As Gerry Myerson comments, there are $F_{2n-1}$ such combinations for $\{1,\dots,n\}$. One way to see this is via a bijection with the Morse code sequences of length $2n-2$ that looks like this:
$$ \begin{align*} &{\cdot}\;{\cdot}\;{\cdot}\;{\cdot}\;{\cdot}\;{\cdot} &&\emptyset\\ &{\cdot}\;{\cdot}\;{\cdot}\;{\cdot}\;- &&(34)\\ &{\cdot}\;{\cdot}\;{\cdot}\;-{\cdot} &&(24)\\ &{\cdot}\;{\cdot}\;-{\cdot}\;{\cdot} &&(23)\\ &{\cdot}-\;{\cdot}\;{\cdot}\;{\cdot} &&(13)\\ &-\;{\cdot}\;{\cdot}\;{\cdot}\;{\cdot} &&(12)\\ &{\cdot}\;{\cdot}\;-\;- &&(23)(34)\\ &{\cdot}\;-\;{\cdot}\;- &&(13)(34)\\ &{\cdot}\;-\;-\;{\cdot} &&(14)\\ &-\;{\cdot}\;{\cdot}\;- &&(12)(34)\\ &-\;-\;{\cdot}\;{\cdot} &&(12)(23)\\ &-\;-\;- &&(12)(23)(34) \end{align*} $$ There are $n-1$ pairs of dots, representing the $n-1$ adjacent numbers $(12)$, $(23)$, ..., $(n\!-\!1\;n)$. Lines joining two pairs combine them, so for example $(12)(23)$ is represented by $--\cdot\;\cdot$, since the first two pairs are activated; whereas $(13)$ is represented by $\cdot-\cdot\cdot\cdot$ since it joins $(12)$ with $(23)$.
It is then easily seen that the Morse code sequences of length $n$ (where a dot has length $1$ and a dash has length $2$) are counted by the Fibonacci numbers $F_{n+1}$, since every length $n$ Morse code sequence is either a dot followed by a length $n-1$ Morse code sequence, or a dash followed by a length $n-2$ Morse code sequence.