Subfield: Algebra


Question: The infinite sequence $a_{0}, a_{1}, a_{2}, \ldots$ of (not necessarily different) integers has the following properties: $0 \leqslant a_{i} \leqslant i$ for all integers $i \geqslant 0$, and

$$
\left(\begin{array}{c}
k \\
a_{0}
\end{array}\right)+\left(\begin{array}{c}
k \\
a_{1}
\end{array}\right)+\cdots+\left(\begin{array}{c}
k \\
a_{k}
\end{array}\right)=2^{k}
$$

for all integers $k \geqslant 0$.

Prove that all integers $N \geqslant 0$ occur in the sequence (that is, for all $N \geqslant 0$, there exists $i \geqslant 0$ with $a_{i}=N$ ).

Solution:
We prove by induction on $k$ that every initial segment of the sequence, $a_{0}, a_{1}, \ldots, a_{k}$, consists of the following elements (counted with multiplicity, and not necessarily in order), for some $\ell \geqslant 0$ with $2 \ell \leqslant k+1$ :

$$
0,1, \ldots, \ell-1, \quad 0,1, \ldots, k-\ell
$$

For $k=0$ we have $a_{0}=0$, which is of this form. Now suppose that for $k=m$ the elements $a_{0}, a_{1}, \ldots, a_{m}$ are $0,0,1,1,2,2, \ldots, \ell-1, \ell-1, \ell, \ell+1, \ldots, m-\ell-1, m-\ell$ for some $\ell$ with $0 \leqslant 2 \ell \leqslant m+1$. It is given that

$$
\left(\begin{array}{c}
m+1 \\
a_{0}
\end{array}\right)+\left(\begin{array}{c}
m+1 \\
a_{1}
\end{array}\right)+\cdots+\left(\begin{array}{c}
m+1 \\
a_{m}
\end{array}\right)+\left(\begin{array}{c}
m+1 \\
a_{m+1}
\end{array}\right)=2^{m+1}
$$

which becomes

$$
\begin{aligned}
\left(\left(\begin{array}{c}
m+1 \\
0
\end{array}\right)+\left(\begin{array}{c}
m+1 \\
1
\end{array}\right)\right. & \left.+\cdots+\left(\begin{array}{c}
m+1 \\
\ell-1
\end{array}\right)\right) \\
& +\left(\left(\begin{array}{c}
m+1 \\
0
\end{array}\right)+\left(\begin{array}{c}
m+1 \\
1
\end{array}\right)+\cdots+\left(\begin{array}{c}
m+1 \\
m-\ell
\end{array}\right)\right)+\left(\begin{array}{c}
m+1 \\
a_{m+1}
\end{array}\right)=2^{m+1},
\end{aligned}
$$

or, using $\left(\begin{array}{c}m+1 \\ i\end{array}\right)=\left(\begin{array}{c}m+1 \\ m+1-i\end{array}\right)$, that

$$
\begin{aligned}
\left(\left(\begin{array}{c}
m+1 \\
0
\end{array}\right)+\left(\begin{array}{c}
m+1 \\
1
\end{array}\right)\right. & \left.+\cdots+\left(\begin{array}{c}
m+1 \\
\ell-1
\end{array}\right)\right) \\
& +\left(\left(\begin{array}{c}
m+1 \\
m+1
\end{array}\right)+\left(\begin{array}{c}
m+1 \\
m
\end{array}\right)+\cdots+\left(\begin{array}{c}
m+1 \\
\ell+1
\end{array}\right)\right)+\left(\begin{array}{c}
m+1 \\
a_{m+1}
\end{array}\right)=2^{m+1} .
\end{aligned}
$$

On the other hand, it is well known that

$$
\left(\begin{array}{c}
m+1 \\
0
\end{array}\right)+\left(\begin{array}{c}
m+1 \\
1
\end{array}\right)+\cdots+\left(\begin{array}{c}
m+1 \\
m+1
\end{array}\right)=2^{m+1}
$$

and so, by subtracting, we get

$$
\left(\begin{array}{c}
m+1 \\
a_{m+1}
\end{array}\right)=\left(\begin{array}{c}
m+1 \\
\ell
\end{array}\right)
$$

From this, using the fact that the binomial coefficients $\left(\begin{array}{c}m+1 \\ i\end{array}\right)$ are increasing for $i \leqslant \frac{m+1}{2}$ and decreasing for $i \geqslant \frac{m+1}{2}$, we conclude that either $a_{m+1}=\ell$ or $a_{m+1}=m+1-\ell$. In either case, $a_{0}, a_{1}, \ldots, a_{m+1}$ is again of the claimed form, which concludes the induction.

As a result of this description, any integer $N \geqslant 0$ appears as a term of the sequence $a_{i}$ for some $0 \leqslant i \leqslant 2 N$.