}((2n-3)5^{n}+2n+3).

This sequence has been submitted to the OEIS as sequence A378728, which starts from index 1 as

0, 1, 12, 98, 684, 4395, 26856, 158692, 915528, 5187989, 28991700,.0,\ 1,\ 12,\ 98,\ 684,\ 4395,\ 26856,\ 158692,\ 915528,\ 5187989,\ 28991700,\ \ldots.

These terms are the partial sums of sequence A014917.

5.3 Differences and schizophrenic numbers

Similar to fi(N,k),f_{i}(N,k), notice that if N{ak+1,ak+2,,(a+1)k}N\in\{ak+1,ak+2,\ldots,(a+1)k\}, then the root and the other vertices of the tree fire the same number of times. Thus, it makes sense to consider the number of fires as a function of Nk.\left\lceil\frac{N}{k}\right\rceil. Therefore, we introduce a new function:

G(m,k)=F(mk,k).G(m,k)=F(mk,k).

Table 2 shows the values of G(m,k)G(m,k) for small values of mm and kk.

kk\mm 1 2 3 4 5 6 7 8 9 10
2 0 1 2 6 7 11 12 23 24 28
3 0 1 2 3 8 9 10 15 16 17
4 0 1 2 3 4 10 11 12 13 19
5 0 1 2 3 4 5 12 13 14 15
6 0 1 2 3 4 5 6 14 15 16
Table 2: Values of G(m,k)G(m,k) for 1m101\leq m\leq 10 and 2k62\leq k\leq 6.

The function of differences proved to be useful in calculating the number of fires for particular vertices. Thus, we introduce a difference function here as well:

D(m,k)=G(m+1,k)G(m,k).D(m,k)=G(m+1,k)-G(m,k).

Table 3 shows D(m,k)D(m,k) for small values of mm and kk.

kk\mm 1 2 3 4 5 6 7 8 9 10
2 1 1 4 1 4 1 11 1 4 1
3 1 1 1 5 1 1 5 1 1 5
4 1 1 1 1 6 1 1 1 6 1
5 1 1 1 1 1 7 1 1 1 1
6 1 1 1 1 1 1 8 1 1 1
Table 3: Values of D(m,k)D(m,k) for 1m101\leq m\leq 10 and 2k62\leq k\leq 6.

For k=2k=2, this is sequence A376132, calculated in [4]. It starts as

1, 1, 4, 1, 4, 1, 11, 1, 4, 1, 11, 1, 4, 1, 26,.1,\ 1,\ 4,\ 1,\ 4,\ 1,\ 11,\ 1,\ 4,\ 1,\ 11,\ 1,\ 4,\ 1,\ 26,\ \ldots.

The distinct values of this sequence are Eulerian numbers, which form sequence A000295, where A000295(n)=2nn1(n)=2^{n}-n-1.

For k=3k=3, we get the sequence D(n,3)D(n,3), which has been submitted to the OEIS as sequence A378962. It starts from index 1 as

1, 1, 1, 5, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5,.1,\ 1,\ 1,\ 5,\ 1,\ 1,\ 5,\ 1,\ 1,\ 5,\ 1,\ 1,\ 18,\ 1,\ 1,\ 5,\ 1,\ 1,\ 5,\ \ldots.

The difference sequences for the total number of fires show a similar pattern to the difference sequences for root fires. The data show that, for a given kk, both sequences have ones in the same places. Moreover, when the difference sequence for the number of root fires has nn in some place, the difference sequence for the total number of fires has the same number, which can be viewed as a function of nn.

The sequences of the unique values are summarized in Table 4. For precision, we note that sequence A000295 is shifted relative to our indexing: A000295(2)=1A000295(2)=1. However, the database has a sequence that matches our indexing: A125128, as A125128(n)=A000295(n+1)A125128(n)=A000295(n+1). Also, excluding index 1, sequence A130103 matches A000295. Sequence A000340 is also shifted, but in the opposite direction: A000340(0)=1A000340(0)=1.

kk\mm 1 2 3 4 5 6 7 A#
2 1 4 11 26 57 120 247 A000295
3 1 5 18 58 179 543 1636 A000340
4 1 6 27 112 453 1818 7279 A014825
5 1 7 38 194 975 4881 24412 A014827
6 1 8 51 310 1865 11196 67183 A014829
7 1 9 66 466 3267 22875 160132 A014830
8 1 10 83 668 5349 42798 342391 A014831
9 1 11 102 922 8303 74733 672604 A014832
10 1 12 123 1234 12345 123456 1234567 A014824
Table 4: Unique values for 2k102\leq k\leq 10.

The last row shows an interesting pattern: the iith number for i<ki<k is a concatenation of the first jj digits. The same is true for sequences in other bases. These sequences start with a(1)=1a(1)=1 and follow the recursion a(n)=ka(n1)+na(n)=ka(n-1)+n.

The square roots of the odd numbers in sequence A014824 are irrational, but they display some interesting patterns that mimic rational numbers.

Example 5.

Consider a(11,10)=12345679011a(11,10)=12345679011. Its square root starts as

111111.11110505555555539054166665767340972160955659283519805.111111.11110505555555539054166665767340972160955659283519805.

There are three blocks of repeated digits here. The number starts with 10 ones, and then shortly after, there is a block of 8 fives and a block of 4 sixes. Compare this with the square root of a(19,10)=1234567901234567899a(19,10)=1234567901234567899, which starts as

1111111111.11111111010555555555555555510054166666666666625487909722222222175.1111111111.11111111010555555555555555510054166666666666625487909722222222175.

There are longer blocks of the same digits and a new block of twos. The blocks with repeated digits decrease in size, and the number of digits between blocks increases.

A schizophrenic number or mock rational number is an irrational number that displays certain characteristics of rational numbers [10], as shown in the example above. Such numbers can be similarly defined in other bases. They all begin with blocks of ones in their corresponding bases and seem to appear rational, but then the pattern breaks, and they appear irrational. Later on, they seem to appear rational again with another digit until the pattern breaks again. The blocks of rationality shrink, and the blocks of irrationality lengthen until, eventually, the whole thing disintegrates into the full chaos of irrational numbers. The schizophrenic numbers in base kk are square roots of odd numbers in the corresponding sequence.

Peter Bala made an interesting comment on sequence A014824 in the OEIS [8], suggesting that the inverse of a schizophrenic number also exhibits schizophrenic patterns. These patterns are even more striking as blocks of repeated digits always contain the same digit: zero. Presumably, the same pattern holds for other bases.

Example 6.

Consider a(11,10)=12345679011a(11,10)=12345679011 and a(19,10)=1234567901234567899a(19,10)=1234567901234567899 as in the previous example. The first few digits of 1a(11,10)\sqrt{\frac{1}{a(11,10)}} and 1a(19,10)\sqrt{\frac{1}{a(19,10)}} are

0.00000900000000049050000004009837500364226906284738141187001561650.0000090000000004905000000400983750036422690628473814118700156165

and

0.0000000009000000000000000008145000000000000011056837500000000016.0.0000000009000000000000000008145000000000000011056837500000000016.

There are blocks of repeated zeros. Ignoring the initial zeros, the block sizes decrease, and the number of digits between blocks increases. The pattern is similar to the schizophrenic pattern, except that the blocks always contain zeros.

We define functions a(n,k)a(n,k) for all k2k\geq 2 as a(n,k)=ka(n1,k)+na(n,k)=ka(n-1,k)+n for n2n\geq 2 and a(1,k)=1a(1,k)=1. This is the same recurrence mentioned in the sequences in Table 4. This recursion explains why the first kk terms of each sequence written in base kk are the concatenations of the first kk digits.

We first get the closed formula for a(n,k)a(n,k).

Lemma 17.

We have that

a(n,k)=kn+1(k1)nk(k1)2.a(n,k)=\frac{k^{n+1}-(k-1)n-k}{(k-1)^{2}}.
Proof.

We prove this with induction on nn. The base case is n=1n=1, and the formula gives a(1,k)=k22k+1(k1)2=1a(1,k)=\frac{k^{2}-2k+1}{(k-1)^{2}}=1, which is true by definition.

Assume a(n,k)=kn+1(k1)nk(k1)2a(n,k)=\frac{k^{n+1}-(k-1)n-k}{(k-1)^{2}}. The recursion tells us that a(n+1,k)=a(n,k)+n+1a(n+1,k)=a(n,k)+n+1. Thus,

a(n+1,k)\displaystyle a(n+1,k) =kn+2(k2k)nk2+(k1)2n+(k1)2(k1)2\displaystyle=\frac{k^{n+2}-(k^{2}-k)n-k^{2}+(k-1)^{2}n+(k-1)^{2}}{(k-1)^{2}}
=kn+2(k1)(n+1)k(k1)2,\displaystyle=\frac{k^{n+2}-(k-1)(n+1)-k}{(k-1)^{2}},

concluding the proof. ∎

We are now ready to prove the observation about the differences mentioned earlier.

5.4 Formulae for differences

In Proposition 10 and Theorem 11, we provided the recurrence and the formula for the function d0d_{0}. In this section, we do the same for function DD.

Proposition 18.

The difference sequence D(m,k)=G(m+1,k)G(m,k)D(m,k)=G(m+1,k)-G(m,k) satisfies the recurrence

D(m,k)={d0(m)+kD(m1k), if m1 is a multiple of k,1, otherwise.D(m,k)=\begin{cases}d_{0}(m)+kD(\frac{m-1}{k}),&\text{ if $m-1$ is a multiple of $k$},\\ 1,&\text{ otherwise}.\end{cases}
Proof.

Assume that the second parameter of all functions is kk. We have

D(m)\displaystyle D(m) =G(m+1)G(m)=F(km+k)F(km)\displaystyle=G(m+1)-G(m)=F(km+k)-F(km)
=f0(km+k)f0(km)+k(F(m)F(m1))\displaystyle=f_{0}(km+k)-f_{0}(km)+k(F(m)-F(m-1))
=d0(m)+k(F(m)F(m1)).\displaystyle=d_{0}(m)+k(F(m)-F(m-1)).

Note that F(m)F(m1)0F(m)-F(m-1)\neq 0 if and only if kk divides m1m-1. In this case, we have

D(m)\displaystyle D(m) =d0(m)+k(F(m)F(m1))\displaystyle=d_{0}(m)+k(F(m)-F(m-1))
=d0(m)+k(G(m1k+1)G(m1k))\displaystyle=d_{0}(m)+k\left(G\left(\frac{m-1}{k}+1\right)-G\left(\frac{m-1}{k}\right)\right)
=d0(m)+kD(m1k).\displaystyle=d_{0}(m)+kD\left(\frac{m-1}{k}\right).

If kk does not divide m1m-1, then F(m)F(m1)=0F(m)-F(m-1)=0 and d0(m)=1d_{0}(m)=1. Therefore,

D(m)=d0(m)+k(F(m)F(m1))=1,D(m)=d_{0}(m)+k(F(m)-F(m-1))=1,

concluding the proof. ∎

Using the above proposition, we can prove the following theorem.

Theorem 19.

We have

D(m,k)=G(m+1,k)G(m,k)=a(d0(m,k),k).D(m,k)=G(m+1,k)-G(m,k)=a(d_{0}(m,k),k).
Proof.

We prove this using induction on the value of d0(m,k)d_{0}(m,k). If d0(m)=1d_{0}(m)=1, then m1m-1 is not a multiple of kk. Thus, D(m,k)=1=a(1,k)D(m,k)=1=a(1,k), as expected. We assume that we have proven the statement for all d0(m)=dd_{0}(m)=d for some value dd, and we need to prove the statement for d0(m)=d+1d_{0}(m)=d+1. This means that m1m-1 is a multiple of kk, so we can substitute:

D(m)=d0(m)+kD(m1k).D(m)=d_{0}(m)+kD\left(\frac{m-1}{k}\right).

Note that d0(m1k)=d0(m)1=dd_{0}(\frac{m-1}{k})=d_{0}(m)-1=d. Therefore, by induction, we know that

D(m1k)=a(d0(m1k,k),k)=a(d,k).D\left(\frac{m-1}{k}\right)=a\left(d_{0}\left(\frac{m-1}{k},k\right),k\right)=a(d,k).

Thus,

D(m)=d+1+ka(d,k)=a(d+1,k)=a(d0(m,K),k).D(m)=d+1+ka(d,k)=a(d+1,k)=a(d_{0}(m,K),k).

Now, we can use Lemma 17 to produce an explicit formula.

Corollary 20.

We have

D(m,k)=G(m+1,k)G(m,k)=kj+1(k1)jk(k1)2,D(m,k)=G(m+1,k)-G(m,k)=\frac{k^{j+1}-(k-1)j-k}{(k-1)^{2}},

where j=νk(mkn1k1)+1j=\nu_{k}(m-\frac{k^{n}-1}{k-1})+1, if mkn1k1m\neq\frac{k^{n}-1}{k-1}. Otherwise, j=nj=n.

6 Acknowledgments

This project was completed as part of the MIT PRIMES STEP program, and we are grateful to the program and its staff.

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