# Matthew sequences

Published:

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I was recently shown a fun problem by Matthew, my younger brother in ninth grade. It goes like this:

What is the probability of answering this question correctly by guessing?

1. 60%
2. 25%
3. 25%
4. 30%

The answer is that there is no answer. The probability of choosing a correct answer by guessing is

$P = \frac{\text{# of ways to choose correct answer}}{4}.$

None of the answers equal the $P$ value that they generate, so none of the answers work.

This problem sounds silly at first, but has some fascinating ideas in it. It is essentially asking us this: Given a sequence of numbers, what numbers are equal to how often they occur in the sequence? The following is my normalized reframing of it.

Definition. A Matthew sequence is a finite sequence of reals in $(0, 1]$ that sum to $1$. For such a sequence $S$ consisting of terms $a_1, \dots, a_n$, let $f_S(x)$ be the proportion of elements of $S$ that equal $x$. That is,

$f_S(x) = \frac{\sum_{k = 1}^n [x = a_k]}{n}.$

A fixed point of $S$ is a fixed point of $f_S$ in $S$. That is, it is a real $x \in S$ such that

$x = \frac{\sum_{k = 1}^n [x = a_k]}{n}.$

The fixed point $1$ of the Matthew sequence $[1]$ is called trivial.

In this language, the problem is asking us to find the fixed point of a given finite sequence.

Examples:

• $1$ is a fixed point of $[1]$.
• $1/3$ is a fixed point of $S = [1 / 2, 1 / 3, 1 / 6]$.
• $2 / 5$ is a fixed point of $[2 / 5, 2 / 5, 1/15, 1/15, 1/15]$.
• The sequence $[1/4, 1/4, 1/3, 1/6]$ has no fixed point.

I think that there are a few natural questions:

1. What numbers can occur as fixed points of a Matthew sequence?
2. How many fixed points can occur in a Matthew sequence of fixed length?
3. Do there exist Matthew sequences with arbitrarily many fixed points?
4. If we know a particular fixed point, can we say anything about the size of the sequence it comes from?

In this post, I shall answer the first three questions.

# Classifying fixed points

Not every number can be a fixed point of a Matthew sequence. Fixed points must be rational, for instance, but we can get even harsher restrictions. For example, it is impossible for $\tfrac{1}{2}$ to be a fixed point of any Matthew sequence. For it to be a fixed point it would need to make up exactly $\tfrac{1}{2}$ of the elements of the sequence. However, since the sequence must sum to $1$, this means that the sequence must be $[1/2, 1/2]$, of which $1/2$ is not a fixed point. This argument generalizes to numbers greater than $\tfrac{1}{2}$ as well.

Lemma. Every nontrivial fixed point of a Matthew sequence is strictly less than $1/2$.

Proof. Suppose that a Matthew sequence $S$ had a fixed point $p \geq 1/2$. Since $2p \geq 1$, the multiplicity of $p$ can be at most $2$. If it is exactly $2$, then we have $2p \leq 1$ by the sum condition, which them implies $p = 1/2$. We must conclude that $S = [1/2, 1/2]$, but then $p = 1/2$ is not a fixed point.

Suppose that there is exactly a single $p$ present in $S$. For $p$ to be a fixed point, we must have $p = 1 / n$ where $n$ is the size of our Matthew sequence. Since $p \geq 1 / 2$, this implies $n \leq 2$, so our sequence is either $[p]$ or $[p, 1 - p]$. The first case is possible only when $p$ is nontrivial, so we must be in the second case. But then $p$ could only be a fixed point if $p = 1/2$, which produces the sequence $[1/2, 1/2]$. Therefore $p$ cannot be a fixed point. $\blacksquare$

Proposition. If $\tfrac{a}{b}$ is a nontrivial fixed point of a Matthew sequence, then $1 \leq a^2 < b$ and $b > 2$.

Proof. Suppose that $a / b$ is nontrivial fixed point with multiplicity $m$ in a Matthew sequence of length $n$. Since $a / b$ is a fixed point, it follows that $an = bm$. We can assume that $a$ and $b$ are coprime, hence $a$ divides $m$, meaning that $a \leq m$. By the sum condition we have $a \leq m \leq \floor{b / a}$, so $a \leq \floor{b / a}$. If $b / a$ is an integer, then $a = 1$ since $a$ and $b$ are coprime. The previous lemma then implies $b > 2$, which gives $a^2 < b$. If $b / a$ is not an integer then we immediately obtain $a^2 < b$. $\blacksquare$

Proposition. If $a / b$ is a rational such that $1 \leq a^2 < b$ and $b > 2$, then $a / b$ is a fixed point of some Matthew sequence.

Proof. We will construct a Matthew sequence with $b$ elements. Begin by placing $a$ copies of $a / b$ into $S$. This gives the partial sum $a^2 / b$, which is strictly less than $1$. The remaining sum to recover is $1 - a^2 / b$. If $a \neq 1$, then

$\frac{1 - a^2 / b}{b - a} \neq \frac{a}{b}.$

Thus, we may add $b - a$ copies of

$\frac{1 - a^2 / b}{b - a}$

to $S$. If $a = 1$, then instead add $1/2 - 1/2b$ and $b - 2$ copies of

$\frac{1 - \frac{1}{b} - \frac{1}{2} + \frac{1}{2b}}{b - 2}.$

This only equals $1 / b$ or $1 / 2 - 1 / 2b$ for non-integer values of $b$. In both cases, $a / b$ is in fact a fixed point. $\blacksquare$

Theorem. The rational $a / b$ is a nontrivial fixed point of some Matthew sequence iff $1 \leq a^2 < b$ and $b > 2$.

Proof. Combine the two previous propositions.

# Counting fixed points

Theorem. Given a positive integer $n$, there exists a Matthew sequence of length $O(n^3)$ with at least $n$ fixed points.

Proof. Let $N$ be a really big, to-be-determined number. Construct $S$ by adding a single $1 / N$, two $2 / N$’s, three $3 / N$’s, and so on; each step of this construction produces a new fixed point if $S$ ends with $N$ elements. We may thus produce $n$ fixed points so long as

$\sum_{k = 1}^n k \frac{k}{N} \leq 1.$

That is, so long as

$n(n + 1)(2n + 1) \leq 6N.$

The left hand side is $O(n^3)$, so taking $N = cn^3$ for a suitable constant $c$ will produce $n$ fixed points in $S$.

To finish the construction we need to ensure that $S$ sums to $1$ and has $N$ elements without ruining too many of the previous fixed points. When done with our initial construction, we will have added $S_n = n(n + 1) / 2$ elements. If we let $p$ be the sum of the first $S_n$ elements, then we could add $N - S_n$ copies of

$\frac{1 - p}{N - S_n}$

to make $S$ sum to $1$. This removes at most one fixed point, so we have at least $n - 1$ fixed points. If we carry out this process from the beginning with $n + 1$ instead of $n$, then we will still obtain a sequence of length $O(n^3)$ with at least $n$ fixed points. $\blacksquare$

Observation. A fixed point of a Matthew sequence is determined by its multiplicity. That is, if $x$ is a fixed point in a sequence of length $n$ that occurs $m$ times, then $x = m / n$. Note that then $x$ contributes $m \cdot m / n = m^2 / n$ to the sum of $S$.

Proposition. A fixed point in a Matthew sequence of length $n$ has multiplicity not exceeding $\sqrt{n}$.

Proof. Let $x$ be a fixed point of the Matthew sequence $S$ of length $n$ with multiplicity $m$. Then $x = m / n$, and $x$ contributes $m^2 / n$ to the sum of $S$. By the sum condition on $S$,

$\frac{m^2}{n} \leq = 1.$

Therefore $m \leq \sqrt{n}$. $\blacksquare$

Theorem. The number of fixed points in a Matthew sequence of length $n$ does not exceed $O(n^{1/3})$.

Proof. Suppose that there are exactly $r$ fixed points in $S$. Since fixed points are determined by their multiplicities, each must have a distinct multiplicity as well, call them $m_1, m_2, \dots, m_r$. The fixed points contribute exactly

$\sum_{k = 1}^r \frac{m_k^2}{n}$

to the sum of $S$, therefore

$\sum_{k = 1}^r \frac{m_k^2}{n} \leq 1$

However, since the multiplicities are distinct integers in $[1, n]$, the smallest that the sum on the left could be occurs when $m_k = k$. This yields

$\sum_{k = 1}^r \frac{k^2}{n} \leq 1,$

which gives, approximately,

$\frac{r^3}{6n} \leq 1.$

Therefore $r \leq (6n)^{1/3}$. $\blacksquare$

# Required Length of Matthew sequences

If $a / b$ is a fixed point of some Matthew sequence, we know that the sequence length must be a multiple of $b$. Can it be anything other than $b$ itself?

As an example, consider $4 / 17$. By one of our previous theorems, we know that $4 / 17$ must be a fixed point of some Matthew sequence. How long is this sequence? If $4 / 17$ has multiplicity $m$ in a sequence of length $n$, then

$n = m \frac{4}{17},$

$17$ divides $n$, and $m \leq \floor{17 / 4}$. These all give us the inequality

$17 \leq n \leq \left( \frac{17}{4} \right)^2 = 18.0625.$

Since $n$ is a multiple of $17$, it follows that $n = 17$.

This argument works because $17$ is not much bigger than $4^2 = 16$. A similar argument under the same hypotheses gives a partial result.

Theorem. If $a / b$ is a fixed point of a Matthew sequence $S$ such that $a$ and $b$ are coprime and $b - a^2 < a$, then $|S| = b$.

Proof. Let $n = |S|$ and $m$ be the multiplicity of $a / b$. By definition, we have

$n = m \frac{b}{a}.$

Since $m \leq \floor{b / a}$ and $b$ divides $n$, we have

$b \leq n \leq \left( \frac{b}{a} \right)^2.$

By a previous lemma, for $a / b$ to be a fixed point we must have $1 \leq a^2 \leq b$. If we write $b = a^2 + k$ for some nonnegative integer $k$, then our inequality becomes

$b \leq n \leq b + k + \frac{k^2}{a^2}.$

Since $k = b - a^2 < a$ and $n$ is an integer, the inequality reduces to

$b \leq n \leq b + k.$

Since $k = b - a^2 < b$, the only multiple of $b$ satisfying this inequality is $b$ itself. Since $n$ is a multiple of $b$, it follows that $n = b$. $\blacksquare$

In general, this question remains open. We could probably think a little harder with our number theory brains to find a better answer, but this seems like a good place to stop for now.