Let $t$, $\lambda$, $v$ be positive integers and let $K$ be a set of positive integers.
A $t$-$(v,K,\lambda)$-design is a pair $(X,\scr{B})$ such that
- $X$ is a set (of so-called vertices) of cardinality $v$,
- $\scr{B}$ is a set of subsets of $X$ (so-calles blocks) such that $|B|\in K$ for each $B\in\scr{B}$,
- Every subset $T$ of $X$ with $|T|=t$ is contained (as a subset) in exactly $\lambda$ elements of $\scr{B}$.
We let $B(t,K,\lambda)$ be the set of all integers $v$ for which a $t$-$(v,K,\lambda)$-design exists.
If $\lambda>1$ and repeated blocks are allowed, then it makes sense to let $\scr{B}$ be a sequence (as opposed to a set) of subsets of $X$; however, here I am only interested in designs with $\lambda=1$ and $t=2$.
Known facts are
$$B(2,\{3\},1)=\{v\in\mathbb{N}\;|\;v\equiv1\text{ or }3\pmod{6}\}\tag{1}$$ $$B(2,\{4\},1)=\{v\in\mathbb{N}\;|\;v\equiv1\text{ or }4\pmod{12}\}\tag{2}$$
By a double counting argument, it can be seen that
$$B(2,\{3,4\},1)\subseteq\{v\in\mathbb{N}\;|\;v\equiv0\text{ or }1\pmod{3}\}$$
so the question is whether this is actually an equality.
A closer look reveals that this can't be the case as there is no $2$-$(6,\{3,4\},1)$-design, but is this perhaps the only exception?
Since $B(2,\{3\},1)$ and $B(2,\{4\},1)$ are subsets of $B(2,\{3,4\},1)$, equations (1) and (2) settle the question for all $v\in\mathbb{N}$ congruent $1$, $3$, $4$, $7$, and $9\pmod{12}$.
For $v\equiv0\pmod{12}$, a $2$-$(v,\{3,4\},1)$-design can be constructed by deleting one vertex from a $2$-$(v+1,\{4\},1)$-design (which exists by equation (2)) and shortening all blocks that are containing it.
Edited on July 6th, 2020:
Let $u=2k+1$ be an odd integer $\ge3$, and let $Y:=\mathbb{Z}_u\times\mathbb{Z}_3$, where $\mathbb{Z}_n$ is the additive Abelian group of integer residues modulo $n$.
Furthermore, for $s\in\mathbb{Z}_u$, $d\in\mathbb{Z}_3$ and $j\in\mathbb{Z}_u$ with $1\le j\le k$, let
$$ \begin{align} G_s&:=\{(s,b)\;|\;b\in\mathbb{Z}_3\} \\ \scr{G}&:=\{G_t\;|\;t\in\mathbb{Z}_u\} \\ B_{s,d}^j&:=\{(s,d),(s+j,d+1),(s+2j,d)\} \\ \scr{B}_s^j&:=\{B_{s,b}^j\;|\;b\in\mathbb{Z}_3\} \\ \scr{B}&:=\bigcup\;\{\scr{B}_t^i\;|\;t\in\mathbb{Z}_u,i\in\mathbb{Z}_u\text{ with }1\le i\le k \} \\ &\;=\{B_{t,b}^i\;|\;t\in\mathbb{Z}_u,b\in\mathbb{Z}_3,i\in\mathbb{Z}_u\text{ with }1\le i\le k\} \end{align} $$
Then $(Y,\scr{G},\scr{B})$ is a so-called group divisible design which means that
- The elements of $\scr{G}$ (so-called groups) form a partition of $Y$.
- Every pair of elements of $Y$ not contained in the same group lies in a unique element of $\scr{B}$ (a so-called block).
- No block intersects a group in more than one elements of $Y$.
From the construction, it can immediately be seen that $(Y,\scr{G}\cup\scr{B})$ is a $2$-$(3u,\{3\},1)$-design as all groups and blocks have size $3$.
Thus, for $v\equiv10\pmod{12}$, a $2$-$(v,\{3,4\},1)$-design can be constructed by adding one vertex to this group divisible design and connecting the new vertex to every group.
This leaves only the integers $v\equiv6\pmod{12}$ in question.
Getting back to our group divisible design constructed above, consider the case that $u$ is divisible by $3$, and that $j$ is chosen as above but with the additional property that $u$ and $j$ are coprime. Let $s\in\{0,1,2\}$. Then
$$\scr{P}_s^j:=\bigcup\;\{\scr{B}_{s+3rj}^j\;|\;0\le r\lt \tfrac{u}{3}\}$$
is a partition of $Y$.
Since we can choose $j$ in $\tfrac{1}{2}\varphi(u)$ ways, $\scr{B}$ contains at least $\tfrac{3}{2}\varphi(u)$ pairwise disjoint partitions of $Y$.
First assume that $u\equiv9\pmod{12}$. Then $\varphi(u)\ge2$, so $\scr{B}$ contains $3$ pairwise disjoint partitions $\scr{P}_1,\scr{P}_2,\scr{P}_3$ of $Y$. Augment $Y$ by three new vertices $w_1,w_2,w_3$ forming one new block, and extend the blocks of $\scr{P}_i$ by $w_i$. This will yield a $2$-$(3u+3,\{3,4\},1)$-design.
If in addition, $u\ge21$, then $\varphi(u)\ge10$, so $\scr{B}$ even contains at least $15$ pairwise disjoint partitions of $Y$. Similarly to the construction above, we can augment $Y$ by a $2$-$(15,\{3\},1)$-design with $15$ new vertices which we adjoin the the elements of the $15$ partitions, and we get a $2$-$(3u+15,\{3,4\},1)$-design.
Finally suppose that $u\equiv3\pmod{12}$, and $u\ge15$. Then $\varphi(u)\ge6$, so $\scr{B}$ contains at least $9$ pairwise disjoint partitions of $Y$. Now augment $Y$ by a $2$-$(9,\{3\},1)$-design with $9$ new vertices, connect these with te $9$ partitions of $Y$ and thus construct a $2$-$(3u+9,\{3,4\},1)$-design.
In summary, this shows that a $2$-$(v,\{3,4\},1)$-design exists for all $v\equiv0\text{ or }1\pmod{3}$ except $6$ and (possibly) $18$ and $42$.
Is anyone aware of designs for the doubtful cases $v=18$ and $v=42$? Or of a proof that such designs do not exist?