Subfield: Combinatorics


Question: A Skolem sequence of order $n$ is a sequence $\left(s_{1}, s_{2}, \ldots, s_{2 n}\right)$ of $2 n$ integers satisfying the conditions:

i) for every $k$ in $\{1,2,3, \ldots, n\}$, there exist exactly two elements $s_{i}$ and $s_{j}$ with $s_{i}=s_{j}=k$, and

ii) if $s_{i}=s_{j}=k$ with $i<j$, then $j-i=k$.


For example, $(4,2,3,2,4,3,1,1)$ is a Skolem sequence of order 4.
Prove that there is no Skolem sequence of order $n$, if $n$ is of the form $4 k+2$ or $4 k+3$, where $k$ is a non-negative integer.

Solution:
Assume that there is a Skolem sequence of order $n$, where $n$ is of the form $4 k+2$ or $4 k+3$. (We will deal with both cases together.)

Let $P_{1}$ be the position number of the left-most 1 in the Skolem sequence. Then $P_{1}+1$ is the position number of the other 1 .

Similarly, let $P_{2}, P_{3}, \ldots, P_{n}$ be the position numbers of the left-most $2,3, \ldots, n$, respectively, in the Skolem sequence. Then $P_{2}+2, P_{3}+3, \ldots, P_{n}+n$ are the position numbers of the other occurrences of $2,3, \ldots, n$, respectively.

Since the Skolem sequence exists, then the numbers $P_{1}, P_{2}, P_{3}, \ldots, P_{n}, P_{1}+1, P_{2}+2$, $P_{3}+3, \ldots, P_{n}+n$ are a rearrangement of all of the position numbers, ie. the numbers 1 , $2, \ldots, 2 n$.

Thus,

$$
\begin{aligned}
P_{1}+P_{2}+\cdots+P_{n}+\left(P_{1}+1\right)+\left(P_{2}+2\right)+\cdots+\left(P_{n}+n\right) & =1+2+\cdots+2 n \\
2\left(P_{1}+P_{2}+\cdots+P_{n}\right)+(1+2+\cdots+n) & =1+2+\cdots+2 n \\
2\left(P_{1}+P_{2}+\cdots+P_{n}\right)+\frac{n(n+1)}{2} & =\frac{2 n(2 n+1)}{2} \\
2\left(P_{1}+P_{2}+\cdots+P_{n}\right)+\frac{n(n+1)}{2} & =n(2 n+1)
\end{aligned}
$$

using the fact that $1+2+\cdots+k=\frac{k(k+1)}{2}$.

We now look at the parities of the terms in equation $(* *)$ in each of our two cases.

If $n=4 k+2$, then $\frac{n(n+1)}{2}=\frac{(4 k+2)(4 k+3)}{2}=(2 k+1)(4 k+3)$ which is the product of two odd numbers, so is odd, and $n(2 n+1)=(4 k+2)(8 k+5)$, which is the product of an even number and an odd number, so is even.

Thus, $(* *)$ is Even + Odd = Even, a contradiction.

If $n=4 k+3$, then $\frac{n(n+1)}{2}=\frac{(4 k+3)(4 k+4)}{2}=(4 k+3)(2 k+2)$ which is the product of an odd number and an even number, so is even, and $n(2 n+1)=(4 k+3)(8 k+7)$, which is the product of two odd numbers, so is odd.

Thus, $(* *)$ is Even + Even $=$ Odd, a contradiction.

Therefore, in either case, a contradiction is reached, so there cannot exist a Skolem sequence of order $n$, if $n$ is of the form $4 k+2$ or $4 k+3$, where $k$ is a non-negative integer.
Assume that there is a Skolem sequence of order $n$, where $n$ is of the form $4 k+2$ or $4 k+3$.

Let $l$ be an integer between 1 and $n$.

If $l$ is even, then the two position numbers in which $l$ is placed differ by an even number (namely $l$ ), and so are either both odd or both even.

If $l$ is odd, then the two position numbers in which $l$ is placed differ by an odd number (namely $l$ ), and so one is odd and the other is even.

Case 1: $n=4 k+2$

Between 1 and $n$, there are $2 k+1$ even numbers $(2,4, \ldots, 4 k+2)$ and $2 k+1$ odd numbers $(1,3, \ldots, 4 k+1)$.

The position numbers in a Skolem sequence of order $n=4 k+2$ are the integers 1,2 , $\ldots, 8 k+4$, so there are $4 k+2$ even position numbers and $4 k+2$ odd position numbers.

Considering the $2 k+1$ odd numbers in the sequence gives us $2 k+1$ odd position numbers and $2 k+1$ even position numbers, ie. an odd number of odd position numbers and an odd number of even position numbers.

Each even number in the sequence will contribute either two odd position numbers or two even position numbers. In other words, all of the even numbers in the sequence contribute an even number of odd position numbers.

Thus, the total number of odd position numbers is odd plus even, which is odd. This is a contradiction, because we know there must be $4 k+2$ odd position numbers.

Case 2: $n=4 k+3$

Between 1 and $n$, there are $2 k+1$ even numbers $(2,4, \ldots, 4 k+2)$ and $2 k+2$ odd numbers $(1,3, \ldots, 4 k+3)$.

The position numbers in a Skolem sequence of order $n=4 k+3$ are the integers $1,2, \ldots, 8 k+6$, so there are $4 k+3$ even position numbers and $4 k+3$ odd position numbers.

Considering the $2 k+2$ odd numbers in the sequence gives us $2 k+2$ odd position numbers and $2 k+2$ even position numbers, ie. an even number of odd position numbers and an even number of even position numbers.

Each even number in the sequence will contribute either two odd position numbers or two even position numbers. In other words, all of the even numbers in the sequence contribute an even number of odd position numbers.

Thus, the total number of odd position numbers is even plus even, which is even. This is a contradiction, because we know there must be $4 k+3$ odd position numbers.

Therefore, in either case, a contradiction is reached, so there cannot exist a Skolem sequence of order $n$, if $n$ is of the form $4 k+2$ or $4 k+3$, where $k$ is a non-negative integer.