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Construct a (13,26,6,3,1) BIBD.

I think there may be some proposition I should use involving prime numbers but I am having a hard time finding it.

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    It would be instructive for those of us who don't know what a BIBD is for you to define the term.2012-01-30
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    I am sorry, it is a Balanced Incomplete Block Design2012-01-30
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    I think this has to do with difference sets.2012-01-30
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    You are looking for a [design](http://en.wikipedia.org/wiki/Block_design) with (v, b, r, k, λ) = (13,26,6,3,1) ?2012-01-30
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    Yes, I am. Do you know how to do this?2012-01-31
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    You either spend a few nights thinking about it, or you do a quick internet search, I suppose. See my answer.2012-01-31

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Wikipedia is always a good start for a search.

You will find out that the $2-(13,3,1)$ design you are looking for a is Steiner triple system, denoted by S(2,3,13) = STS(13). According to wikipedia there are two non-isomorphic such designs and both can be found in the database of t-designs.

According to Mathworld Wolfram one of these designs is a finite hyperbolic plane, but I wouldn't know about that.

(edit) Actually, here is a book, describing constructions for these Steiner triples --- and some discussion showing it it not embeddable in a (Desarguesian) projective plane.

(edit2) As you, yourself, point out in the comments below, this problem can actually be solved by considering difference sets. This explains the construction from the book, by noting that $\{1,2,5\}$ and $\{1,6,8\}$ is a difference family for $\mathbb Z/13\mathbb Z$ with $s=2$. I'm guessing that's what you should be looking for, after noticing that a difference set won't do the trick, because the parameters don't match, i.e. they don't satisfy $k^2-k = \lambda (v-1)$.

$$ \begin{array}{c|ccc} - &1&2&5\\ \hline 1&0&1&4\\2&12&0&3\\5&9&10&0\end{array} \text{ and } \begin{array}{c|ccc} - &1&6&8\\ \hline 1&0&5&7\\6&8&0&2\\8&6&11&0\end{array} $$

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    I am confused as to where the 2- comes from? I was thinking of computing the difference set. Then adding i from 0 to t-1 to the first set to get every block.2012-01-31
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    @Jackson: the 2 comes from the generalization to $t-(v,k,\lambda)$ designs. The case where $t=2$ are precisely the BIBD's.2012-01-31
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    I do not see anything in my books about t-(v,k,lambda) designs; Have you seen difference sets? I think we are suppose to use those. I think it worked because i got 26 different blocks2012-01-31
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    @Jackson: you should have specified that while asking your question. Always post as much relevant information as possible.2012-01-31
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    @Myself: If you scroll up, I did say it in a comment about an hour ago, right after posting the question.2012-01-31
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    But my mistake, for not putting it in the question. I am sorry.2012-01-31
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    I could say D_i = {a^i, a^2t+i, a^4t+i} for i = 0,1,..., t-12012-01-31
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    I have deleted some off-topic comments. Let's all stick to the math and not malign other users.2012-01-31
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    @Jackson I have expanded my answer a bit. But if this is not what you are looking for, then you should edit your question and try to be (a lot) more specific.2012-01-31