Comma Sequence: Finite Paths, Landmines and the Immortal 20

This article explores the "comma sequence," a mathematical concept developed by Eric Angelini, and its intriguing properties, including its finite nature in most bases and the possibility of an infinite sequence under specific conditions.

The Comma Sequence: Definition and Derivation

The comma sequence is built upon the idea of "comma separators." Given any sequence of numbers, a comma separator is formed by taking the digit immediately to the left of a comma and the digit immediately to the right. For example, in the sequence 1, 4, 9, 16, 25:

• The comma separator between 1 and 4 is 14.
• The comma separator between 4 and 9 is 49.
• The comma separator between 9 and 16 is 91 (taking the last digit of 9 and the first digit of 16).
• The comma separator between 16 and 25 is 62.

This process can be applied to any sequence.

The Angelini's Challenge: Comma Sequence Equals Difference Sequence

A common way to analyze sequences is to look at the differences between consecutive terms. For the sequence of squares (1, 4, 9, 16, 25), the difference sequence is:

• 4 - 1 = 3
• 9 - 4 = 5
• 16 - 9 = 7
• 25 - 16 = 9

This results in the sequence of odd numbers (3, 5, 7, 9).

Eric Angelini posed a challenge: find a sequence where the comma sequence derived from it is identical to its difference sequence.

Constructing Such a Sequence

Let's try to build such a sequence, starting with 1, and always choosing the smallest, simplest positive number without repeats.

1. Start with 1: The sequence begins with 1.
2. First Term: We need to find the next term. Let's try 12.
   • The difference between 1 and 12 is 11.
   • The comma separator between 1 and 12 is 12.
   • These don't match. The goal is for the difference to equal the comma separator.

Let's re-evaluate the definition. The difference between two terms must be equal to the comma separator formed by those two terms.

Let the sequence be $S_1, S_2, S_3, \dots$. The difference is $S_{n+1} - S_n$. The comma separator is formed by the last digit of $S_n$ and the first digit of $S_{n+1}$.

Let's try again, starting with 1. 1. Start with 1: $S_1 = 1$. 2. Finding $S_2$: * The difference $S_2 - S_1$ must equal the comma separator formed by $S_1$ and $S_2$. * If $S_2 = 12$, then $S_2 - S_1 = 11$. The comma separator is 12. These don't match.

The example provided in the transcript starts with 1, and the next term is 12. The difference is 11. The comma separator is 12. This seems to be a slight misstatement in the explanation, or a different interpretation of the rule. The core rule, as clarified later, is that the difference between two terms must be equal to the comma separator.

Let's follow the example given: 1. Start with 1. 2. Next term is 12. * Difference: $12 - 1 = 11$. * Comma separator (from 1 and 12): 12. * The example then states: "The reason I chose 12 is that the difference here is 11. And that matches the comma sequence. Good. That must always happen." This implies that the difference is the comma sequence, not the comma separator. This is a crucial distinction.

Let's assume the rule is: the difference between $S_{n+1}$ and $S_n$ is equal to the comma separator formed by the last digit of $S_n$ and the first digit of $S_{n+1}$.

Let's restart the construction with this understanding:

1. Start with 1.
2. Find $S_2$:
   • We need $S_2 - 1 = \text{comma separator}(1, S_2)$.
   • If $S_2 = 12$, then $S_2 - 1 = 11$. The comma separator is 12. This doesn't work.

The example in the transcript seems to imply that the comma sequence (the sequence of comma separators) is the same as the difference sequence.

Let's follow the example's construction: Sequence: 1, 12, 35, 94, 135, 186...

1. 1 to 12:
   • Difference: $12 - 1 = 11$.
   • Comma separator (from 1 and 12): 12.
   • The speaker says, "The reason I chose 12 is that the difference here is 11. And that matches the comma sequence." This suggests the comma sequence is the difference sequence.

Let's assume the rule is: $S_{n+1} - S_n = \text{comma separator}(S_n, S_{n+1})$.

1. Start with 1.
2. Find $S_2$:
   • $S_2 - 1 = \text{comma separator}(1, S_2)$.
   • The comma separator will be $1X$ where $X$ is the first digit of $S_2$.
   • If $S_2 = 12$, then $12 - 1 = 11$. The comma separator is 12. This still doesn't match.

The example then proceeds: * $S_1 = 1$. * $S_2 = 12$. Difference is 11. Comma separator is 12. * The speaker then says, "What are we going to put after the 12? Well, one thing for sure is the comma separator is going to be 20some." This implies the comma separator is formed by the last digit of $S_n$ and the first digit of $S_{n+1}$. * So, for $S_2=12$, the last digit is 2. The next comma separator will be $2X$. * The difference $S_3 - S_2$ must be $2X$. * The speaker guesses the next term begins with 3, making the comma separator 23. * So, $S_3 - 12 = 23 \implies S_3 = 35$. * Check: $35 - 12 = 23$. The comma separator from 12 and 35 is 23 (last digit of 12 is 2, first digit of 35 is 3). This works!

Let's continue: 1. $S_3 = 35$. 2. Find $S_4$: * Last digit of $S_3$ is 5. * The speaker guesses the next term begins with 9, making the comma separator 59. * So, $S_4 - 35 = 59 \implies S_4 = 94$. * Check: $94 - 35 = 59$. The comma separator from 35 and 94 is 59. This works!

The sequence is generated by the rule: $S_{n+1} = S_n + \text{comma separator}(S_n, S_{n+1})$, where $\text{comma separator}(S_n, S_{n+1})$ is formed by the last digit of $S_n$ and the first digit of $S_{n+1}$.

This sequence is: 1. 1 2. 12 (Difference 11, but the rule is applied from the previous term. The first term is a starting point.) 3. 35 ($12 + 23 = 35$. Comma separator from 12 and 35 is 23.) 4. 94 ($35 + 59 = 94$. Comma separator from 35 and 94 is 59.) 5. 135 ($94 + 41 = 135$. Comma separator from 94 and 135 is 41.) 6. 186 ($135 + 51 = 186$. Comma separator from 135 and 186 is 51.)

The comma separator is always a two-digit number (between 1 and 99), meaning the jumps between terms are relatively small. This keeps the sequence "under control." If a number ends in 0, say 30, and the next number starts with 9, the comma separator would be 09, a single-digit number.

The Kangaroo and Landmines: When the Sequence Dies

The analogy of a kangaroo hopping along a number line is used to describe the sequence. The kangaroo takes small steps (jumps of 1 to 99). However, there are "landmines" – numbers where the sequence cannot continue.

A landmine occurs when the last two digits of a number add up to 9, and all other preceding digits are nines. For example, 999999. The last two digits are 9 and 9, which add to 18, not 9. The actual rule for a landmine is: when the last digit of $S_n$ and the first digit of $S_{n+1}$ form a comma separator, and there is no valid $S_{n+1}$ that satisfies the rule.

More precisely, a landmine is a number $N$ such that if $N$ is $S_n$, there is no valid $S_{n+1}$ that can be generated. This happens when the last digit of $N$ (let's call it $L$) and any possible first digit of $S_{n+1}$ (let's call it $F$) form a two-digit number $LF$ such that $N + LF$ cannot be a valid $S_{n+1}$ (i.e., its first digit is not $F$).

Example of a landmine: If we start with 3: 1. 3 2. Find $S_2$: * Last digit of $S_1$ is 3. * If we assume the next term starts with 3, the comma separator is 33. * $S_2 = 3 + 33 = 36$. * Check: $36 - 3 = 33$. Comma separator from 3 and 36 is 33. This works. 3. $S_2 = 36$. Find $S_3$: * Last digit of $S_2$ is 6. * We need $S_3 - 36 = \text{comma separator}(36, S_3)$. * The comma separator will be $6X$. * If $X=1$, comma separator is 61. $S_3 = 36 + 61 = 97$. The first digit of 97 is 9, not 1. So 61 is not a valid comma separator. * If $X=2$, comma separator is 62. $S_3 = 36 + 62 = 98$. First digit is 9, not 2. * ... * If $X=9$, comma separator is 69. $S_3 = 36 + 69 = 105$. First digit is 1, not 9. * No valid $X$ exists. Therefore, 36 is a landmine. The sequence dies after two steps.

The specific landmine condition is when the last digit of $S_n$ is $L$, and for any possible first digit $F$ of $S_{n+1}$, the number $S_n + LF$ does not start with $F$. This happens when $L+F$ is 9, and $S_n$ is a number like $X9$, where $X$ is any number of nines. For example, 999999.

The Fate of Sequences in Base 10

• Starting with 1: The sequence goes for 2,137,453 steps before hitting a landmine.
• Starting with 3: The sequence dies after 2 steps (at 36).
• Starting with 2: The sequence goes for $2 \times 10^{14}$ terms before hitting a landmine.
• Starting with 4, 5, 6: These also go for a long time but eventually hit landmines.

It has been proven that all comma sequences in base 10 eventually die. This proof was extremely complex, initially done for base 3 by computer-aided methods, and later generalized by two graduate students to base 10 and other bases up to 643.

The Immortal Kangaroo: A Variation

While all sequences in base 10 eventually die under the original rules, a slight variation allows for an "immortal kangaroo" – an infinite sequence.

The original rule states that for each step, there is usually only one possible next term. However, in rare cases, there might be two possible next terms (never more than two). When this happens, the original rule dictates choosing the smallest of the two options.

The variation is: if there are two choices for the next term, we are allowed to choose the larger one.

Example: Starting with 14. 1. 14 2. Find $S_2$: * Last digit of 14 is 4. * Choice 1 (smallest): Assume $S_2$ starts with 5. Comma separator is 45. * $S_2 = 14 + 45 = 59$. * Check: $59 - 14 = 45$. Comma separator from 14 and 59 is 45. This is a valid path. * Choice 2 (larger): Assume $S_2$ starts with 6. Comma separator is 46. * $S_2 = 14 + 46 = 60$. * Check: $60 - 14 = 46$. Comma separator from 14 and 60 is 46. This is also a valid path.

If we are allowed to choose between 59 and 60, we can potentially "dodge" landmines. Even if one path leads to a landmine, the other might continue.

The Axiom of Choice and the Immortal Path

It has been proven that there exists at least one infinite path in base 10 if we are allowed to make these choices. This proof relies on König's Lemma, which in turn depends on the Axiom of Choice. This means we know an infinite path exists, but we don't have a constructive algorithm to find it. We don't know the specific sequence of choices (smaller or larger) that leads to immortality. This sequence of choices is itself a sequence in the OEIS (Online Encyclopedia of Integer Sequences), but its full form is unknown.

The Magic Starting Number: 20

Despite the non-constructive proof, extensive computational analysis has revealed a surprising fact: * If you start with any number other than 20 (between 1 and 99), the sequence will eventually die, even with the option to choose the larger path. * Starting with 20 is the only way to achieve an immortal sequence in base 10.

If you start with 20, and make the "right" choices at each branching point, the sequence can continue indefinitely. The first few steps for starting with 20 are: 1. 20 2. Find $S_2$: * Last digit of 20 is 0. * Assume $S_2$ starts with 2. Comma separator is 02. * $S_2 = 20 + 02 = 22$. * Check: $22 - 20 = 02$. Comma separator from 20 and 22 is 02. This works. * (There might be other choices later on, but for this initial step, 22 is the only valid next term.)

The research on comma sequences is a collaborative effort involving Eric Angelini, Michael Branicki, Giovani Rester, David Wilson, and others. While many questions remain open, such as a simple formula for path length or a general proof for bases greater than two, the existence of the immortal kangaroo sequence starting with 20 in base 10 is a fascinating discovery.