ClearerIt might be clearer to say that you have a set of points and $t$ blocks each of size $s$ (I suppose no two totally identical). What can be said about the range of possible pairs $(a,b)$ where $a$ is the size of the union of the blocks and $b$ is the number of point pairs which are in at least one block? I’ll generalize that to
What arecan be said about the range of possible quadruples $(s,t,a,b)?$$(s,t,a,b)$ where $a$ is the size of the union of the blocks and $b$ is the number of point pairs which are in at least one block? Specifically, fix one or more and ask what the range is for the other(s).
I didn’t name the number of points since you don’t use it. I’m not sure what you mean by $k(n).$
I’ll look at this question since you mention it : Certainly $t \leq \binom{a}s.$ When there is equality, also $b = \binom{a}2.$ That is clearly the largest $b$ can be given$b \leq \binom{a}2.$
Q: What is the range on $t$ be given that $a.$$b = \binom{a}2?$
How small could $t$ be given that $b = \binom{a}2?$
It turns out that the answer is
Certainly$$\frac{a(a-1)}{s(s-1)} \leq t \leq \binom{a}{s}=\frac{a}{s}\frac{a-1}{s-1}\frac{a-2}{s-2}\cdots\frac{a-s+1}{1}$$
$$t \geq \frac{\binom{a}2}{\binom{s}2}=\frac{a(a-1)}{s(s-1)}$$ With equality The lower bound occurs in the event that every pair is in exactly one block (which of course requires $t$$\frac{a(a-1)}{s(s-1)}$ to be an integer.) One also sees that the number of blocks containing a given point is $$\frac{a-1}{s-1}$$ which must also be an integer.
Then we have what is called a Balanced incompleteIncomplete Block Design BIBD with parameters $(v,k,\lambda)=(a,s,1)$ Also called a $2-(a,s)$ Steiner System. It turns out that, for fixed $s,$ this is possible provided $a$ is large enough and the two integrality conditions are met.
A BIBD$(a,3,1)$ exists for $a \equiv1,3 \bmod 6.$ One of these$-(a,3,1)$ is also called a Steiner Triple System. and exists for $a \equiv 1,3 \bmod 6.$
Projective planes are BIBD$(q^2+q+1,q+1,1)$and$-(q^2+q+1,q+1,1)$ and Affine planes are $BIBD(q^2,q,1)$$BIBD-(q^2,q,1)$
So this was a matter of fixing $a$ and $s$, letting $b=\binom{a}{2}$ and looking at the possible sizes of $t.$
A question which might be easy, but isn't totally immediateobvious to me is:
Given $s,t$, how small can $a$ and/or $b$ be?
Given $s,t$, how small can $a$ and/or $b$ be?
It might , in fact be tricky, I didn't think about it too much. The cases
If $t$ small$t=2$ then we can have $a=s+1$ in which case $b=\binom{s}2+s-1=\frac{(s+2)(s-1)}2$ which are both minimums.
For $3 \le t \le s+1$ we can have $a=s+1$ and $s$ large might be different than$b=\binom{s+1}2.$
The minimum $s$ large$a$ is $a=s+c$ with $c$ minimal subject to $\binom{s+c}s \ge t$ and then one has $t$ small.$\binom{s+c-1}2 \lt b \le \binom{s+c}2.$