Properties of the cycles that contain all vectors of weight $\le k$

Abstract

We study the sequences whose one period contains all the n-binary vectors of Hamming weight $\le k$ exactly once. It is well known that such sequences exist for any n and $0\le k\le n$. However, their many basic properties and even their numbers are still unknown. A classical method for constructing such sequences is by joining the cycles generated by pure circulating registers, pure summing registers or complementing summing registers. In this paper, we show that, when $k=2$ such sequences can all be constructed by joining cycles of pure circulating registers, but for $n\ge 4$ and $k\ge 3$ this is not the case any more. We also show that for $n\ge 7$ and $k\ge 3$, the sequences constructed by joining cycles of pure circulating registers are different from those constructed by joining cycles of pure summing registers or complementing summing registers. Besides, we do some experiments and determine the numbers of such sequences for some small n and k.

Appendix: A general bound for \(\mu _n(w)\)

In this section, we given an improved bound for $\mu _n(w)$. Firstly we show a property for combinatorial numbers.

Lemma 9

Let n and w be two integers such that $n>4$ and $w<n$. Suppose $d=\gcd (n,w)>1$. Let p be the smallest prime divisor of d. Then we have

$$\begin{aligned} \left( {\begin{array}{c}n\\ w\end{array}}\right) >2(d-1)\left( {\begin{array}{c}n/p\\ w/p\end{array}}\right) . \end{aligned}$$

Proof

If $\frac{w}{p}=1$, then we have $w=d=p$ and

$$\begin{aligned} \left( {\begin{array}{c}n\\ w\end{array}}\right) =\left( {\begin{array}{c}n\\ p\end{array}}\right) \ge 2n>2(p-1)\frac{n}{p}=2(d-1)\left( {\begin{array}{c}n/p\\ w/p\end{array}}\right) . \end{aligned}$$

If $\frac{w}{p}\ge 2$, then we have

$$\begin{aligned} \left( {\begin{array}{c}n\\ w\end{array}}\right)&=\prod _{i=0}^{w-1}\frac{n-i}{w-i} =\prod _{i=0}^{w/p-1}\prod _{j=0}^{p-1}\frac{n-(ip+j)}{w-(ip+j)} =\prod _{i=0}^{w/p-1}\prod _{j=0}^{p-1}\frac{(n/p-i)p-j}{(w/p-i)p-j}\\&=\prod _{i=0}^{w/p-1}\frac{(n/p-i)p}{(w/p-i)p}\cdot \prod _{i=0}^{w/p-1}\prod _{j=1}^{p-1}\frac{(n/p-i)p-j}{(w/p-i)p-j}\\&\ge \left( {\begin{array}{c}n/p\\ w/p\end{array}}\right) \prod _{i=0}^{w/p-1}\frac{(n/p-i)p-(p-1)}{(w/p-i)p-(p-1)}. \end{aligned}$$

Because

$$\begin{aligned} \prod _{i=0}^{w/p-1}\frac{(n/p-i)p-(p-1)}{(w/p-i)p-(p-1)}&\ge \prod _{i=w/p-1,w/p-2}\frac{(n/p-i)p-(p-1)}{(w/p-i)p-(p-1)}\\&=(n-w+1)\cdot \frac{n-w+p+1}{p+1}, \end{aligned}$$

for the proof of this lemma, we just need to show that $(n-w+1)\cdot \frac{n-w+p+1}{p+1}>2(d-1)$. Since $d=\gcd (n,w)=\gcd (n,n-w)$ we know that $d\le n-w$. If $p=d=n-w$ then

$$\begin{aligned} (n-w+1)\cdot \frac{n-w+p+1}{p+1}=(p+1)\frac{p+p+1}{p+1}=2p+1>2(d-1). \end{aligned}$$

If $p<n-w$ then

$$\begin{aligned} (n-w+1)\cdot \frac{n-w+p+1}{p+1}\ge (d+1)\frac{p+1+p+1}{p+1}>2(d-1). \end{aligned}$$

This completes the proof. $\square$

Theorem 4

For $n>4$ and $w<n$, we have $\mu _n(w)<\frac{3}{2n}\left( {\begin{array}{c}n\\ w\end{array}}\right)$.

Proof

Denote by p the smallest prime divisor of $d=\gcd (n,w)$. Then

$$\begin{aligned} \mu _n(w)&= \frac{1}{n}\sum _{r\mid d}\phi (r)\left( {\begin{array}{c}n/r\\ w/r\end{array}}\right) \\&=\frac{1}{n}\left( {\begin{array}{c}n\\ w\end{array}}\right) +\frac{1}{n}\sum _{r\mid d,r\ne 1}\phi (r)\left( {\begin{array}{c}n/r\\ w/r\end{array}}\right) \\&\le \frac{1}{n}\left( {\begin{array}{c}n\\ w\end{array}}\right) +\frac{d-1}{n}\left( {\begin{array}{c}n/p\\ w/p\end{array}}\right) \\&< \frac{1}{n}\left( {\begin{array}{c}n\\ w\end{array}}\right) +\frac{d-1}{n}\cdot \frac{1}{2(d-1)}\left( {\begin{array}{c}n\\ w\end{array}}\right) \\&=\frac{3}{2n}\left( {\begin{array}{c}n\\ w\end{array}}\right) . \end{aligned}$$

The last inequality is valid because of Lemma 9. $\square$