Chip-Firing on Infinite -ary Trees
Abstract
We use an infinite -ary tree with a self-loop at the root as our underlying graph. We consider a chip-firing process starting with 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 as a function of 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 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 -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 -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 -ary tree with a self-loop at the root, and our chip-firing game starts with chips at the root.
In Section 3, we show that if we start with chips at the root of our tree, the stable configuration will form a finite tree of height , where is the integer such that . The number of chips on a vertex in the stable configuration is dependent only on its layer and is a function of the -ary expansion of .
In Section 4, we calculate the number of times each vertex fires, given the total number of chips (). In Section 4.2, we use the formula from the previous section to explore the number of root fires in more detail. Given , the number of root fires as a function of —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 , 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 -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 has exactly one parent vertex , defined as the first vertex after in the shortest path from to the root. The vertex is said to be a child of the vertex . Any vertex other than vertex on the shortest path from to root is an ancestor of , while is a descendant of each of its ancestors.

We define a vertex as being on layer if the shortest path from the root to traverses edges. By this definition, the root 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 -ary tree, every non-leaf vertex has children, and all leaves are the same distance from the root. In other words, all leaves are located on the same layer , where 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 chips. When a vertex fires, it sends one chip along each edge. We consider the chip-firing game where the initial state has 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 chips.
The following theorem, proved in [2], describes when the chip-firing game may stabilize. It applies to any connected undirected graph , where is the set of vertices and is the set of edges. We denote the number of vertices and edges by and , respectively.
Theorem 1 (Theorem 2.3 in [2]).
Let be the total number of chips. We have:
-
•
If , then the game is infinite and never reaches a stable configuration.
-
•
If , then the game is finite, reaching a stable configuration.
-
•
If , 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 -ary tree with a finite number of chips; therefore, we have , 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
Our formulae related to -ary trees can be better expressed in base . Hence, we use the notation or to indicate that a string of digits is written in base . For example, . Correspondingly, we will write to denote the string of digits representing a decimal integer in base . Thus, is equivalent to .
We denote by the largest power of that divides . In other words, is the number of trailing zeros in . When is prime, the value is called the -adic valuation of . We will extend the terminology to composite numbers as well. In other words,
3 Stable configurations
In Proposition 3.3 from [7], the stable configurations are described for the case . We describe the stable configurations for any .
Proposition 3.
If we start with chips at the root, where , then the vertices containing chips in the stable configuration form a perfect -ary tree with height . Furthermore, every vertex on the same layer has the same number of chips.
Proof.
We prove this proposition by induction on . When , the number of chips at the root ranges from to , which is already a stable configuration, so this is a perfect -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 ; we will prove that it holds for . Suppose we start with chips at the root, where . We deploy the following firing strategy.
-
1.
Fire the root repeatedly until it cannot fire anymore.
-
2.
Fire the root’s children and their subtrees in parallel.
-
3.
Whenever the children of the root fire, fire the root.
-
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 and chips. In step 3, when the children of the root fire, they each send 1 chip to the root. After that, the root gets 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 and , 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 -ary tree with height . As the firing processes in all 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 -ary tree with height , 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 chips at the root, where , then for , the resulting stable configuration has chips on each vertex on layer , where is the base expansion of with possible leading zeros.
Proof.
We know that all vertices with chips form a perfect -ary tree, where each vertex with chips has from 1 to chips, inclusive, and each vertex in the same layer has the same number of chips. Suppose each vertex on layer has chips. As there are vertices on layer , the total number of chips is , which equals . Thus, the distribution of chips into layers in the final configuration implies representing in base with digits between and . 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 . When we place chips on each vertex, we are placing a total of
chips, as desired. ∎
We see that numbers of the form play a special role, as the stable configuration reaches layer for the first time when we start with the number of chips in this form placed at the root.
From now on, let be the base expansion of with possible leading zeros. When we need to emphasize the dependency of these digits on the base and the number , we will use the notation instead of . Similar to [7, 4], for the stable configuration starting with chips at the root in a -ary tree, let us denote the number of chips at each vertex on layer as for . Proposition 4 states that . When the values of and are fixed, we will use the notations and to simplify the calculations.
The number also plays a special role in our paper. It was defined through inequalities
After multiplying every term by and adding 1, we get
Taking the log base of both sides gives us
Thus, we can define as
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 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 -ary tree.
4.1 The number of fires for a vertex on layer
Let denote the total number of fires of each vertex on layer . When and are fixed, we will sometimes use the notation to simplify the calculations. We first look at the difference in the number of fires between layers and .
Lemma 5.
The difference in the number of fires between vertices on layers and equals the number of chips in the stable configuration belonging to the descendants of one particular vertex on layer divided by . In other words,
where .
Proof.
The number of times a vertex fires equals the total number of chips sent to all children of the vertex divided by . 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 times the number of fires of each child. Thus, the difference between the number of fires of a vertex and its child is the number of chips belonging to the descendants of in the stable configuration divided by .
Let vertex be on layer , and let . The number of vertices on layer that are the descendants of is . Each descendant contains chips by Proposition 4. Adding them up across all of the vertices, the number of chips placed at the descendants of is , and the result follows after dividing this by . ∎
Example 1.
As the chips on the last layer, , never fire, we have . From Lemma 5, we get .
Now, we can calculate the number of fires for each vertex on layer .
Theorem 6.
Given the total number of chips and the index , the number of fires for each vertex on layer is:
Proof.
We prove this formula using backward induction, starting with . As the last layer never fires, we have . Our formula gives us the same: . Thus, the base of the induction is established.
Suppose the formula is true for a given . We want to calculate the number of fires for . In other words, we want to find assuming that .
From Lemma 5, we have that
Thus, after reordering terms and keeping track of indices, we get
concluding the proof. ∎
The following corollary allows us to express the number of fires of each vertex on layer through the number of fires of the root.
Corollary 7.
We have
where .
Proof.
The formula for the number of fires of each vertex on layer is
This matches up with the total number of fires for the root of a tree if we start with chips, where is the first digits of . In other words,
In this case,
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 and the index , the number of times the root fires is
Example 2.
Suppose we have a ternary tree , and the total number of chips is 9 . This means that we have index . The ternary representation of is , implying that and . This means that the total number of times the root fires is times.
As in [4], we are looking for a recursive formula.
Corollary 9.
We have
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
The case of is important throughout our paper, as the stable configuration reaches layer for the first time when we start with chips in this form placed at the root. Moreover, if we start with chips at the root, then for . Thus, the number of fires for a vertex on level is
Plugging in , we get
matching the answer from [7].
The total number of root fires when is
Plugging in , we see that for chips on a binary tree, the root fires 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
When we plug in , we get that there are root fires, which corresponds to sequence A000340 in the OEIS with a shifted index. Similarly, when we plug in and , we get and 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 , 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 . Therefore, we introduce a new set of functions:
Table 1 shows values of for and .
\ | 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 |
The sequence has been submitted to the OEIS as sequence A378724 and starts as
Consider the difference sequence . In this section, we are interested in the case . The following proposition describes the recursion for .
Proposition 10.
The difference sequence satisfies the recurrence
Proof.
We know that
by Corollary 9. Now, since depends only on , if we have , it means that is not a multiple of . This implies that . If is a multiple of , then we need to calculate . We have and . Therefore, , and , as claimed. ∎
Now, we can state the formula.
Theorem 11.
The difference sequence satisfies the formula
Proof.
If , then we can prove the formula based on the recurrence and using induction on . When , we have and ,. Thus, the base of the induction is established. Assume that . By the recurrence from Proposition 10, we have
Now, if cannot be represented as for any integer , then we can prove the formula based on the recurrence and using induction. Our base case is when . Therefore, we see that is not a multiple of . The recurrence gives us . Thus, the base of induction is established.
Assume for all , such that is true, then satisfies the recurrence. Consider some such that . Since , we know that must be a multiple of . By the recurrence we have . Since , we have . Notice that . Thus, we have that . ∎
In other words, excluding the first digit, if ends with ones, we have .
Remark.
If , then the number of trailing ones in the base representation of , without the first digit, is equal to the number of trailing zeros of in base or . This is the same as .
Example 3.
The difference sequence 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 into the binary representation of . Sequence A091090 starts from index 1 as
The difference sequence has been submitted to the OEIS as sequence A378725 and starts as
One can notice that these sequences all have a similar structure due to a similar repetition pattern. Each sequence begins with ones. Afterward, there are ones between every two numbers that are not equal to 1. Furthermore, for any , if we remove all numbers less than , the resulting sequence starts with occurrences of , followed by occurrences of between every two numbers that are not equal to .
In other words, we can construct the difference sequence in the following way. Start by creating a sequence of all ’s. Then, replace every th with a beginning with the th . Next, replace every th with a , starting with the th . Continue this process indefinitely, replacing every th occurrence of with , beginning with the th . This fractal structure of sequences 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 -ary tree with a self-loop at the root. Let denote the total number of fires in a -ary tree when we start with chips placed at the root. As we have done previously, we sometimes remove the parameters in the function when and are fixed.
Theorem 12.
The total number of fires is
where .
Proof.
Rearranging and replacing with , we get
This concludes the proof. ∎
For , 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
For , the sequence of the total number of fires starts from index 1 as
This sequence has been submitted to the OEIS as sequence A378726. We have a recursive formula for function , similar to the formula for found in Corollary 9.
Proposition 13.
We have
Proof.
The total number of fires is equal to the number of root fires plus the number of fires in all of the subtrees with roots on level 2. Each subtree receives chips from the root, so each would fire times. ∎
5.2 Special case of
If we start with chips at the root, we have for . This case is special, so we want to look at the total number of fires for such separately.
Proposition 14.
We have
Proof.
Using the formula for the total number of fires, we get that there are a total of
fires. ∎
For , we get
This matches Corollary 3.7 in [7], and it is sequence A045618 in the OEIS, with an index shift.
For , we get
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 , our sequence A045618 is the sum of the first terms of sequence A000337, which is a much older sequence that is simpler. Similarly, for , the sequence A212337 is the sum of the first terms of A014915.
Here, we make a general statement.
Proposition 15.
The term is the sum of the first terms of the sequence , which is defined as and .
Proof.
First, we show that the closed form of is
The closed form satisfies , so it is left to prove that it satisfies the recurrence. Indeed, the expression is equal to
The partial sums of sequence are given by
Now, we have
Substituting this gives
This simplifies to
which is for . This is the same as the sequences we study for index . ∎
For , we get
This sequence has been submitted to the OEIS as sequence A378727. Starting from index 1, the first few terms are
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 is divisible by for any non-negative .
Proof.
By Proposition 14,
We want to prove that the above is divisible by . We have
Dividing the numerator by gives
We want to prove that this expression is divisible by . It would be sufficient to consider this expression as a function of and to prove that this expression and its derivative equals 0 at . When , this expression is equal to . The derivative of this expression is
When , the derivative is
finishing the proof. ∎
For our next example, we consider .
Example 4.
For , we get
This sequence has been submitted to the OEIS as sequence A378728, which starts from index 1 as
These terms are the partial sums of sequence A014917.
5.3 Differences and schizophrenic numbers
Similar to notice that if , 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 Therefore, we introduce a new function:
Table 2 shows the values of for small values of and .
\ | 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 |
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:
Table 3 shows for small values of and .
\ | 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 |
For , this is sequence A376132, calculated in [4]. It starts as
The distinct values of this sequence are Eulerian numbers, which form sequence A000295, where A000295.
For , we get the sequence , which has been submitted to the OEIS as sequence A378962. It starts from index 1 as
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 , both sequences have ones in the same places. Moreover, when the difference sequence for the number of root fires has in some place, the difference sequence for the total number of fires has the same number, which can be viewed as a function of .
The sequences of the unique values are summarized in Table 4. For precision, we note that sequence A000295 is shifted relative to our indexing: . However, the database has a sequence that matches our indexing: A125128, as . Also, excluding index 1, sequence A130103 matches A000295. Sequence A000340 is also shifted, but in the opposite direction: .
\ | 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 |
The last row shows an interesting pattern: the th number for is a concatenation of the first digits. The same is true for sequences in other bases. These sequences start with and follow the recursion .
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 . Its square root starts as
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 , which starts as
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 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 and as in the previous example. The first few digits of and are
and
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 for all as for and . This is the same recurrence mentioned in the sequences in Table 4. This recursion explains why the first terms of each sequence written in base are the concatenations of the first digits.
We first get the closed formula for .
Lemma 17.
We have that
Proof.
We prove this with induction on . The base case is , and the formula gives , which is true by definition.
Assume . The recursion tells us that . Thus,
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 . In this section, we do the same for function .
Proposition 18.
The difference sequence satisfies the recurrence
Proof.
Assume that the second parameter of all functions is . We have
Note that if and only if divides . In this case, we have
If does not divide , then and . Therefore,
concluding the proof. ∎
Using the above proposition, we can prove the following theorem.
Theorem 19.
We have
Proof.
We prove this using induction on the value of . If , then is not a multiple of . Thus, , as expected. We assume that we have proven the statement for all for some value , and we need to prove the statement for . This means that is a multiple of , so we can substitute:
Note that . Therefore, by induction, we know that
Thus,
∎
Now, we can use Lemma 17 to produce an explicit formula.
Corollary 20.
We have
where , if . Otherwise, .
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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