For every natural number $c \geq 2$, let $f(c)$ denote the least natural number $f$ with the following property : every union-closed family of sets with at least $f$ members has $c$ members whose intersection is nonempty. If I'm not wrong, it is easy to prove that $f(c)$ always exists. Frankl's conjecture amounts to say that $f(c) \leq 2c-1$.

If $c-1$ is a power of $2$, then $f(c) \geq 2c-1$ (look at the family of all subsets of a set with cardinality $k$, where $c-1 = 2^{k-1}$). Thus, if $c-1$ is a power of $2$, then Frankl's conjecture (if true) is optimal.

But if $c-1$ is not a power of $2$, it can happen that $f(c) < 2c-1$, so that Frankl's conjecture is not optimal in these cases.

For example, if I'm not wrong, $f(4) = 6$ and $f(6) = 10$. So, I searched the Oeis site for a sequence $A$ with the following properties :

$A(1+2^{k-1}) = 1+2^{k}$; $A(4) = 6$; $A(6) = 10$.

A188163 ( http://oeis.org/search?q=3%2C5%2C6%2C9%2C10&language=english&go=Search) :

1, 3, 5, 6, 9, 10, 11, 13, 17, 18, 19, 20, 22, 23, 25, 28, 33, 34, 35, 36, 37, 39, 40, 41, 43, 44, 46, 49, 50, 52, 55, 59, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 77, 78, 79, 81, 82, 84, 87, 88, 89, 91, 92, 94, 97, 98, 100, 103, 107, 108, 110, 113, 117, 122

seems a good candidate, since (numbering $A(n)$from $A(1) = 1$)

$A(3) = 5$, $A(5) = 9$, $A(9) = 17$ and so on, thus $A(1+2^{k-1}) = 1+2^{k}$ in the limits of the table (I don't know if this is always true, I discover this sequence today)

and also $A(4) = 6$; $A(6) = 10$.

(For $c = 1$, we have $f(c) = 2 \neq A(c) = 1$, but it doesn't seem very important to me.)

Is there literature about a possible relation between $f(c)$ and $A188163(c)$ ?