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From wikipedia, I know that Gold codes have length $2^n-1$, where $n$ is number of LFSR elements and that there are $2^n-1$ different Gold sequences for this case. E.g. assuming that $n=6$, I have a Gold sequence with a length of $63$ bits. But how many users can I differentiate with this?

On the one hand I would have said that I can differentiate $2^{2^n-1}$, i.e. in this case $2^{63}$ users because there are $63$ bits that can either be $0$ or $1$. But on the other hand, wikipedia says there are only $2^n-1$ Gold sequences with that length which would imply that I can also serve $2^n-1$ users.

Or the question asked the other way around: If I have to address $K$ users, each by means of an unique code, how many bits for the Gold codes do I need?

Edit: Moreover, I found this page that says that there are only $6$ different Gold sequences with a length of $63$ bits.

Dilip Sarwate
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bonanza
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  • @MattL. thanks for pointing me to this. Does it mean that if I want to address $2^m+1$ user, I need to spend $2^m-1$ bits for each code? – bonanza Dec 17 '15 at 13:51
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    Yes, the sequence length is $2^m-1$ and your set size is $2^m+1$, where $m$ is the length of the shift register. – Matt L. Dec 17 '15 at 14:02
  • I don't think that this question is a duplicate of the previous one. Yes, there is the (minor) issue of $2^n+1$ versus $2^n-1$ but the main question here is: Why can't we distinguish $2^{63}$ users with $63$ bits of a Gold sequence instead of just $65 = 2^{\log_2{63+1}}+1$. Incidentally, I have edited the question to change gold to Gold. The sequences are named after a person, not the metal. – Dilip Sarwate Dec 17 '15 at 16:10
  • That should have been $65 = 2^{\log_2(63+1)}+1 = 2^n+1$. – Dilip Sarwate Dec 17 '15 at 16:27
  • @DilipSarwate yes, that is one yet unanswered question. My current assumption is that, theoretically, this would be possible, but DS-CDMA codes need special correlation properties and that is why we need to spend (much) more bits. What do you think? Is that reasonable? – bonanza Dec 18 '15 at 11:30

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