On network codes and partial spreads. (English) Zbl 1432.51005
Optimal codes and related topics, OC 2013. Proceedings of the seventh international workshop, Albena, Bulgaria, September 6–12, 2013. Dedicated to Stefan Dodunekov (1945–2012). Sofia: Institute of Mathematics and Informatics, Bulgarian Academy of Sciences. 77-78 (2013).
From the text: For \(r = 1\) we have the following result.
Theorem. If \(n = kl + 1 \ge 2l + 1\) where \(k, l\in\mathbb N\) then
\[ \mathcal A_q(n,2l,l) = q^{n-l} + q^{n-2l} + \ldots + q^{n-(k-1)l} + 1 = \left\lfloor \frac{q^n - 1}{q^l - 1}\right\rfloor - (q - 1). \]
Unfortunately, we do not know the exact value of \(\mathcal A_q(n,2l,l)\) if \(r > 1\). The counting argument used to prove the Theorem does not work for \(r > 1\). However, based on the Theorem above and a result of El-Zanati et al. in which \(n = 3k + 2\ge 8\) (hence \(r = 2)\) we may ask the
Question. Do we always have
\[ \begin{aligned} \mathcal A_q(n,2l,l) &= q^{n-l} + q^{n-2l} + \ldots + q^{n-(k-1)l} + q^{r-1} \\ &= \left\lfloor \frac{q^n - 1}{q^l - 1}\right\rfloor - q^{r-1}(q - 1) \end{aligned} \]
if \(n = kl + r\) where \(k\ge 2\) and \(0 < r < l\) ?
The Theorem above answers a question posed by T. Bu in [Discrete Math. 31, 79–83 (1980; Zbl 0445.15004)] positively for \(r = 1\). However, [S. El-Zanati et al., Des. Codes Cryptography 54, No. 2, 101–107 (2010; Zbl 1200.51008)] shows that the answer is negative for \(r > 1\).
For the entire collection see [Zbl 1321.00113].
Theorem. If \(n = kl + 1 \ge 2l + 1\) where \(k, l\in\mathbb N\) then
\[ \mathcal A_q(n,2l,l) = q^{n-l} + q^{n-2l} + \ldots + q^{n-(k-1)l} + 1 = \left\lfloor \frac{q^n - 1}{q^l - 1}\right\rfloor - (q - 1). \]
Unfortunately, we do not know the exact value of \(\mathcal A_q(n,2l,l)\) if \(r > 1\). The counting argument used to prove the Theorem does not work for \(r > 1\). However, based on the Theorem above and a result of El-Zanati et al. in which \(n = 3k + 2\ge 8\) (hence \(r = 2)\) we may ask the
Question. Do we always have
\[ \begin{aligned} \mathcal A_q(n,2l,l) &= q^{n-l} + q^{n-2l} + \ldots + q^{n-(k-1)l} + q^{r-1} \\ &= \left\lfloor \frac{q^n - 1}{q^l - 1}\right\rfloor - q^{r-1}(q - 1) \end{aligned} \]
if \(n = kl + r\) where \(k\ge 2\) and \(0 < r < l\) ?
The Theorem above answers a question posed by T. Bu in [Discrete Math. 31, 79–83 (1980; Zbl 0445.15004)] positively for \(r = 1\). However, [S. El-Zanati et al., Des. Codes Cryptography 54, No. 2, 101–107 (2010; Zbl 1200.51008)] shows that the answer is negative for \(r > 1\).
For the entire collection see [Zbl 1321.00113].
