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I am looking for ways to construct an infinite family of designs with parameters $2-(v,3,3)$ and apart from some doubling-type recursive constructions (such as in this paper) I haven't found anything in the literature.

So, are there "explicit" ways to construct a family of such designs?

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  • $\begingroup$ When you say "design", do you mean a $2$-design? Usually designs have four parameters, and are then called $t$-$(v,k,\lambda)$ designs. $\endgroup$ – Tom De Medts Dec 10 '13 at 9:59
  • $\begingroup$ @TomDeMedts Yes, $2$-designs. I made it clear now in the question. Thanks. $\endgroup$ – Felix Goldberg Dec 10 '13 at 10:00
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    $\begingroup$ Three copies of a Steiner triple system?? Or is that cheating too. $\endgroup$ – Gordon Royle Dec 10 '13 at 13:02
  • $\begingroup$ @GordonRoyle Definitely not cheating but wouldn't work for my ulterior aims... $\endgroup$ – Felix Goldberg Dec 10 '13 at 14:54
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An elementary counting argument shows that $2$-$(v,3,3)$ exists only if $v$ is odd (or, more precisely, for $\lambda \equiv 3 \pmod{6}$ a $2$-$(v,3,\lambda)$ exists only if $v \equiv 1 \pmod{2}$). This necessary condition is sufficient.

Arguably the simplest direct construction for the case $\lambda = 3$ that covers all odd $v$ is to use commutative quasigroups. Let $Q = (V, \otimes)$ be a commutative idempotent quasigroup of order $v \equiv 1 \pmod{2}$. A perpendicular array of strength $3$ is a ${{v+1}\choose{2}}\times 3$ array $L$ whose rows are the ordered triples $(x, y, x\otimes y)$ for $x, y \in V$ and $x \leq y$. Because $Q$ is idempotent, $L$ contains $v$ rows of the form $(x,x,x)$. Deleting such rows and ignoring the ordering on the rows, we obtain a set of triples. Every pair $\{a,b\}$ of distinct elements $a, b \in V$ occurs exactly three times in this set of triples, coming from the ordered triples $(a,b,a\otimes b)$, $(a,c,b)$ when $a \otimes c = c \otimes a = b$, and $(d,b,a)$ when $d\otimes b = b \otimes d =a$. Thus, it suffices to show that a commutative idempontet quasigroup exists for all odd $v = 2n+1$, which is true because we can define $\otimes$ as the binary operation $i\otimes j = (n+1)(i+j) \pmod{v}$ with $V = \{0,1,\dots,v-1\}$.

So, by defining $Q = (V, \otimes)$ as above, the construction is to take $(x,y,x\otimes y)$ for $x\leq y$, throw away $(x,x,x)$, and drop the ordering.

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  • $\begingroup$ Thanks! I was actually going to post this myself, having spent a cozy afternoon with Colbourn & Rosa's Triple Systems. :) $\endgroup$ – Felix Goldberg Dec 10 '13 at 18:16
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    $\begingroup$ @Felix Ha ha. Looks like I beat you to it! This kind of construction is nicely explained in a very accessible way in "Design Theory" by Lindner and Rodger too if you're interested: books.google.com/books/about/… $\endgroup$ – Yuichiro Fujiwara Dec 11 '13 at 5:46
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Yuichiro's is the best answer, but here's another "cheat" answer (which I nonetheless think is kind of neat):

Take a PBD$(v,\{3,5\})$, triple each block of size three, and replace each block of size five by the complete design $\binom{[5]}{3}$. Various PBD$(v,\{3,5\})$ can be found by cyclotomy.

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{i,i+j,i+2j} for $i \in Z_{v}$ and $j=1,2,\ldots \frac{v-1}{2}$ should work.

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    $\begingroup$ This seems to be a simplification of Yuichiro Fujiwara's answer. $\endgroup$ – S. Carnahan Jan 10 '14 at 0:23

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